 # Centre of Mass, Linear Momentum, Collision

## Concept Of Physics

### H C Verma

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.

##### Solution : $m_1$ = $1$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.

##### Solution : 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 $52^o$, $$$y_1 = 0 \\ m_2 = 1kg, x_2= -\big( 0.96* 10^{-10} \big)sin 52^o,$$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.

##### Solution : 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

##### Solution : 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$$R^2 \times T \times \rho \\ m_2 = \pi \big(2R\big) ^2 | T \times \rho \\ where T = Thickness of the two disc \\ \rho = Density of the two disc \\ \therefore the position of the center of the mass \Bigg( \frac{m_1x_1 + m_2x_2 } { m_1 + m_2 } , \frac{m_1y_1 + m_2y_2} { m_1 + m_2} \Bigg) \\ x_1 = R, y_1 = 0 \\ x_2 = 0, y_2 = 0 \\ \Bigg( \frac{ \pi R^2 T \rho R + 0} {\pi R^2 T \rho + \pi \big(2R\big)^2 T \rho} , \frac{0} { m_1 + m_2} \Bigg) =\Bigg( \frac{ \pi R^2 T \rho R} { 5\pi R^2 T \rho } , 0 \Bigg) =\Bigg( \frac{R}{5},0 \Bigg) 5 A disc of radius R is cut out from a larger disc of radius 2R in such a way that the edge of the hole touches the edge of the disc. Locate the center of mass of the residual disc. ##### Solution : Let 'O' be the origin of the system. \\ R = Radius of the smaller disc \\ 2R = Radius of the bigger disc \\ The smaller disc is cut out from bigger disc \\ As from the figure \\ m_1= \pi$$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.

##### Solution : 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 $(9-E3)$.

##### Solution :  $m_1$=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?

##### Solution :

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 ?

##### Solution :

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 ?

##### Solution : 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 ?

##### Solution : $m_1$ = $60$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.

##### Solution : 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 = $L$.

##### Solution : 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.

##### Solution :

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.

##### Solution :

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.

##### Solution :

$m_1v_1$ = $m_2v_2$ $\\$ $\Rightarrow$ $50 \times 1.8$ = $6 \times 10^{24} \times v_2$ $\\$ $\Rightarrow$ $v_2$ = $\frac{50 \times 1.8}{6 \times 10^24}$ = $1.5 \times 10^{-23}$ 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.

##### Solution : 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 ?

##### Solution : 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.

##### Solution : 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 $R2$ $(figure 9-E5)$

##### Solution : 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.

##### Solution : $P_{Incidence} = (h/\lambda)cos\theta \vec{i}-(h/ \lambda)sin \theta \vec{j}$ $\\$ $P_{Reflected} = (h/\lambda)cos\theta \vec{i}-(h/ \lambda)sin \theta \vec{j}$ $\\$ 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

##### Solution : 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 ?

##### Solution :

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 $900 kg/m^3$.

##### Solution :

$d = 1cm \qquad v = 20m/s \qquad u = 0 \qquad \rho = 900kg/m^3 = 0.9 gm/cm^3$ $\\$ $volume = (4/3)\pi r^3 = (4/3) \pi (0.5)^3 = 0.5238cm^3$ $\\$ $\therefore$ 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.

##### Solution :

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 ?

##### Solution :

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.

##### Solution :

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

##### Solution :

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.

##### Solution : 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 ?

##### Solution :

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 ?

##### Solution :

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.

##### Solution :

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.

##### Solution :

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

##### Solution :

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.

##### Solution :

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.

##### Solution :

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.

##### Solution : 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 ?

##### Solution : 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.

##### Solution : $u = \sqrt{2gh} =$ 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.

##### Solution : 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.

##### Solution : 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).

##### Solution : 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 ?

##### Solution : 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 ?

##### Solution : 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.

##### Solution : $h =5m, \qquad \theta = 45^o, \qquad e=(\frac{3}{4})$ $\\$ 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

##### Solution : 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.

##### Solution :

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.

##### Solution :

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.

##### Solution : 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.

##### Solution : 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.

##### Solution : 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.

##### Solution : 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.

##### Solution : 