**1.** Three particles of masses $1.0$ kg, $2.0$ kg and $3.0$ kg are
placed at the corners A, B and C respectively of an
equilateral triangle ABC of edge $1$ m. Locate the center
of mass of the system.

$m_1$ = $1$kg, $ $ $m_2$ = $2$kg, $ $ $m_3$ = $3$kg, $\\$ $x_1$ = $0$, $ $ $x_2$ = $1$, $ $ $x_3$ = $\frac{1}{2}$ $\\$ $y_1$ = $0$, $ $ $y_2$ = $0$, $ $ $y_3$ = $\frac{\sqrt{3}}{2}$ $\\$ The position of center of mass is $\\$ C.M = $\Bigg($ $\frac{m_1x_1 + m_2x_2 + m_3x_3} { m_1 + m_2 + m_3}$ , $\frac{m_1y_1 + m_2y_2 + m_3y_3} { m_1 + m_2 + m_3}$ $\Bigg)$ $\\$ = $\Bigg($ $\frac{\big(1\times0 \big) + \big(2\times1 \big) + \big(3\times \frac{1}{2}\big)} { 1+2+3}$ , $\frac{\big(1\times0 \big) + \big(2\times0 \big) + \big(3\times \big(\frac{\sqrt{3}}{2}\big)\big)} { 1+2+3}$ $\Bigg)$ $\\$ =$\Bigg($ $\frac{7}{12}$, $\frac{3\times\sqrt{3}}{2}$ $\Bigg)$ from the point $B$.

**2.** The structure of a water molecule is shown in figure
$(9-E1)$. Find the distance of the center of mass of the
molecule from the center of the oxygen atom.

Let $\theta$ be the origin of the system $\\$ In the above figure $\\$ $m_1$ = $1$kg, $ $ $x_1$= -$\big($ 0.96* $10^{-10}$ $\big)$sin $52^o$, $ $$y_1$ = $0$ $\\$ $m_2$ = $1$kg, $ $ $x_2$= -$\big($ 0.96* $10^{-10}$ $\big)$sin $52^o$, $ $$y_2$ = $0$ $\\$ $x_3$ = $0$, $ $ $y_3$= -$\big($ 0.96* $10^{-10}$ $\big)$cos $52^o$, $\\$ The position of center of mass $\Bigg($ $\frac{m_1x_1 + m_2x_2 + m_3x_3} { m_1 + m_2 + m_3}$ , $\frac{m_1y_1 + m_2y_2 + m_3y_3} { m_1 + m_2 + m_3}$ $\Bigg)$ $\\$ = $\Bigg($ $\frac{-\big( 0.96\times 10^{-10} \big)sin 52^o + \big( 0.96\times 10^{-10} \big)sin 52^o+ \big(16\times0\big)} { 1+1+16}$ , $\frac{0+0+16y_3} {18}$ $\Bigg)$ $\\$ = $\big($ 0,$\frac{8}{9} \big(0.96\times 10^{-10} \big)cos 52^o$ $\big)$

**3.** Seven homogeneous bricks, each of length L, are
arranged as shown in figure $(9-E2)$. Each brick is
displaced with respect to the one in contact by $\frac{L}{10}$.
Find the x-coordinate of the center of mass relative to
the origin is shown.

Let 'O' (0,0) be the origin of the system, $\\$ Each brick is mass 'M' & length 'L' $\\$ Each brick is displaced w.r.t. one in contact by $\frac{L}{10}$ $\\$ $\therefore$ The X coordinate of the center of mass $\\$ $X_{cm}$ = $\frac{m\big(\frac{L}{2}\big)+m\big(\frac{L}{2} + \frac{L}{10}\big) +m\big(\frac{L}{2} + \frac{2L}{10}\big)+m\big(\frac{L}{2} + \frac{3L}{10}\big) +m\big(\frac{L}{2} + \frac{3L}{10} - \frac{L}{10}\big) +m\big(\frac{L}{2} + \frac{L}{10}\big)+ m\big(\frac{L}{2}\big)}{7m}$ $\\$ = $\frac{\frac{L}{2} +\frac{L}{2} + \frac{L}{10}+\frac{L}{2}+\frac{L}{5}+\frac{L}{2}+\frac{3L}{10}+\frac{L}{2}+\frac{L}{5}+\frac{L}{2}+\frac{L}{10}+\frac{L}{2}}{7}$ $\\$ =$\frac{\frac{7L}{2}+\frac{5L}{10}+\frac{2L}{5}}{7}$ = $\frac{35L+5L+4L}{10\times7}$ = $\frac{44L}{70}$ =$\frac{11}{35}L$

**4.** A uniform disc of radius $R$ is put over another uniform
disc of radius $2R$ of the same thickness and density. The peripheries of the two discs touch each other. Locate the
center of mass of the system

Let the center of the bigger disc be the origin $\\$ $2R$ = Radius of the bigger disc $\\$ $R$ = Radius of the smaller disc $\\$ $m_1$ = $\pi$$R^2$ $\times$ $T$ $\times$ $\rho$ $\\$ $m_2$ = $\pi$ $\big($ 2R$\big)$$^2$$ $|$ $T$ $\times$ $\rho$

Let the center of the bigger disc be the origin $\\$ $2R$ = Radius of the bigger disc $\\$ $R$ = Radius of the smaller disc $\\$ $m_1$ = $\pi$$R^2$ $\times$ $T$ $\times$ $\rho$ $\\$ $m_2$ = $\pi$ $\big($2R$\big)$ $^2$ $|$ $T$ $\times$ $\rho$ $\\$ where $T$ = Thickness of the two disc $\\$ $\rho$ = Density of the two disc $\\$ $\therefore$ the position of the center of the mass $\Bigg($ $\frac{m_1x_1 + m_2x_2 } { m_1 + m_2 }$ , $\frac{m_1y_1 + m_2y_2} { m_1 + m_2}$ $\Bigg)$ $\\$

Let the center of the bigger disc be the origin $\\$ $2R$ = Radius of the bigger disc $\\$ $R$ = Radius of the smaller disc $\\$ $m_1$ = $\pi$$R^2$ $\times$ $T$ $\times$ $\rho$ $\\$ $m_2$ = $\pi$ $\big($2R$\big)$ $^2$ $|$ $T$ $\times$ $\rho$ $\\$ where $T$ = Thickness of the two disc $\\$ $\rho$ = Density of the two disc $\\$ $\therefore$ the position of the center of the mass $\Bigg($ $\frac{m_1x_1 + m_2x_2 } { m_1 + m_2 }$ , $\frac{m_1y_1 + m_2y_2} { m_1 + m_2}$ $\Bigg)$ $\\$ $x_1$ = $R$, $ $ $y_1$ = $0$ $\\$ $x_2$ = $0$, $ $ $y_2$ = $0$ $\\$ $\Bigg($ $\frac{ \pi R^2 T \rho R + 0} {\pi R^2 T \rho + \pi \big(2R\big)^2 T \rho}$ , $\frac{0} { m_1 + m_2}$ $\Bigg)$ =$\Bigg($ $\frac{ \pi R^2 T \rho R} { 5\pi R^2 T \rho }$ , 0 $\Bigg)$ =$\Bigg($ $\frac{R}{5}$,0 $\Bigg)$

**5.** A disc of radius $R$ is cut out from a larger disc of radius
$2R$ in such a way that the edge of the hole touches the
edge of the disc. Locate the center of mass of the residual
disc.

Let 'O' be the origin of the system. $\\$ $R$ = Radius of the smaller disc $\\$ $2R$ = Radius of the bigger disc $\\$ The smaller disc is cut out from bigger disc $\\$ As from the figure $\\$ $m_1$= $\pi$$R^2$$T$$\rho$, $ $ $x_1$=$R$, $ $ $y_1$=0 $\\$ $m_2$= $\pi$ $\big($2R $\big)$ $^2$ $T$ $\rho$, $ $ $x_2$=0, $ $ $y_2$=0 $\\$ The position of the C.M. = $\Bigg($ $\frac{ -\pi R^2 T \rho R + 0} {-\pi R^2 T \rho + \pi \big(2R\big)^2 T \rho}$ , $\frac{0} { m_1 + m_2}$ $\Bigg)$ $\\$ =$\Bigg($ $\frac{ -\pi R^2 T \rho R} { 3\pi R^2 T \rho }$ , 0 $\Bigg)$ =$\Bigg($ $\frac{-R}{3}$,0 $\Bigg)$

**6.** A square plate of edge d and a circular disc of diameter
d are placed touching each other at the midpoint of an
edge of the plate as shown in figure $(9-Q2)$. Locate the
center of mass of the combination, assuming same mass
per unit area for the two plates.

Let m be the mass per unit area.$\\$ $\therefore$ Mass of the square plate = $M_1$ = $d^2$m $\\$ Mass of the circular disc = $M_2$ = $\frac{\pi d^2}{4}$m $\\$ Let the center of the circular disc be the origin of the system. $\\$ $\therefore$ Position of the center of mass = $\Bigg($ $\frac{d^2md + \pi \big( d^2/4 \big)m \times 0 }{d^2md + \pi \big( d^2/4 \big)m}$, $\frac{0+0}{M_1+M_2}$ $\Bigg)$ = $\Bigg($ $\frac{d^3m}{d^2m \big( 1 +\frac{\pi}{4} \big)}$, 0 $\Bigg)$ = $\Bigg($ $\frac{4d}{\pi+4}$,0 $\Bigg)$ $\\$ The new center of mass is $\Bigg($ $\frac{4d}{\pi+4}$ $\Bigg)$ right of the center of the circular disc.

**7.** Calculate the velocity of the centre of mass of the system
of particles shown in figure $(9-E3)$.

$m_1$=1kg, $\qquad$ $\vec{v_1}$= $-1.5$co37$\vec{i}$ $-$ 1.55sin$\vec{i}$= $-1.2$ $\vec{i}$ $-$ 0.9$\vec{j}$ $\\$ $m_2$=1.2kg, $\qquad$ $\vec{v_2}$=0.4$\vec{j}$ $\\$ $m_3$=1.5kg, $\qquad$ $\vec{v_3}$=$-$0.8$\vec{i}$ + 0.6$\vec{j}$ $\\$ $m_4$=0.5kg, $\qquad$ $\vec{v_4}$=3$\vec{i}$ $\\$ $m_5$=1kg, $\qquad$ $\vec{v_5}$=1.6$\vec{i}$ $-$ 1.2$\vec{j}$ $\\$ So, $\quad$ $\vec{v_c}$= $\frac{m_1\vec{v_1} + m_2\vec{v_52} + m_3\vec{v_3}+m_4\vec{v_4}+m_5\vec{v_5}}{m_1+m_2+m_3+m_4+m_5}$ $\\$ = $\frac{1\big(-1.2\vec{i} - 0.9\vec{j}\big) + 1.2\big(0.4\vec{j}\big) +1.5\big(-0.8\vec{i} - 0.6\vec{j}\big)+0.5\big(3\vec{i}\big) +1\big(1.6\vec{i} - 1.2\vec{j}\big)}{5.2}$ $\\$ = $\frac{-1.2\vec{i} - 0.9\vec{j} + 4.8\vec{j} +1.5\big(-0.8\vec{i} - 0.6\vec{j}\big)+0.5\big(3\vec{i}\big) +1\big(1.6\vec{i} - 1.2\vec{j}\big)}{5.2}$ $\\$ =$\frac{0.7\vec{i}}{5.2}$ - $\frac{0.72\vec{j}}{5.2}$

**8.** Two blocks of mares $10$kg and $20$ kg are placed on the
X-axis. The first mass is moved on the axis by a distance
of $2$ cm. By what distance should the second mass be
moved to keep the position of the centre of mass
unchanged?

Two masses $m_1$ & $m_2$ are placed on X-axis, $\\$ $m_1$ = $10$kg, $ $ $m_2$ = $20$ kg $\\$ The first mass is displaced by a distance of 2cm $\\$ $\therefore$ $\vec{X}_{cm}$ = $\frac{m_1x_1 + m_2x_2} { m_1 + m_2}$ =$\frac{10 \times2 + 20x_2 }{30}$ $\\$ $\Rightarrow$ $0$ $\Rightarrow$ $\frac{20 + 20x_2 }{30}$ $\Rightarrow$ 20 + 20$x_2$ = 0 $\\$ $\Rightarrow$ 20 =$-$20$x_2$ $\Rightarrow$ $x_2$ =$-1$ $\\$ $\therefore$ The $2^{nd}$ mass should be displaced by 1cm towards left so as to kept the position of center of mass unchanged.

**9.** Two blocks of masses $10$ kg and $30$ kg are placed along
a vertical line. The first block is raised through a height
of $7$ cm. By what distance should the second mass be
moved to raise the centre of mass by $1$ cm ?

Two masses $m_1$ & $m_2$ are kept in the vertical line, $\\$ $m_1$ = $10$kg, $ $ $m_2$ = $30$ kg $\\$ The first block is rised through a height of $7$cm $\\$ The center of mass is raised by $1$ cm, $\\$ $\therefore$ $1$ = $\frac{m_1y_1 + m_2y_2} { m_1 + m_2}$ = $\frac{10 \times7 + 30y_2 }{40}$ $\\$ $\Rightarrow$ $1$ $\Rightarrow$ $\frac{10 \times7 + 30y_2 }{40}$ $\Rightarrow$ $10 \times7$ + 30$y_2$ = 40 $\Rightarrow$ 30$y_2$ =$-30$ $\Rightarrow$ $y_2$ =$-1$ $\\$ The mass of 30 kg body should be displaced 1 cm downword inorder to raised cener of mass through 1cm.

**10.** Consider a gravity-free hall in which a tray of mass M,
carrying a cubical block of ice of mass m and edge L, is
at rest in the middle $(figure 9-E4)$. If the ice melts, by
what distance does the centre of mass of "the tray plus
the ice" system descend ?

As the hall is gravity-free, after the ice melts, it would tend to acquire the spherical shape. But, there is no external force acting on the system.So, the center of mass of the system would not move.

**11.** Mr. Verma $(50 kg)$ and Mr. Mathur $(60 kg)$ are sitting
at the two extremes of a 4 m long boat $(40 kg)$ standing
still in water. To discuss a mechanics problem, they
come to the middle of the boat. Neglecting friction with
water, how far does the boat move on the water during
the process ?

$m_1$ = $60$kg, $\qquad$ $m_2$ = $40$kg, $\qquad$ $m_3$ = $50$kg, $\\$ Let $A$ bethe origin of the system. $\\$ Initially Mr. Verma & Mr. Mathur are at extreme position of the boat. $\\$ $\therefore$ The center of mass will be at a distance = $\frac{60 \times 0 + 40 \times2 + 50\times 4}{150}$ = $\frac{280}{150}$ = $1.87$m from $'A'$. $\\$ When they come to the mid point of the boat the CM lies at $2$m from $'A'$ $\\$ $\therefore$ The shift in CM = $2-1.87$ = $0.13$m towards right. $\\$ But as there is no external force in longitudinal direction their CM would not shift.$\\$ So, the boat moves $0.13$m or $13$cm towards right.

**12.** A cart of mass M is at rest on a frictionless horizontal
surface and a pendulum bob of mass m hangs from the
roof of the cart $(figure 9-E6)$. The string breaks, the bob
falls on the floor, makes several collisions on the floor
and finally lands up in a small slot made in the floor.
The horizontal distance between the string and the slot
is $L$. Find the displacement of the cart during this
process.

Let the bob fall at $A$. The mass of bob = m $\\$ The mass of cart = M $\\$ Initially, their center of mass will be at $\frac{m \times L + M\times0}{M+m}$ = $\Bigg($ $\frac{m}{M+m}$ $\Bigg)$L $\\$ Distance from P, when the bob falls in the slot the CM is at distance 'O' from P. $\\$ Shift in CM = 0 - $\frac{mL}{M+m}$ = $-\frac{mL}{M+m}$ towards left. $\\$ $\qquad$ = $\frac{mL}{M+m}$ towards right. $\\$ But there is no external force in horiontal direction. $\\$ So the cart displaces a distance $\frac{mL}{M+m}$ towards right. $\\$

**13.** The balloon, the light rope and the monkey shown in
figure $(9-E7)$ are at rest in the air. If the monkey reaches
the top of the rope, by what distance does the balloon
descend ? Mass of the balloon = $M$, mass of the
monkey = $m$ and the length of the rope ascended by the
monkey = $L$.

Initially, the monkey $y$ & balloon are at rest. $\\$ So the CM is at 'P' $\\$ When the monkey descends through a distance 'L' $\\$ The CM will shift $t_0$ = $\frac{m \times L + M \times 0}{M+m}$ = $\frac{mL}M+m{}$ from P. $\\$ So, the balloon descends through a distance $\frac{mL}{M+m}$

**14.** Find the ratio of the linear momenta of two particles of
masses $1.0$ kg and $4.0$ kg if their kinetic energies are
equal.

Let the mass of the two particles be $m_1$ & $m_2$ respectively $\\$ $m_1$ = $1$kg, $\qquad$ $m_1$ = $4$kg, $\\$ $\therefore$ According to question $\frac{1}{2}$$m_1v_1$$^2$ = $\frac{1}{2}$$m_2v_2$$^2$ $\\$ $\Rightarrow$ $\frac{m_1}{m_2}$ = $\frac{{v_2}^2}{{v_1}^2}$ $\Rightarrow$ $\frac{v_2}{v_1}$ = $\sqrt{\frac{m_1}{m_2}}$ = $\frac{v_1}{v_2}$ = $\sqrt{\frac{m_2}{m_1}}$ $\\$ Now, $\frac{m_1v_1}{m_2v_2}$ = $\frac{m_1}{m_2}$ $\times$ $\sqrt{\frac{m_2}{m_1}}$ = $\frac{\sqrt{m_1}}{\sqrt{m_2}}$ = $\frac{\sqrt{1}}{\sqrt{4}}$ = $\frac{1}{2}$ $\\$ $\Rightarrow$ $\frac{m_1v_1}{m_2v_2}$ = 1:2.

**15.** A uranium-238 nucleus, initially at rest, emits an alpha
particle with a speed of $1.4 \times 10 7$ m/s. Calculate the
recoil speed of the residual nucleus thorium-234. Assume
that the mass of a nucleus is proportional to the mass
number.

As the uranium-238 nucleus emits a $\alpha$-particle with a speed of $1.4 \times 10^7$m/sec. Let $v_2$ be the speed of the residual nucleus thorium-234. $\\$ $\therefore$ $m_1v_1$ = $m_2v_2$ $\\$ $\Rightarrow$ $4 \times 1.4 \times 10^7$ = $234 \times v_2$ $\\$ $\Rightarrow$ $v_2$ = $\frac{4 \times 1.4 \times 10^7}{234}$ = $2.4 \times 10^5$ m/sec.

**16.** A man of mass $50$ kg starts moving on the earth and
acquires a speed of $1.8 $m/s. With what speed does the
earth recoil ? Mass of earth = $6 \times 10^{24}$ kg.

$m_1v_1$ = $m_2v_2$ $\\$ $\Rightarrow$ $50 \times 1.8 $ = $6 \times 10^{24} \times v_2$ $\\$ $\Rightarrow$ $v_2$ = $\frac{50 \times 1.8}{6 \times 10^24}$ = $1.5 \times 10^{-23}$ m/sec. $\\$ So, the earth will recoil at a speed of $1.5 \times 10^{-23}$ m/sec. $\\$

**17.** A neutron initially at rest, decays into a proton, an
electron and an antineutrino. The ejected electron has
a momentum of $1.4 \times 10^ {-26}$ kg-m/s and the antineutrino $6'4 \times 10^{-27}$kg-m/s. Find the recoil speed of the proton
(a) if the electron and the antineutrino are ejected along
the same direction and (b) if they are ejected along
perpendicular directions. Mass of the proton
= $1.67 \times 10 ^{-27}$ kg.

Mass of proton = $1.67 \times 10^{-27}$ $\\$ Let '$V_p$' be the velocity of proton $\\$ Given momentum of electron = $1.4 \times 10^{-26}$kg m/sec $\\$ Given momentum of antineutrino = $6.4 \times 10^{-27}$kg m/sec $\\$ a) The electron & the antineutrino are ejected in the same direction. As the total momentum is considered the proton should be ejected in hte opposite direction. $\\$ $1.67 \times 10^{-27} \times V_p$ = $1.4 \times 10^ {-26} + 6.4 \times 10^{-27} = 20.4 \times 10^{-27}$ $\\$ $\Rightarrow$ $V_p = (\frac{20.4}{1.67} )= 12.2$ m/sec in the opposite direction. $\\$ b) The electron & the antineutrino are ejected $\perp$ to each other. $\\$ Total momentum of electron and antineutrino, $\\$ = $\sqrt{(14^2)+(6.4^2) \times 10^{-27}}$kg m/s = $15.4 \times 10^{-27} $kg m/sec $\\$ Since, $1.67 \times 10^{-27}V_p = 15.4 \times 10^{-27} $ kg m/sec. $\\$ So, $V_p$ = 9.2 m/sec.

**18.** A man of mass M having a bag of mass m slips from
the roof of a tall building of height H and starts falling
vertically $(figure 9-E8)$. When at a height h from the
ground, he notices that the ground below him is pretty
hard, but there is a pond at a horizontal distance x from
the line of fall. In order to save himself he throws the
bag horizontally (with respect to himself) in the direction
opposite to the pond. Calculate the minimum horizontal
velocity imparted to the bag so that the man lands in
the water. If the man just succeeds to avoid the hard
ground, where will the bag land ?

Mass of man = M, $\qquad$ Initially velocity = 0, $\qquad$ Mass of bad = m $\\$ Let the man throws the bag towards left with a velocity $v$ towards left. So, there is no external force in the horizontal direction. $\\$ The momentum will be conserved. Let he goes right with the velocity m = MV $\Rightarrow$ $\frac{mv}{M}$ $\Rightarrow$ v = $\frac{MV}{m}$ $\\$ Let the total time he will take to reach ground = $\sqrt{2H/g} = t_1$ $\\$ Let the total time he will take to reach the height $h$ = $t_2$ = $\sqrt{2(H-h)/g}$ $\\$ Then the time of his flying = $t_1-t_2 = \sqrt{2H/g}- \sqrt{2(H-h)/g} = \sqrt{2/g}(\sqrt{H}- \sqrt{H-h})$ $\\$ Within in this time he reaches the ground in the pond covering a horizontal distance x $\Rightarrow$ $x = V \times t $ $\Rightarrow$ $V = x / t $ $\\$ $\therefore$ $v = \frac{M}{m} \frac{x}{t} = \frac{M}{m} \times \frac{\sqrt{g}}{\sqrt{2}(\sqrt{H}-\sqrt{H-h})}$ $\\$ As there is no external force in horizontal direcion, the x-coordinate of CM will remain at that position. $\\$ $0 = \frac{M \times (x) + m \times x}{M+m} \Rightarrow x_1 = \frac{-M}{m}x $ $\\$ $\therefore$ The bag will reach the bottom at a distance (M/m) $\times$ towards left of the line it falls.

**19.** A ball of mass $50$ g moving at a speed of $2.0$ m/s strikes
a plane surface at an angle of incidence $45^°$. The ball is
reflected by the plane at equal angle of reflection with
the same speed. Calculate (a) the magnitude of the
change in momentum of the ball (b) the change in the
magnitude of the momentum of the ball.

Mass $= 50g = 0.05$Kg $//$ $v = 2cos45^o \vec{i} - 2sin45^o \vec{j}$ $\\$ $v_1 = 2cos45^o \vec{i} - 2sin45^o \vec{j}$ $\\$ a) change in momentum = $ m\vec{v} - m \vec{v_1}$ $\\$ $=0.05(2cos45^o \vec{i} - 2sin45^o \vec{j})- 0.05(-2cos45^o \vec{i} - 2sin45^o \vec{j})$ $\\$ $=0.1cos45^o \vec{i} - 0.1 sin45^o + 0.1cos 45^o\vec{i}+0.1sin45^o\vec{j}$ $\\$ $=0.2 cos45^o \vec{i}$ $\\$ $\therefore magnitude = \sqrt{\Bigg(\frac{0.2}{\sqrt{2}}\Bigg)^2}$ = $ \frac{0.2}{\sqrt{2}}$ = $0.14 kg m/s.$ $\\$c) The change in magnitude of the momentum of the ball $-\mid \vec{P_i}\mid -\mid \vec{P_f}\mid = 2\times0.5 - 2 \times 0.5 = 0.$

**20.** Find the centre of mass of a uniform plate having
semicircular inner and outer boundaries of radii $R$, and
$R2$ $(figure 9-E5)$

The center of mass of the plate will be on the symmetrical axis. $\\$ $\Rightarrow$ $\vec{y_{cm}}$ = $\frac{ \big(\frac{\pi {R_2}^2}{2}\big)\big(\frac{4{R_2}}{3\pi}\big) - \big(\frac{\pi {R_1}^2}{2}\big)\big(\frac{4{R_1}}{3\pi}\big) }{\frac{\pi {R_2}^2}{2} - \frac{\pi {R_1}^2}{2}}$ $\\$ = $\frac{ \big(2/3\big){R_2}^3 - \big(2/3\big){R_1}^3 }{\pi/2\big({R_2}^2 - {R_1}^2\big)}$ = $\frac{4}{3\pi}$ $\frac{\big(R_2 -R_1\big)\big({R_2}^2 + {R_1}^2 + R_1R_2\big)}{\big(R_2 -R_1\big)\big(R_2+R_1\big)}$ $\\$ =$\frac{4}{3\pi}$ $\frac{\big({R_2}^2 + {R_1}^2 + R_1R_2\big)}{ R_!+R_2}$ above the center.

**21.** Light in certain cases may be considered as a stream of
particles called photons. Each photon has a linear
momentum hIX where h is the Planck's constant and
is the wavelength of the light. A beam of light of
wavelength A. is incident on a plane mirror at an angle
of incidence $O$. Calculate the change in the linear
momentum of a photon as the beam is reflected by the
mirror.

$P_{Incidence} = (h/\lambda)cos\theta \vec{i}-(h/ \lambda)sin \theta \vec{j} $ $\\$ $ P_{Reflected} = (h/\lambda)cos\theta \vec{i}-(h/ \lambda)sin \theta \vec{j} $ $\\$ The change in momentum will be only in the x-axis direction i.e. $\\$ $\mid \Delta P \mid = (h/\lambda)cos\theta - ((h/ \lambda)cos \theta) = (2h/ \lambda)cos \theta$

**22.** A block at rest explodes into three equal parts. Two
parts start moving along $X$ and $Y$ axes respectively with
equal speeds of $10$ m/s. Find the initial velocity of the
third part

As the block is exploded only due to its internal energy. So net external force during this process is 0. So the center mass will not change. $\\$ Let the body while exploded was at the origin of the coordinate system.If the two bodies of equal mass is moving at a speed of $10$ m/s in +x & +y axis directions respectively, $\\$ $\sqrt{10^2 + 10^2 + 210.10cos90^o} = 10\sqrt{2} m/s \quad 45^o w.r.t. +x axis$ $\\$ If the center mass is at rest, then the third mass which has equal mass with other two, will move in the opposite direction (i.e. $1135^o w.r.t. +x axis$) of the resultant at the same velocity.

**23.** Two fat astronauts each of mass $120$ kg are travelling
in a closed spaceship moving at a speed of $15$ km/s in
the outer space far removed from all other material
objects. The total mass of the spaceship and its contents
including the astronauts is $660$ kg. If the astronauts do
slimming exercise and thereby reduce their masses to
$90$ kg each, with what velocity will the spaceship move ?

Since the spaceship is removed from any material object & totally isolated from the surrounding, the missions by astronauts couldn't slip away from the spaceship. So the total mass of the spaceship remain unchanged and also it's velocity.

**24.** During a heavy rain, hailstones of average size $1.0$ cm
in diameter fall with an average speed of $20$ m/s.
Suppose $2000$ hailstones strike every square meter of a
$10 m \times 10$ m roof perpendicularly in one second and
assume that the hailstones do not rebound. Calculate
the average force exerted by the falling hailstones on
the roof. Density of a hailstone is $900 kg/m^3$.

$d = 1cm \qquad v = 20m/s \qquad u = 0 \qquad \rho = 900kg/m^3 = 0.9 gm/cm^3$ $\\$ $ volume = (4/3)\pi r^3 = (4/3) \pi (0.5)^3 = 0.5238cm^3 $ $\\$ $\therefore$ mass = $v\rho = 0.5238 \times 0.9 = 0.4714258gm$ $\\$ $\therefore$ mass of 2000 hailstone = $2000 \times 0.4714 = 947.857$ $\\$ $ \therefore$ Rate of change in momentum per unit area = $947.857 \times 2000 = 19 N/m^3$ $\\$ $\therefore$ Total force exerted = $19 \times 100 = 1900 N$

**25.** A ball of mass m is dropped onto a floor from a certain
height. The collision is perfectly elastic and the ball
rebounds to the same height and again falls. Find the
average force exerted by the ball on the floor during a
long time interval.

A ball of mass m is dropped onto a floor from a certain height let 'h'. $\\$ $\therefore$ $\quad$ $v_1 = \sqrt{2gh}$, $\qquad$ $v_1 = 0$, $\qquad$ $ v_2 = -\sqrt{2gh}$ & $v_2 = 0$ $\\$ $\therefore$ Rate of change of velocity :- $\\$ $F = \frac{m \times 2\sqrt{2gh}}{t}$ $\\$ $\therefore v = \sqrt{2gh}, \qquad s = h, \qquad v=0, $ $\\$ $\Rightarrow v = u+at$ $\\$ $\Rightarrow \sqrt{2gh} = gt \Rightarrow t = \sqrt{\frac{2h}{g}}$ $\\$ $\therefore $ Total time $\quad 2\sqrt{\frac{2h}{t}} $ $\\$ $ \therefore F = \frac{m \times 2\sqrt{2gh}}{2\sqrt{\frac{2h}{g}}} = mg$

**26.** A railroad car of mass M is at rest on frictionless rails
when a man of mass m starts moving on the car towards
the engine. If the car recoils with a speed v backward
on the rails, with what velocity is the man approaching
the engine ?

A railroad car of mass M is at rest on frictionless rails when a man of mass m starts moving on the car towards the engine. The car recoils with a speed v backward on the rails. $\\$ Let the mass is moving with a velocity x w.r.t. the engine.$\\$ $\therefore$ The velocity of the mass w.r.t. earth is $(x-v)$ towards right. $\\$ $V_{cm} = 0 \quad $( Initially at rest) $\\$ $\therefore 0 = -Mv + m(x-v)$ $\\$ $\Rightarrow Mv = m(x-v) \Rightarrow mx = Mv + mv \Rightarrow x =\Bigg( \frac{M+m}{m}\Bigg)v \Rightarrow x = \Bigg(1 + \frac{M}{m} \Bigg)v$

**27.** A gun is mounted on a railroad car. The mass of the car,
the gun, the shells and the operator is $50$ m where m is
the mass of one shell. If the velocity of the shell with
respect to the gun (in its state before firing) is $200$ m/s,
what is the recoil speed of the car after the second shot ?
Neglect friction.

A gun is mounted on a railroad car. The mass of the car, the gun, the shells and the operator is 50m where m is the mass of the one shell. The muzzle velocity of the shells is 200m/s. $\\$ Initial, $\quad V_{cm} = 0$ $\\$ $\therefore 0 = 49 m \times V + m \times 200 \Rightarrow V = \frac{-200}{49}$m/s $\\$ $\therefore \frac{200}{49} m/s $ towards left $\\$ When another shel is fired, then the velocity of the car, with respect to the platform is, $\\$ $\Rightarrow V = \frac{200}{49} m/s$ towards left $\\$ When another shel is fired, then the velocity of the car, with respect to the platform is, $\\$ $\Rightarrow V = \frac{200}{48} m/s$ towards left $\\$ $\therefore$ Velocity of the car w.r.t. the earth is $\Bigg( \frac{200}{49} + \frac{200}{48} \Bigg)m/s$ towards left

**28.** Two persons each of mass m are standing at the two
extremes of a railroad car of mass M resting on a smooth
track $(figure 9-E10)$. The person on left jumps to the left
with a horizontal speed u with respect to the state of
the car before the jump. Thereafter, the other person
jumps to the right, again with the same horizontal speed
u with respect to the state of the car before his jump.
Find the velocity of the car after both the persons have
jumped off

Two persons each of mass m are standing at the two extremes of a railroad car of mass m resting on a smooth track. $\\$ Case - I $\\$ Let the velocity of the railroad car w.r.t. the earth is V after the jump of the man.$\\$ $\therefore 0 = -mu + (M+m)V$ $\\$ $\Rightarrow V = \frac{mu}{M+m}$ towards right $\\$ Case - II $\\$ When the man on the right jumps, the velocity of it w.r.t. the car is u. $\\$ $\therefore 0 = mu = Mv'$ $\\$ $\Rightarrow v' = \frac{mu}{M}$ $\\$ (V' is the change in velocity of the platform when platform itself is taken as reference assuming the car to be at rest) $\\$ $\therefore$ So, net velocity towards left (i.e. the velocity of the car w.r.t. the earth) = $\frac{mv}{M}- \frac{mv}{M+m} = \frac{mMu + m^2v-Mmu}{M(M+m)} = \frac{m^2v}{M(M+m)}$

**29.** $Figure (9-E11)$ shows a small block of mass m which is
started with a speed v on the horizontal part of the
bigger block of mass M placed on a horizontal floor. The
curved part of the surface shown is semicircular. All the
surfaces are frictionless. Find the speed of the bigger
block when the smaller block reaches the point A of the
surface.

A small block of mass m which is started with a velocity V on the horizontal part of the bigger block of mass M placed on a horizontal floor. $\\$ Since the small body of mass m is started with a velocity V in the horizontal direction, so the total initial momentum at the initial position in the horizontal direction will remain same as the total final momentum at the point A on the bigger block in the horizontal direction. $\\$ From L.C. K. $m: mv + M \times O = (m+M)v \Rightarrow v' = \frac{mv}{M+m}$

**30.** In a typical Indian Bugghi (a luxury cart drawn by
horses), a wooden plate is fixed on the rear on which
one person can sit. A bugghi of mass $200$ kg is moving
at a speed of $10$ km/h. As it overtakes a school boy
walking at a speed of $4$ km/h, the boy sits on the wooden
plate. If the mass of the boy is $25$ kg, what will be the
new velocity of the bugghi ?

Mass of the bugghi = $200$kg, $\qquad$ $V_B = 10km/hour$ $\\$ $\therefore$ Mass of the body = $2.5$kg & $V_{Boy} = 4km/hour$ $\\$ If we take the boy and bugghi as a system then total momentum before the process of sitting will remain constant after the proces of sitting. $\\$ $\therefore m_bV_b = m_{boy}V_{boy} = (m_b + m_{boy})v$ $\\$ $\Rightarrow 200 \times 10 + 25 \times 4 = (200 + 25 ) \times v$ $\\$ $ \Rightarrow v = \frac{2100}{225} = \frac{28}{3} = 9.3 m/sec $

**31.** A ball of mass $0.50$ kg moving at a speed of $5.0$ m/s
collides with another ball of mass $1.0$ kg. After the
collision the balls stick together and remain motionless.
What was the velocity of the $1.0$ kg block before the
collision ?

Mass of the ball = $m_1$ = $0.5$kg, the velocity of the ball = $5$ m/s $\\$ Mass of the another ball $m_2$ = $1$kg. $\\$ Let it's velocity = v' m/s. Using the law of conservation of momentum, $\\$ $0.5 \times 5 + 1 \times v' = 0 \Rightarrow v' = -2.5$ $\\$ $\therefore$ Velocity of second ball is 2.5m/s opposite to the direction of motion of $1^{st}$ ball.

**32.** A $60$ kg man skating with a speed of $10$ m/s collides
with a $40$ kg skater at rest and they cling to each other.
Find the loss of kinetic energy during the collision.

Mass of the man =$ m_1$ = $60$kg $\\$ Speed of the man = $v_1$ = $10$kg $\\$ Mass of the skater =$ m_2$ = $40$kg $\\$ Let the velocity = v' $\\$ $\therefore$ $60 \times 10 + 0 = 100 \times v' \Rightarrow v' = 6 m/s $ $\\$ Loss in K.E. = $ (1/2)60 \times (10)^2 = (1/2) \times 100 \times 36 = 1200$ J.

**33.** Consider a head-on collision between two particles of
masses m1 and m2. The initial speeds of the particles
are u1 and u2 in the same direction. The collision starts
at t = 0 and the particles interact for a time interval At.
During the collision, the speed of the first particle
varies as $\\$
$v(t) = u_1 + \frac{t}{\Delta t}(v_1-u_1)$ $\\$
Find the speed of the second particle as a function of
the time during the collision.

Using law of conservation of momentum $\\$ $m_1u_1 + m_2u_2 = m_1v(t) + m_2v'$ $\\$ Where v' = speed of $2^{nd}$ paricle during collision. $\\$ $\Rightarrow m_1u_1 + m_2u_2 = m_1u_1 + m_1 + (t/ \Delta t)(v_1 - u_1) + m_2v'$ $\\$ $ \Rightarrow \frac{m_2u_2}{m^2}- \frac{m_1}{m_2} \frac{t}{\Delta t} (v_1-u_1)v'$ $\\$ $\therefore v' = u_2 - \frac{m_1}{m_2} \frac{t}{\Delta t }(v_1 - u)$

**34.** A bullet of mass m moving at a speed v hits a ball of
mass M kept at rest. A small part having mass m' breaks
from the ball and sticks to the bullet. The remaining
ball is found to move at a speed v1 in the direction of
the bullet. Find the velocity of the bullet after the
collision

Mass of the Bullet = m and speed = v $\\$ Mass of the ball = M $\\$ m' = frictional mass from the ball $\\$ Using law of conservation of momentum, $\\$ $\qquad$ $mv + 0 = (m' + m)v' + (M - m')v_1$ $\\$ where v' = final velocity of the bullet + frictional mass $\\$ $\Rightarrow v' =\frac{mv - (M+m')V_1}{m+m'}$

**35.** A ball of mass m moving at a speed v makes a head-on
collision with an identical ball at rest. The kinetic energy
of the balls after the collision is three fourths of the
original. Find the coefficient of restitution.

Mass of the $1^{st}$ ball = m and speed = v $\\$ Mass of the $2^{nd}$ ball = m $\\$ Let final velocities of $1^{st}$ and $2^{nd}$ ball are $v_1$ and $v_2$ respectively. $\\$ Using law of conservation of momentum, $\\$ $m_1(v_1+v_2) = mv$ $\\$ $\Rightarrow v_1 + v_2 =v $ ........... (1)$\\$ Also, $\quad$ $v_1 - v_2 =ev$ ............ (2)$\\$ Given that final K.E. = $3/4 $ Initial K.E. $\\$ $\Rightarrow 1/2 mv_1^2 + 1/2 mv_2^2 = 3/4 \times 1/2 mv^2$ $\\$ $\Rightarrow v_1^2 + v_2^2 = 3/4 v^2$ $\\$ $\Rightarrow \frac{(v_1 + v_2)^2 + (v_1 - v_2 )^2}{2} = \frac{3}{4}v^2$ $\\$ $\Rightarrow \frac{(1 +e^2)v^2}{2} = \frac{3}{4}v^2 \Rightarrow 1 + e^2 = \frac{3}{2} \Rightarrow e^2 = \frac{1}{2} \Rightarrow e = \frac{1}{\sqrt{2}}$

**36.** A block of mass $2.0$ kg moving at $2.0$ m/s collides head
on with another block of equal mass kept at rest.
(a) Find the maximum possible loss in kinetic energy
due to the collision. (b) If the actual loss in kinetic
energy is half of this maximum, find the coefficient of
restitution.

Mass of the block = $2$kg and speed = $2m/s$ $\\$ Mass of the $2^{nd}$ block = $2kg$ $\\$ Let final velocities of $2^{nd}$ block =v $\\$ Using law of conservation of momentum, $\\$ $2 \times 2 = (2+2)v \Rightarrow v = 1m/s $ $\\$ $ \therefore$ Loss in K.E. in inelastic collision $\\$ $ -(\frac{1}{2}) \times 2 \times (2)^2v - \frac{1}{2}(2+2) \times (1)^2 -4 -2 -2J $ $\\$ b) Actual Loss = $\frac{maximum \quad loss }{2} =1J$ $\\$ $ -(\frac{1}{2}) \times 2 \times (2)^2- (\frac{1}{2})2 \times v_1^2 + (\frac{1}{2}) \times 2 \times v_2^2 = 1$ $\\$ $\Rightarrow 4 - (v_1^2 + v_2^2) = 1$ $\\$ $\Rightarrow 4 - \frac{(1 + e^2) \times 4}{2} =1 $ $\\$ $ 2(1+e^2) =3 \Rightarrow 1+e^2 = \frac{3}{2} \Rightarrow e^2 =\frac{1}{2} \Rightarrow e = \frac{1}{\sqrt{2}}$

**37.** A particle of mass 100 g moving at an initial speed u
collides with another particle of same mass kept initially
at rest. If the total kinetic energy becomes 0.2 J after
the collision, what could be the minimum and the
maximum value of u.

Final K.E. = $0.2 J$ $\\$ Initial K.E. = $1/2 mv_1^2 + 0 = 1/2 \times 0.1u^2 = 0.05 u^2$ $\\$ $ mv_1 = mv_2 = mu$ $\\$ Where $v_1$ and $v_2$ are final velocities of $1^{st}$ and $2^{nd} block\quad respectively.$ $\\$ $\Rightarrow v_1 + v_2 = u \qquad ...........(1)$ $\\$ $(v_1 - v_2) + l(1_1- u_2) = 0 \Rightarrow la = v_2 - v_1 \qquad ............(2)$ $\\$ $u_2 =0, \qquad u_1 = u.$ $\\$ Adding Eq.(1) and Eq.(2) $\\$ $2v_2 = (1 +l)u \Rightarrow v_2 = (u/2)(1+l)$ $\\$ $\therefore v_1= u-\frac{u}{2} -\frac{u}{2}l$ $\\$ $ v_1 = \frac{u}{2}(1-l)$ $\\$ $Given (1/2)mv_1^2 + (1/2)mv_2^2 = 0.2 $ $\\$ $\Rightarrow v_1^2 + v_2^2 = 4$ $\\$ $\Rightarrow \frac{u^2}{4} (1-l)^2 +\frac{u^2}{4} (1+l)^2 =4 \qquad \Rightarrow \frac{u^2}{2}(1-l^2) = 4 \qquad u^2 = \frac{8}{1-l^2}$ $\\$ For maximum value of u, denominator should be minimum, $\\$ $\Rightarrow l =0 $ $\\$ $ \Rightarrow u^2 = 8 \Rightarrow u =2\sqrt{2} m/s $ $\\$ For maximum value of u, denominator should be maximum, $\\$ $\Rightarrow l =1 $ $\\$ $ \Rightarrow u^2 = 4 \Rightarrow u =2 m/s $

**38.** Two friends A and B (each weighing 40 kg) are sitting
on a frictionless platform some distance d apart. A rolls
a ball of mass 4 kg on the platform towards B which B
catches. Then B rolls the ball towards A and A catches
it. The ball keeps on moving back and forth between A
and B. The ball has a fixed speed of 5 m/s on the
platform. (a) Find the speed of A after he rolls the ball
for the first time. (b) Find the speed of A after he catches
the ball for the first time. (c) Find the speeds of A and
B after the ball has made 5 round trips and is held by
A. (d) How many times can A roll the ball ? (e) Where
is the centre of mass of the system "A + B + ball" at the
end of the nth trip ?

Two friends A & B (each 40kg) are sitting on a frictionless platform some distance d apart A rols a ball of mass 4 kg on the platform towards B, which B catches. Then B rolls the ball towards A and A catches it. The ball keeps on moving back & forth between A and B. The ball has a fixed velocity 5 m/s. $\\$ a) Case -I :- Total momentum of the man A & the ball will remain constant. $\\$ $\therefore 0 = 4 \times 5 - 40 \times v \qquad \Rightarrow v= 0.5 m/s $ towards left. $\\$ b) Case -II :- When B catches the ball , the momentum between the B & the ball will remain constant. $\\$ $\Rightarrow 4 \times 5 = 44v \Rightarrow v=(20/44)m/s$ $\\$ Case-III :- When B throws the ball, then L.C.L.M.,$\\$ $ \Rightarrow 44 \times (20/44) = -4 \times 5 +40 \times v \qquad \Rightarrow v = 1m/s (towards \quad right)$ $\\$ Cases-IV :- When A catches the ball , then applying L.C.L.M. $\\$ $\Rightarrow -4 \times 5 + (-0.5 )\times 40 = -44v \qquad \Rightarrow v = \frac{10}{11}$ m/s towards left $\\$

c) Case-V :- When A throws the ball, by applying L.C.L.M. $\\$ $ 44 \times (10/11)= 4\times 5 -40 \times V \qquad \Rightarrow V= 60/40 = 3/2$m/s towards left $\\$ Case -VI:- When B receives the ball then by applying L.C.L.M. $\\$ $ 40 \times 1 + 4 \times 5 = 44 \times v \qquad \Rightarrow V= 60/44$ m/s towards right $\\$ Case-VII:- When B throws the ball, then applying L.C.L.M. $\\$ $ 44 \times (66/44) = -4 \times 5 +40 \times V \qquad \Rightarrow V= 80/40 = 2$m/s towards right. $\\$ Case-VIII:- When A catches the ball then applying L.C.L.M. $\\$ $\Rightarrow -4 ]times 5 -40 \times (3/2) = -44v \qquad \Rightarrow v =(80/44) = (20/11)$m/s towards left $\\$ Similarly after 5 round trips $\\$ The velocity of A will be (50/110) & velocity of B will be 5 m/s. $\\$d) Since after 6 round trip, the velocity of A is 60/11 i.e. >5m/s So, it can't catch the ball. So it can only roll the ball six.$\\$ e)Let the ball & the body A at the initial position be at origin.$\\$ $\therefore X_c = \frac{40 \times 0 + 40 \times 0 + 40 \times d}{ 40 + 40 + 4} = \frac{10}{11}d$

**39.** A ball falls on the ground from a height of 2.0 m and
rebounds up to a height of 1.5 m. Find the coefficient of
restitution.

$u = \sqrt{2gh} =$ velocity on the ground when ball approaches the ground. $\\$ $\Rightarrow u = \sqrt{2 \times 9.8 \times 2}$ $\\$ v = velocity of ball when it separates from the ground.$\\$ $ \vec{v} + l\vec{u} = 0$ $\\$ $\Rightarrow l\vec{u} = -\vec{v} \Rightarrow l = \frac{\sqrt{2 \times 9.8 \times 1.5}}{\sqrt{2 \times 9.8 \times 2}} = \sqrt{\frac{3}{4}} = \frac{ \sqrt{3}}{2} $ $\\$

**40.** In a gamma decay process, the internal energy of a
nucleus of mass M decreases, a gamma photon of energy
E and linear momentum E I c is emitted and the nucleus
recoils. Find the decrease in internal energy.

K.E. of nucleus =$(1/2)mv^2 = (1/2)m\Bigg(\frac{E}{mc}\Bigg)^2 = \frac{E^2}{2mc^2}$ $\\$ Energy limited by Gamma photon = E $\\$ Decrease in internal energy = E + \frac{E^2}{2mc^2}

**41.** A block of mass 2.0 kg is moving on a frictionless
horizontal surface with a velocity of 1.0 m/s (figure
9-E12) towards another block of equal mass kept at rest.
The spring constant of the spring fixed at one end is 100
N/m. Find the maximum compression of the spring.

Mass of each block $M_A$ and $M_B$ = 2kg. $\\$ Initial velocity of the $1^{st}$ block (V) = 1m/s. $\\$ $ V_A = 1m/s \qquad V_B = 0m/s $ $\\$ Spring constant of the spring = 100N/m $\\$ The block A strikes the spring with a velocity 1m/s $\\$ After the collision it's velocity decreases continuously and at a instant the whole system (Block A + the compund spring + Block B) move together with a common velocity.$\\$ Let the velocity be V.$\\$ Using Conservation of energy, $(1/2)M_AV_A^2 + (1/2)M_BV_B^2 = (1/2)M_{AV^2} + (1/2)M_{BV^2} = (1/2)kx^2. $ $\\$ $(1/2) \times 2(1)^2 + 0 = (1/2) \times 2\times v^2 + (1/2) \times 2 \times v^2 + (1/2) x^2 \times 100$ $\\$ (Where x = max compression of the spring ) $\\$ $\Rightarrow 1= 2v^2 + 50 x^2 \qquad .......(1)$ $\\$ As there is no external force in the horizontal direction, the momentum should be conserved. $\\$ $\Rightarrow M_AV_A + M_BV_B = (M_A +M_B)V$ $\\$ $\Rightarrow 2\times 1 = 4\times v$ $\\$ $\Rightarrow V = (1/2)$m/s $ \qquad ........(2)$ $\\$ Putting in eq.(1) $\\$ 1 = $2 \times (1/4) + 50x +2$ $\\$ $(1/2) = 50x^2$ $\\$ $x^2 = 1/100m^2$ $\\$ $x = (1/10) m = 0.1m = 10cm.$

**42.** A bullet of mass 20 g travelling horizontally with a speed
of 500 m/s passes through a wooden block of mass 10.0
kg initially at rest on a level surface. The bullet emerges
with a speed of 100 m/s and the block slides 20 cm on
the surface before coming to rest. Find the friction
coefficient between the block and the surface
(figure 9-E13).

Mass of bullet m = 0.02kg $\\$ Initial velocity of bullet $V_1$ = 500m/s $\\$ Mass of block M = 10kg. $\\$ Initial velocity of block $u_2$ = 0 $\\$ Final velocity of bullet = 100m/s = v $\\$ Let the final velocity of block when the bullet emerges out, if block =v'. $\\$ $mv_1 + Mu_2 = mv + Mv'$ $\\$ $\Rightarrow 0.02 \times 500 = 0.02 \times 100 + 10 \times v'$ $\\$ $\Rightarrow v' = 0.8m/s $ $\\$ After moving a distance 0.2m it stops. $\\$ $\Rightarrow$ change in K.E. = Work done $\\$ $0- (1/2) \times 10 \times (0.8)^2 = -\mu \times 10 \times 10 \times 0.2 \Rightarrow \mu =0.16. $

**43.** A projectile is fired with a speed u at an angle 0 above
a horizontal field. The coefficient of restitution of
collision between the projectile and the field is e. How
far from the starting point, does the projectile makes its
second collision with the field ?

The projected velocity = u. $\\$ The angle of projection = $\theta$ $\\$ When the projectile hits the ground for the $1^{st}$ time, the velocity would be the same i.e. u.$\\$ Here the Component of velocity parallel to ground, $ucos\theta$ should remain constant.But the vertical component of the projectile undergoes a change after the collision. $\\$ $\Rightarrow e = \frac{usin\theta}{v} \Rightarrow v = eu sin\theta$ $\\$ Now for the $2^{nd}$ projectile motion, $\\$ U = velocity of projection = $\sqrt{(ucos\theta)^2 + (eusin\theta)^2}$ and Angle of projetion = $\alpha = tan^{-1}\Bigg(\frac{eusin\theta}{acos\theta}\Bigg) = tan^{-1}(e tan\theta)$ $\\$ or $tan\\alpha = etan\theta \qquad ......(2)$ $\\$ Because, $y= xtan \alpha - \frac{gx^2sec^2\alpha}{2u^2} \qquad ......(3) $ $\\$ Here, $y = 0, \qquad tan\alpha = etan\theta, \qquad sec^2\alpha = 1 +e^2 tan^2\theta $ $\\$ And $u^2 =u^2cos\theta + e^2sin^2\theta$ $\\$ Putting the above values in the equation (3) $\\$ $x e tan\theta = \frac{gx^2(1 + e^2tan^2\theta)}{2u^2(cos^2\theta +e^2sin^2\theta)} $ $\\$ $\Rightarrow x =\frac{2eu^2tan\theta(cos^2\theta +e^2sin^2\theta}{g(1 + e^2tan^2\theta)} $ $\\$ $\Rightarrow x = \frac{2eu^2tan\theta - cos^2\theta}{g} = \frac{eu^2sin2\theta}{g} $ $\\$ $\Rightarrow $ So, From the starting point O, it will fall at a distance = $ \frac{u^2sin2\theta}{g} + \frac{eu^2sin2\theta}{g} = \frac{u^2sin2\theta}{g}(1+e)$ $\\$

**44.** A ball falls on an inclined plane of inclination 0 from a
height h above the point of impact and makes a perfectly
elastic collision. Where will it hit the plane again ?

Angle inclination of the plane = $\theta$ $\\$ M the body falls through a height of h, $\\$ The Striking velocity of the projectile with the inclined plane v = $\sqrt{2gh}$ $\\$ Now, the projectile makes on angle $(90^o -2\theta)$ $\\$ Velocity of projection = $ u = \sqrt{2gh}$ $\\$ Let AB = L $\\$ So, $ x= lcos\theta, \quad y =-lsin\theta $ $\\$ From equation of trajectory, $ y = x tan\alpha - \frac{gx^2sec^2\alpha}{2u^2}$ $\\$ $-lsin\theta = lcos\theta tan(90^o - 2\theta) - \frac{g \times l^2cos^2\theta sec^2(90^o -2\theta)}{2 \times 2gh} $ $\\$ $\Rightarrow -lsin \theta = lcos\theta cot2 \theta - \frac{g \times l^2cos^2\theta cosec^22\theta}4{gh} $ $\\$ So, $\frac{l^2cos^2\theta cosec^22\theta}{4h} = sin\theta +cos\theta cot2\theta$ $\\$ $ \Rightarrow l = \frac{4h}{cos^2\theta cosec^22\theta}(sin\theta + cos\theta cot2\theta) = \frac{4h \times sin^2 2\theta}{cos^2\theta}\Bigg( sin\theta + cos\theta \times \frac{cos2\theta}{sin2\theta} \Bigg)$ $\\$ $\frac{4h \times 4sin^2 \theta cos^2\theta}{cos^2\theta}\Bigg( \frac{sin\theta \times sin2\theta + cos\theta cos2\theta}{sin 2\theta} \Bigg) = 16 h sin^2 \theta \times \frac{cos\theta}{2sin\theta cos\theta} = 8hsin\theta $

**45.** Solve the previous problem if the coefficient of restitution
is e. Use 0 = 45°, $e = \frac{3}{4}$
and h = 5 m.

$h =5m, \qquad \theta = 45^o, \qquad e=(\frac{3}{4}) $ $\\$ Here the velocity with ehih it would strike = v =$\sqrt{2g \times 5} = 10$m/sec $\\$ After collision, let it make anangle $ \beta$ with hoeizontal . The horizontal component of velocity $10 cos 45^o$ will remain unchanged and the velocity in the perpendicular direction to the plane after willsine.$\\$ $ \Rightarrow V_y = e \times 10sin45^o$ $\\$ $ = (3/4) \times 10 \times \frac{1}{\sqrt{2}} = (3.75)\sqrt{2} $m/sec $\\$ $ V_x = 10cos 45^o = 5\sqrt{2}$m/sec $\\$ So, $ u = \sqrt{V_x^2 + V_y^2} = \sqrt{50 + 28.125} = \sqrt{78.125} = 8.83 $ m/sec $\\$ Angle of reflection from the wall $\beta = tan^{-1}\Bigg( \frac{3.75\sqrt{2}}{5\sqrt{2}}\Bigg) = tan^{-1} \Bigg( \frac{3}{4}\Bigg) = 37^o$ $\\$ $ \Rightarrow $ Angle of projection $\alpha = 90 - (0 + \beta ) = 90 - (45^o + 37^o) = 8^o $ $\\$ let the distance where it falls = $L $ $\\$ $\Rightarrow x = L cos \theta, y = -L sin \theta$ $\\$ Angle of projection $(\alpha) = -8^o $ $\\$ Using equation of trajectory, $ y =xtan \alpha - \frac{gx^2 sec^2 \alpha}{2u^2} $ $\\$ $ \Rightarrow -lsin \theta = l cos \theta \times tan 8^o - \frac{g}{2} \times \frac{lcos^2 \theta sec^28^o}{u^2} $ $\\$ $ \Rightarrow -sin45^o = cos45^o -tan 8^o - \frac{10 cos^2 45^o sec 8^o}{(8.83)^2} (l) $ $\\$ Solving the above equation we get $ l =18.5 m$

**46.** A block of mass 200 g is suspended through a vertical
spring. The spring is stretched by 1.0 cm when the block
is in equilibrium. A particle of mass 120 g is dropped
on the block from a height of 45 cm. The particle sticks
to the block after the impact. Find the maximum
extension of the spring. Take g = 10 m/s 2

Block of the particle = m = 120gm = 0.12kg $\\$ In the equillibrium condition, the spring is streched by a distance x = 1.00 cm = 0.01m. $\\$ $\Rightarrow 0.2 \times g =K. x.$ $\\$ $\Rightarrow 2= K ]times 0.01 \Rightarrow K = 200 N/m$ $\\$ The velocity with the particle m will strike M is given by u$\\$ $ = \sqrt{2 \times 10 \times 0.45} = \sqrt{9} = 3m/sec$ $\\$ So after the collision, the velocity of the particle and the block is $\\$ $ v = \frac{0.12 \times 3}{0.32} = \frac{9}{8} m/sec$ $\\$ Let the spring be streched through an extra deflection of $\delta$. $\\$ $0 - (1/2) \times 0.32 \times (81/64) = 0.32 \times 10 \times \delta - (1/2 \times 200 \times (\delta + 0.1)^2 -(1/2) \times 200 \times (0.01)^2) $ $\\$ Solving the above equation we get $\\$ $\delta = 0.045 = 4.5 cm $

**47.** A bullet of mass 25 g is fired horizontally into a ballistic
pendulum of mass 5.0 kg and gets embedded in it
$(figure 9-E14)$. If the centre of the pendulum rises by a
distance of 10 cm, find the speed of the bullet.

Mass of the bullet = 25g = 0.025kg. $\\$ Mass of pendulum = 5kg. $\\$ The vertical displacement h =10cm = 0.1m $\\$ Let it strike the pendulum with a velocity u.$\\$ Let the final velocity be v, $\\$ $\Rightarrow mu = (M + m)v $ $\\$ $\Rightarrow v = \frac{m}{M+m}u = \frac{0.025}{5.025}\times u = \frac{u}{201}$ $\\$ Using conservation of energy, $\\$ $ 0 - (1/2) (M + m )V^2 = -(M +m)g \times h \quad \Rightarrow \frac{u^2}{(201)^2} = 2 \times 10 \times 0.1 = 2 $ $\\$ $\Rightarrow u = 201 \times \sqrt{2} = 280 m/sec.$

**48.** A bullet of mass 20 g moving horizontally at a speed of
300 m/s is fired into a wooden block of mass 500 g
suspended by a long string. The bullet crosses the block
and emerges on the other side. If the centre of mass of
the block rises through a height of 20.0 cm, find the
speed of the bullet as it emerges from the block.

Mass of the bullet = M = 20 gm = 0.02kg.$\\$ Mass of wooden block M = 500gm = 0.5Kg $\\$ Velocity of the bullet with ehich it strikes u = 300 m/sec$\\$ Let the bullet emerges out with velocity V and vlocity of block = V' $\\$ As per law of conservation of momentum. $\\$ $mu = Mv' + mv \qquad ...........(1)$ $\\$ Again applying work-enrgy principle for the block after the collision,$\\$ $0-(1/2)M \times V'^2 = -Mgh$ (where h =0.2m)$\\$ $\Rightarrow V^2 = 2gh $ $\\$ $V' = \sqrt{2gh} = \sqrt{20 \times 0.2} = 2$m/sec $\\$ Substituting the value of V' in equation (1), we get $\\$ $ 0.02 \times 300 = 0.5 \times 2 + 0.2 \times V$ $\\$ $\Rightarrow V = \frac{6.1}{0.02} = 250 $m/sec.

**49.** Two masses m, and m2 are connected by a spring of
spring constant k and are placed on a frictionless
horizontal surface. Initially the spring is stretched
through a distance $x_o$ when the system is released from
rest. Find the distance moved by the two masses before
they again come to rest.

Mass of the two blocks are $m_1, m_2 .$ $\\$ Initially the spring is stretched by $x_o$ $\\$ Spring constant K. $\\$ For the blocks to come to rest again, $\\$ Be $x_1$ and $x_2$ towards right and left respectively, $\\$ As o external force acts in horizontal direction, $\\$ $m_1x_1 = m_2x_2 \qquad ..........(1)$ $\\$ Again , the energy would be conserved in the spring. $\\$ $\Rightarrow (1/2)k \times x^2 = (1/2) k (x_1 +x_2 -x_0)^2$ $\\$ $\Rightarrow x_0 = x_1 + x_2 -x_0$ $\\$ $\Rightarrow x_1 +x_2 = 2x_0 \qquad ................(2)$ $\\$ $\Rightarrow x_1 =2x_0 -x_2 \quad similarly \quad x_1 = \Bigg( \frac{2m_2}{m_1+m_2}\Bigg)x_0 $ $\\$ $\Rightarrow m_1(2x_0 -x_2) = m_2x_2 \qquad \Rightarrow 2m_1x_0 - m_1x_2 = m_2x_2 \qquad \Rightarrow x_2= \Bigg( \frac{2m_1}{m_1 + m_2} \Bigg)x_0$

**50.** Two blocks of masses m, and m2 are connected by a
spring of spring constant k (figure 9-E15). The block of
mass m2 is given a sharp impulse so that it acquires a
velocity vo towards right. Find (a) the velocity of the
centre of mass, (b) the maximum elongation that the
spring will suffer.

a) $\therefore$ velocity of cw=enter of mass = $\frac{m_2 \times v_0 + m_1 \times 0}{m_1 +m_2} = \frac{m_2v_0}{m_1 +m_2}$ $\\$ b) The spring will attain maximum elongation when both velocity of two blocks will attain the velocity of center of mass. $\\$ d) x $\rightarrow$ maximum elongation of the spring. $\\$ Change of Kinetic energy = Potential stored in spring.$\\$ $\Rightarrow (1/2)m_2v_0^2 -(1/2)(m_1 + m_2) \Bigg( \frac{m_2v_0}{m_1 +m_2}\Bigg)^2 = (1/2) kx^2$ $\\$ $\Rightarrow m_2v_0^2 \Bigg( 1- \frac{m_2}{m_1 +m_2} \Bigg) = kx^2 \qquad \Rightarrow x = \Bigg( \frac{m_2}{m_1 +m_2} \Bigg)^{1/2} \times v_0$

**51.** Consider the situation of the previous problem. Suppose
each of the blocks is pulled by a constant force F instead
of any impulse. Find the maximum elongation that the
spring will suffer and the distances moved by the two
blocks in the process.

If both the blocks are pulled by some force, they suddenly move with some acceleration and instantaneously stop at same position where the elongation of spring is maximum $\\$ $\therefore $ Let $x_1, x_2 \rightarrow $ extension by block $m_1$ and $m_2$ $\\$ Total work done =$Fx_1 + Fx_2 \qquad ......(1)$ $\\$ $\therefore$ Increase the potential energy of spring = $(1/2) K (x_1 + x_2)^2 \qquad .........(2)$ $\\$ Equation (1) and (2) $\\$ $F(x_1 +x_2) = (1/2) K(x_1 + x_2)^2 \quad \Rightarrow (x_1+x_2) = \frac{2F}{K}$ $\\$ Since the net external force of the two blocks is zero thus same force act on opposite direction. $\\$ $\therefore$ $m_1x_1 = m_2x_2 \qquad ..........(3) $ $\\$ And $(x_1 + x_2) = \frac{2F}{K}$ $\\$ $\therefore x_2 = \frac{m_1}{m_2} \times 1 $ $\\$ Substituting $\frac{m_1}{m_2} \times 1 + x_1 = \frac{2F}{K}$ $\\$ $\Rightarrow x_1\Bigg( 1+ \frac{m_1}{m_2} \Bigg) = \frac{2F}{K} \qquad \Rightarrow x_1 = \frac{2F}{K} \frac{m_2}{m_1 + m_2}$ $\\$ Similarly, x_2 = $\Rightarrow x_1 = \frac{2F}{K} \frac{m_2}{m_1 + m_2}$

**52.** Consider the situation of the previous problem. Suppose
the block of mass m, is pulled by a constant force F, and
the other block is pulled by a constant force F2. Find
the maximum elongation that the spring will suffer.

Acceleration of mass $m_1$ = $\frac{F_1 - F_2}{m_1 + m_2}$ $\\$ Similarly Acceleration of mass $m_2 = \frac{F_1 - F_2}{m_1 + m_2}$ $\\$ Due to $F_1$ and $F_2$ block of mass $m_1$ and $m_2$ will experience different acceleration and experience inertia force. $\\$ $\therefore$ Net force on $ m_1 = F_1 -m_1a$ $\\$ $= F_1 - m_1 \times \frac{F_1 - F_2}{m_1 + m_2} = \frac{m_1F_1 +m_2F_2 -m_1F_1 +F_2m_1}{m_1+m_2} = \frac{m_2F_1 + m_1F_2}{m_1 +m_2}$ $\\$ Similarly Net force on $m_2 = F_2 -m_2a$ $\\$ $= F_2 - m_2 \times \frac{F_2 - F_1}{m_1 + m_2} = \frac{m_1F_2 +m_2F_2 -m_2F_2 +F_1m_2}{m_1+m_2} = \frac{m_1F_2 + m_2F_2}{m_1 +m_2} $ $\\$ If $m_1$ displaced by a distance $x_1$ and $x_2$ by $m_2$ the maximum extension of the spring is $x_1 +m_2$ $\\$ $Work done by the blocks = energy stored in the spring.,$ $\\$ $\Rightarrow \frac{m_2F_1 + m_1F_2}{m_1 +m_2} \times x_1 + \frac{m_2F_1 + m_1F_2}{m_1 +m_2} \times x_2 = (1/2)K (x_1 + x_2)^2$ $\\$ $x_1 +x_2 = \frac{2}{K} \frac{m_1F_2 + m_2F_2}{m_1 +m_2} $

**53.** Consider a gravity-free hall in which an experimenter
of mass 50 kg is resting on a 5 kg pillow, 8 ft above the
floor of the hall. He pushes the pillow down so that it
starts falling at a speed of 8 ft/s. The pillow makes a
perfectly elastic collision with the floor, rebounds and
reaches the experimenter's head. Find the time elapsed
in the process.

Mass of the man $(M_m)$ is 50 kg. $\\$ Mass of the pillow $(M_p)$ is 5 kg. When the pillow is pushed by the man, the pillow will go down while the man goes up. It because of external force on the system which is zero. $\\$ $\Rightarrow$ acceleration of center of mass is zero.$\\$ $\Rightarrow$ velocity of center of mass is zero $\\$ $\therefore$ As the initial velocity of the system is zero. $\\$ $\therefore M_m \times V_m = M_p \times V_p \qquad ......(1)$ $\\$ Given the velocity of pillow is $80$ft/s $\\$ Which is relative velocity of pillow w.r.t. man.$\\$ $\vec{V}_{p/m} = \vec{V}_p - \vec{V}_m = \vec{V}_p - (-\vec{V}_m ) = \vec{V}_p +\vec{V}_m \qquad \Rightarrow \vec{V}_p = \vec{V}_{p/m}-\vec{V}_m $ $\\$ Putting in equation (1) $\\$ $M_m \times V_m = M_p (V_{p/m}-V_m )$ $\\$ $\Rightarrow 50 \times V_m = % \times (8 - V_m)$ $\\$ $\Rightarrow 10 \times V_m = 8 - V_m \quad \Rightarrow V_m = \frac{8}{11} = 0.721 $m/s $\\$ $\therefore$ Absolute velocity of pillow = $8 - 0.727 = 7.2$ ft/sec $\\$ $\therefore$ Time taken to reach the floor = $\frac{S}{V} = $\frac{8}{7.2} = 1.1 sec$ $\\$ As the mass of ball >>> then pillow $\\$ The velocity of block before the collision = velocity after the collision $\\$ $\Rightarrow$ Time of ascent = 1.11 sec $\\$ $\therefore$ Total time taken = $1.11 + 1.11 = 2.22$sec.

**54.** The track shown in figure (9-E16) is frictionless. The
block B of mass 2m is lying at rest and the block A of
mass m is pushed along the track with some speed. The
collision between A and B is perfectly elastic. With what
velocity should the block A be started to get the sleeping
man awakened ?

Let the velocity of A =$u_1$ $\\$ Let the final velocity when reaching B becomes collision = $v_1$ $\\$ $\therefore (1/2)mv_1^2 - (1/2)mu_1^2 =mgh $ $\\$ $v_1^2 - u_1^2 = 2gh \qquad \Rightarrow v_1 = \sqrt{2gh-u_1^2} \qquad .....(1)$ $\\$ When the block B reached at the upper man's head, the velocity of B is just zero.$\\$ $\therefore (1/2)\times 2m \times 0^2 - (1/2) \times 2m \times v^2 =mgh \qquad \Rightarrow v =\sqrt{2gh}$ $\\$ $\therefore$ Before collision velocity of $u_A = v_1 \qquad u_b = 0$ $\\$ After collision velocity of $v_A = v \quad (say) \qquad v_B = \sqrt{2gh} $ $\\$ Since it is an elastic collision the momentum and K.E. should be conserved $\\$ $\therefore m \times v_1 + 2m \times 0 = m \times v +2m \times \sqrt{2gh}$ $\\$ $\Rightarrow v_1 -v = 2 \sqrt{2gh}$ $\\$ Also, $(1/2) \times m \times v_1^2 + (1/2) 2m \times 0^2 = (1/2) \times m \times v^2 +(1/2) \times 2m \times (\sqrt{2gh})^2 $ $\\$ $\Rightarrow v_1^2 -v^2 = 2\sqrt{2gh} \times \sqrt{2gh} \qquad ......(2)$ $\\$ Dividing (1) by (2), $ \frac{(v_1 +v)(v_1 -v)}{v_1 +v} = \frac{2 \times \sqrt{2gh} \times \sqrt{2gh}}{2 \times \sqrt{2gh}} \rightarrow v_1 +v = \sqrt{2gh} \qquad ......(3)$ $\\$ Adding (1) and (3), $2v_1 =3\sqrt{2gh} \Rightarrow v_1 =(3/2)\sqrt{2gh}$ $\\$ But $v_1 = \sqrt{2gh + u^2} = \Bigg( \frac{3}{2} \Bigg) \sqrt{2gh} $ $\\$ $\Rightarrow 2gh +u^2 = \frac{9}{4} \times 2gh$ $\\$ $u = 2.5 \sqrt{2gh}$ $\\$ So the block will travel with the velocity greater than $2.5\sqrt{2gh}$ so awake the man by B.

**55.** A bullet of mass 10 g moving horizontally at a speed of
5017 m/s strikes a block of mass 490 g kept on a frictionless track as shown in figure (9-E17). The bullet
remains inside the block and the system proceeds
towards the semicircular track of radius 0.2 m. Where
will the block strike the horizontal part after leaving the
semicircular track ?

Mass of the block = 490 gm. $\\$ Mass of the bullet = 10 gm. $\\$ Since the bulet embedded inside the block, it is an plastic collision. $\\$ Initially velocity of bullet $v_1 = 50 \sqrt{7}$ m/sec $\\$ Velocity of the block is $v_2 = 0$ $\\$ Let the final velocity of both = v $\\$ $\therefore 10 \times 10^{-3} \times 50 \times \sqrt{7} +10^{-3} \times 190 | 0 = (490 +10) \times 10^{-3} \times V_A$ $\\$ $V_A = \sqrt{7} $m/s $\\$ When the block lossess the contact at 'D' the component mg will act on it.$\\$ $\frac{m(V_B)^2}{r} = mg sin\theta \Rightarrow (V_B)^2 = gr sin \theta \qquad ....(1)$ $\\$ Prutting work energy principle $\\$ $(1/2)m \times (V_B)^2 - (1/2) \times m \times (V_A)^2 = -mg(0.2 + 0.2sin \theta)$ $\\$ $\Rightarrow (1/2) \times gr sin \theta - (1/2) \times (\sqrt{7})^2 = -mg(0.2 + 0.2sin \theta) $ $\\$ $\Rightarrow 3.5 - (1/2) \times 9.8 \times 0.2 \times sin \theta = 9.8 \times 0.2(1+ sin \theta) $ $\\$ $ \Rightarrow 3.5 - 0.98 sin \theta = 1.96 + 1.96 sin \theta $ $\\$ $ sin \theta = (1/2) \quad \Rightarrow \theta = 30^o $ $\\$ Angle of projection =$ 90^o -30^o = 60^o$ $\\$ $\therefore$ time of reaching the ground = $\sqrt{\frac{2h}{g}} \quad = \sqrt{\frac{2 \times (0.2 + 0.2 \times sin 30^o)}{9.8} } = 0.247$ sec $\\$ $\therefore$ Distance travelled in horizontal direction.$\\$ $s =V cos \theta \times t = \sqrt{gr sin \theta} \times t = \sqrt{9.8 \times 2 \times (1/2)} \times 0.247 = 0.196$m. $\\$ Total distance = $ (0.2 -0.2 cos 30^o) + 0.196 = 0.22 $m.

**56.** Two balls having masses m and 2m are fastened to two
light strings of same length 1 (figure 9-E18). The other
ends of the strings are fixed at 0. The strings are kept
in the same horizontal line and the system is released
from rest. The collision between the balls is elastic. (a)
Find the velocities of the balls just after their collision.
(b) How high will the balls rise after the collision ?

Let the velocity of m reaching at lower end = $v_1$ $\\$ from work energy principle, $\\$ $\therefore (1/2) \times m \times (1/2) \times m \times 0^2 = mg l $ $\\$ $\Rightarrow v_1 = \sqrt{2gl}$ $\\$ Similarly velocity of heavy block will be $v_2 = \sqrt{2gh}$ $\\$ $v_1 = v_2 = u (say)$ $\\$ Let the final velocity of m and 2m $v_1$ and $v_2$ respectively, $\\$ According to law of conservation of momentum. $\\$ $ m \times x_1 + 2m \times V_2 = mv_1 + 2mv_2 $ $\\$ $ \Rightarrow m \times u -2 m u = mv_1 + 2mv_2 $ $\\$ $ \Rightarrow v_1 +2v_2 = -u \qquad.........(1)$ $\\$ Again, $v_1 - v_2 = - (V_1- V_2)$ $\\$ $\Rightarrow v_1 -v_2 = -[u -(-v)] = -2V \qquad.....(2)$ $\\$ Subtracting, $\\$ $3v_2 = u \Rightarrow v_2 = \frac{u}{3} = \frac{\sqrt{2gl}}{3}$ $\\$ Substituting in (2) $\\$ $v_1 - v_2 = -2u \Rightarrow v_1 = -2u +v_2 = -2u + \frac{u}{3} = -\frac{5}{3}u = -\frac{5}{3} \times \sqrt{2gl} = -\frac{\sqrt{50gl}}{3} $ $\\$ b) Putting the work energy principle, $\\$ $(1/2) \times 2m \times 0^2 - (1/2) \times 2m \times (v_2)^2 = -2m \times g \times h$ $\\$ [h $\rightarrow$ height gone by heavy ball] $\\$ $\Rightarrow (1/2) \frac{2g}{9} = l \times h \qquad h = \frac{l}{9}$ $\\$ Similarly, $(1/2) \times m \times 0^2 -(1/2) \times m \times v_1^2 =m \times g \times h_2 $ $\\$ [ height reached by small ball ]$\\$ $\Rightarrow (1/2) \times \frac{50gl}{9} = g \times h_2 \qquad \Rightarrow h_2 = \frac{25l}{9}$ $\\$ Some $h_2$ is more than $2l$, the velocity at height point will not be zero and the 'm' will rise by a distance $2l$.

**57.** A uniform chain of mass M and length L is held
vertically in such a way that its lower end just touches
the horizontal floor. The chain is released from rest in
this position. Any portion that strikes the floor comes to
rest. Assuming that the chain does not form a heap on
the floor, calculate the force exerted by it on the floor
when a length x has reached the floor.

Let us consider a small element at a distance 'x' from the floor of length 'dy'.$\\$ So, $dm = \frac{M}{L}dx$ $\\$ $\\$ So, the velocity with which the element will strike the floor is $ v = \sqrt{2gx}$ $\\$ $\therefore$ So, the momentum transferred to the floor is, $\\$ $M = (dm)v = \frac{M}{L} \times dx \times \sqrt{2gx} $ [because the element comes to rest] $\\$ S0, the force exerted on the floor change in momentum is given by, $F_1 =\frac{dM}{dt} = \frac{M}{L} \times \frac{dx}{dt} \times \sqrt{2gx} $ $\\$ Because, $v = \frac{dx}{dt} = \sqrt{2gx}$( for the chain element) $\\$ $F_1 = \frac{M}{L} \times \sqrt{2gx} \times \sqrt{2gx} = \frac{M}{L} \times 2gx = \frac{2Mgx}{L} $ $\\$ Again, the force exerted due to 'x' length of chain on the floor due to its own weight is given by, $\\$ $W = \frac{M}{L}(x) \times g = \frac{Mgx}{L}$ $\\$ So, the total force exerted is given by, $F = F_1 + W = \frac{2Mgx}{L} + \frac{Mgx}{L} = \frac{3Mgx}{L}$

**58.** The blocks shown in figure (9-E19) have equal masses.
The surface of A is smooth but that of B has a friction
coefficient of 0.10 with the floor. Block A is moving at
a speed of 10 m/s towards B which is kept at rest. Find
the distance travelled by B if (a) the collision is perfectly
elastic and (b) the collision is perfectly inelastic. Take g
= 10 mls 2.

$V_1 = 10$m/s $\qquad V_2 =0$ $\\$ $V_1, v_2 \rightarrow$ velocity of ACB after collision. $\\$a) If the edlision is perfectly elastic, $\\$ $mV_1 + mV_2 = mv_1 + mv_2$ $\\$ $\Rightarrow 10 + 0 = v_1 + v_2$ $\\$ $\Rightarrow v_1 + v_2 = 10 \qquad.......(1)$ $\\$ Again, $v_1 -v_2 = - (u_1 - v_2) = -(10- 0) = -10 \qquad .......(2)$ $\\$ Subtracting (2) from (1)$\\$ $2v_2 = 20 \quad \Rightarrow v_2 = 10 $m/s $\\$ The deacceleration of B = $\mu g $ $\\$ Putting work energy principle, $\\$ $\therefore (1/2) \times m \times 0^2 - (1/2) \times m \times v_2^2 = -m \times a \times h $ $\\$ $\Rightarrow -(1/2) \times 10^2 = -\mu g \times h \qquad \Rightarrow h = \frac{100}{2 \times 0.1 \times 10 } = 50m$ $\\$ b) If the collision perfectly elastic. $\\$ $m \times u_1 + m \times u_2 = (m+m)\times v$ $\\$ $\Rightarrow m \times 10 + m \times 0 = 2m \times v \qquad \Rightarrow v = \frac{10}{2} = 5m/s$ $\\$ The two blocks will move together sticking to each other $\\$ Putting work energy princilple, $\\$ $(1/2) \times 2m \times 0^2 - (1/2) \times v^2 = 2m \times \mu g \times s$ $\\$ $\Rightarrow \frac{5^2}{0.1 \times 10 \times 2} =s \qquad \Rightarrow s = 12.5m$

**59.** The friction coefficient between the horizontal surface
and each of the blocks shown in figure (9-E20) is 0.20.
The collision between the blocks is perfectly elastic. Find
the separation between the two blocks when they come
to rest. Take $g = 10 m/s^2$

Let the velocity of 2kg block on reaching the 4 kg block before collision = $u_1.$ $\\$ Given, $V_2 = 0$(velocity of 4kg block). $\\$ $\therefore$ From work energy principle, $\\$ $(1/2) m \times u_1^2 - (1/2)m \times 1 = -m \times ug \times s$ $\\$ $\Rightarrow \frac{u_1^2 -1}{2} = -2 \times 5 \qquad \Rightarrow -16 = \frac{u_1^2 -1}{4}$ $\\$ $64 \times 10^{-2} = u_1^2 -1 \qquad \Rightarrow u_1 = 6 m/s$ $\\$ Since it is a perfectly elastic collision, $\\$ Let $V_1 , V_2 \rightarrow$ velocity of 2kg & 4 kg block after collision, $\\$ $m_1V_1 + m_2V_2 = m_1v_1 + m_2v_2$ $\\$ $\Rightarrow 2 \times 0.6 +4 \times 0 = 2v_1 +4v_2 \Rightarrow v_1 + 2v_2 = 0.6 \qquad ...(1)$ $\\$Again, $V_1 -V_2 = -(u_1 -u_2) = -(0.6-0)=-0.6$ $\\$ Subtracting (2) from (1)$\\$ $3v_2 = 1.2 \qquad \Rightarrow v_2 = 0.4m/s$ $\\$ $\therefore v_1 = 0.6 + 0.4 = -0.2 m/s$ $\\$ $\therefore$ Putting work energy principle for $1^{st}$ 2kg block when come to rest.$\\$ $(1/2) \times 4 \times 0^2 - (1/2) \times 4 \times (0.4)^2 = -4 \times 0.2 \times 10 \times s$ $\\$ $2 \times 0.4 \times 0.4 = 4 \times 0.2 \times 10 \times s \qquad \Rightarrow S_2 = 4cm $ $\\$ Distance between 2kg & 4 kg block = $S_1 +S_2 = 1+4 =5cm$

**60.** A block of mass m is placed on a triangular block of
mass M, which in turn is placed on a horizontal surface
as shown in figure (9-E21). Assuming frictionless surfaces find the velocity of the triangular block when
the smaller block reaches the bottom end

The block 'm' will slide down the inclined plane of mass M with acceleration $a_1g sin \alpha$ (relative) to the inclined plane. $\\$ The horizontal component of $a_1$ will be, $a_x = g sin \alpha cos \alpha,$ for which the block M will accelerate towards left. Let, the acceleration be $a_2$. $\\$ According to the concept of center of mass, (in horizontal direction external force is zero). $\\$ $ma_x = (M+ m)a_2$ $\\$ $\Rightarrow a_2 = \frac{ma_x}{M + m} = \frac{mgsin \alpha cos \alpha}{ M+m } \qquad ....(1)$ $\\$ So, the absolute (Resultant) acceleration of 'm' on the block 'M' along the direction of incline will be $\\$ $a = g sin \alpha - a_2 cos \alpha = g sin \alpha -\frac{mgsin \alpha cos^2 \alpha}{ M+m } = g sin \alpha \Bigg( 1- \frac{mcos^2 \alpha}{M+m} \Bigg)$ $\\$ $= g sin \alpha \Bigg(\frac{M +m -mcos^2 \alpha}{M+m} \Bigg)$ $\\$So, $a = gsin \alpha \Bigg(\frac{M + msin^2 \alpha}{M+m} \Bigg)$ $\\$ Let the time taken by the block 'm' to reach the bottom end br 't', $\\$ Now, $S= ut + (1/2) at^2$ $\\$ $\Rightarrow \frac{h}{ sin \alpha} = (1/2) at^2 \qquad \Rightarrow t = \sqrt{\frac{2}{a sin \alpha}}$ $\\$ So, the velocity of the bigger block after time 't' will be. $\\$ $V_m = u +a_2t = \frac{mg sin \alpha cos \alpha}{ M+m}\sqrt{\frac{2h}{a sin \alpha}} = \sqrt{\frac{2m^2g^2h sin^2 \alpha cos^2 \alpha}{(M+m)^2 a sin \alpha}}$ $\\$ Now. Subtracting the value of a from equation (2) we get, $\\$ $V_M = \Bigg( \frac{2m^2g^2h sin^2 \alpha cos^2 \alpha}{(M+m)^2 a sin \alpha} \times \frac{(M+m)}{g sin \alpha (M+m sin^2 \alpha)} \Bigg)^{1/2}$ $\\$ or, $\\$ $V_M = \Bigg( \frac{2m^2g^2h cos^2 \alpha}{(M+m)(M+m sin^2 /alpha)} \Bigg)^{1/2}$

**61.** Figure (9-E22) shows a small body of mass m placed
over a larger mass M whose surface is horizontal near
the smaller mass and gradually curves to become
vertical. The smaller mass is pushed on the longer one
at a speed v and the system is left to itself. Assume that
all the surfaces are frictionless. (a) Find the speed of the
larger block when the smaller block is sliding on the
vertical part. (b) Find the speed of the smaller mass
when it breaks off the larger mass at height h. (c) Find
the maximum height (from the ground) that the smaller
mass ascends. (d) Show that the smaller mass will again
land on the bigger one. Find the distance traversed by
the bigger block during the time when the smaller block
was in its flight under gravity.

The mass 'm' is given a velocity 'v' over the largest mass M. $\\$a) When the smaller block is travelling on the vertical part, let the velocity of the bigger block be $v_1$ towards left. $\\$ From law of conservation of momentum, (in the horizontal direction) $\\$ $mv = (M+m)v_1$ $\\$ $\Rightarrow v_1 = \frac{mv}{M+m}$ $\\$ b) When the smaller block breaks off, let its resultant velocity is $v_2$. $\\$ From law of conservation of energy, $\\$ $(1/2)mv^2 = (1/2)Mv_1^2 + (1/2)mv_2^2 +mgh$ $\\$ $\Rightarrow v_2^2 = v^2 - \frac{M}{m}v_1^2 -2gh \qquad .....(1)$ $\\$ $\Rightarrow v_2^2 = V^2 \Bigg(1- \frac{M}{m} \times \frac{m^2}{(M+m)^2} \Bigg) -2gh $ $\\$ $\Rightarrow = \Bigg( \frac{(m^2 + Mm + m^2 )}{(M+m)^2}v^2 -2gh \Bigg)^{1/2}$ $\\$c) Now the vertical component of the velocity $v_2$ of mass 'm' is given by, $v_y^2 = v_2^2 = v_1^2$ $\\$ $= \frac{M^2 + Mm +m^2}{(M+m)^2}v^2 - 2gh - \frac{m^2v^2}{(M+m)^2}$ [$\therefore v_1 = \frac{mv}{M+v}$ ] $\\$ $\Rightarrow v_y^2 = \frac{M^2 + Mm +m^2 -m^2}{(M+m)^2}v^2 - 2gh $ $\\$ $\Rightarrow v_y^2 = \frac{Mv^2}{(M+m)}v^2 - 2gh \qquad .......(2)$ $\\$ To find the maximum height (from the ground), let us assume the body rises to the height 'h' over and above 'h', $\\$ now, $(1/2)mv_y^2 = mgh_1 \Rightarrow h_1 = \frac{V_y^2}{2g} \qquad .....(3)$

$\\$ So, Total height = $h+h_1 = h + \frac{V_y^2}{2g} = h+ \frac{Mv^2}{(M+m)2g} - h $ $\\$[From equation (2) and (3)] $\\$ $\Rightarrow H = \frac{Mv^2}{(M+m)2g}$ $\\$ d) Because, the smaller mass has also got a horizontal component of velocity $v_1$ at the time it breaks off from 'M' (which has a velocity v_1), the block 'm' will again land on the block 'M'(bigger one). $\\$ Let us find out the time of flight of block 'm' after it breaks off.$\\$ During the upwards motion (BC),$\\$ $0 = v_y -gt_1 $ $\\$ $\Rightarrow t_1 = \frac{v_y}{g} = \frac{1}{g}\Bigg( \frac{Mv^2}{(M+m)}-2g \Bigg)^{1/2} \qquad ...(4)$ [from the equation (2)]$\\$ So, the time for which the smaller blocks was in its flight is given by, $\\$ $T = 2t_1 = \frac{2}{g}\Bigg( \frac{Mv^2 -2(M+m)gh}{(M+m)} \Bigg)^{1/2}$ $\\$ So, the distance travelled by the bigger block during this time is, $\\$ $S = v_1T = \frac{mv}{M+m} \times \frac{2}{g} \frac{(Mv^2 -2(M+m)gh)^{1/2}}{(M+m)^{1/2}}$ $\\$ or, $S = \frac{2mv(Mv^2 -2(M+m)gh)^{1/2}}{g(M+m)^{3/2}}$

**62.** A small block of superdense material has a mass of
$3 \times 10^24$ kg. It is situated at a height h (much smaller
than the earth's radius) from where it falls on the earth's
surface. Find its speed when its height from the earth's
surface has reduced to h /2. The mass of the earth is
$6 \times 10^24$ kg.

Given h <<< R $\\$ $G_{mass} =6 | 10^{24}kg.$ $\\$ $M_b = 3 \times 10^{24}kg.$ $\\$ Let $V_a \rightarrow$ velocity of earth.$\\$ $V_b \rightarrow$ velocity of thr block$\\$ The two block are attracted by gravitational force of attraction. The gravitational potential energy stored will be the K.E. of two blocks. $\\$ $\vec{G}^{plm} \Bigg( \frac{1}{R+(h/2)} - \frac{1}{R+h}\Bigg) = (1/2)m_e \times v_e^2 + (1/2)m_b \times v_b^2$ $\\$ Again as the an internal forces acts $\\$ $M_eV_e = m_bV_b \qquad V_e = \frac{m_bV_b}{M_e} \qquad...........(2)$ $\\$ Putting in equation (1) $\\$ $G_{me} \times m_b \Bigg( \frac{2}{2R +h} - \frac{1}{R+h} \Bigg)$ $\\$ $= (1/2) \times M_e \times \frac{m_b^2v_b^2}{M_e^2} \times v_e^2 + (1/2)M_b \times V_b^2$ $\\$ $=(1/2) \times m_b \times V_b^2 \Bigg( \frac{M_b}{M_e} +1 \Bigg) $ $\\$ $\Rightarrow GM\Bigg( \frac{2R +2h -2R -h}{(2R+h)(R+h)} \Bigg) = (1/2) \times V_b^2 \times \Bigg( \frac{3 \times 10^{24}}{6 \times 10^{24}} +1 \Bigg) \Rightarrow \Bigg[\frac{GM \times h}{2R^2 + 3Rh +h^2}\Bigg] = (1/2) \times V_b^2 \times (3/2) $ $\\$ As h <<< R, if can be neglected,$\\$ $\frac{GM \times h}{2R^2} = (1/2) \times V_b^2 \times (3/2) \quad \Rightarrow V_b = \sqrt{\frac{2gh}{3}}$

**63.** A body of mass m makes an elastic collision with another
identical body at rest. Show that if the collision is not
head-on, the bodies go at right angle to each other after
the collision.

Since it is not an head on collision, th two bodies move in different dimensions. Let $V_1, V_2 \rightarrow$ velocities of the bodies vector collision. Since, the collision is elastic. Applying law of conservation of momentum on X-direction. $\\$ $mu_1 + mx = mv_1 cos \alpha + mv_2 cos \beta$ $\\$ $\Rightarrow v_1 cos a + v_2 cos b = u_1 \qquad ......(1)$ $\\$ Putting law of conservation of momentum in y direction.$\\$ $0 = mv_1 sin \alpha - mv_2 sin \beta$ $\\$ $\Rightarrow v_1 sin \alpha = v_2 sin \beta \qquad ......(2)$ $\\$ Again $(\frac{1}{2})mu_1^2 + 0 = (\frac{1}{2})mv_1^2 + (\frac{1}{2})mxv_2^2 $ $\\$ $\Rightarrow u_1^2 = v_1^2 + v_2^2 \qquad .........(3)$ $\\$ Squaring equation(1) $\\$ $u_1^2 = v_1^2 cos^2 \alpha + v_2^2 cos^2 \beta + 2 v_1v_2 cos \alpha cos \beta $ $\\$ Equating (1) & (3) $\\$ $v_1^2 + v_2^2 = v_1^2 cos^2 \alpha + v_2^2 cos^2 \beta + 2v_1v_2 cos \alpha cos \beta $ $\\$ $\Rightarrow v_1^2 sin^2 \alpha + v_2^2 sin^2 \beta = 2v_1v_2 cos \alpha cos \beta $ $\\$ $\Rightarrow 2v_1^2 sin^2 \alpha = 2v_1 \times \frac{v_1 sin \alpha}{sin \beta} \times cos \alpha cos \beta $ $\\$ $\Rightarrow sin \alpha sin \beta = cos \alpha cos \beta \qquad \Rightarrow cos \alpha cos \beta = sin \alpha sin \beta =0 $ $\\$ $\Rightarrow cos (\alpha + \beta)=0=cos 90^o \qquad (\alpha + \beta)=90^o $

**64.** A small particle travelling with a velocity u collides
elastically with a spherical body of equal mass and of
radius r initially kept at rest. The centre of this spherical
body is located a distance p(< r) away from the direction
of motion of the particle (figure 9-E23). Find the final
velocities of the two particles.$\\$[Hint : The force acts along the normal to the sphere
through the contact. Treat the collision as one-
dimensional for this direction. In the tangential direction
no force acts and the velocities do not change].

Let the mass of both the particle and the spherical body be 'm'. The particle velocity 'v' has two components, $v cos \alpha $ normal to the sphere and $v sin \alpha $ tangential to sphere. $\\$ After the collision, they will exchange their velocities. So, the spherical body will have a velocity $v cos \alpha $ and the particle will not have any component of velocities in this direction $\\$ [The collision will due to the component $v cos \alpha $ in the normal direction. But, the tangential velocity, of the particle $v sin \alpha $ will be unaffected ].$\\$ So, the velocity of the sphere = $v cos \alpha = \frac{v}{r} \sqrt{r^2 - \rho^2} $[from (fig-2)] $\\$ And the velocity of the particle = $v sin \alpha = \frac{v\rho}{r} $