**1** **Give the magnitude and direction of the net force acting on$\\$ (a) a drop of rain falling down with a constant speed,$\\$ (b) a cork of mass 10 g floating on water,$\\$ (c) a kite skillfully held stationary in the sky,$\\$ (d) a car moving with a constant velocity of $30 km/h$ on a rough road,$\\$ (e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.**

(a) Zero net force$\\$ The rain drop is falling with a constant speed. Hence, it acceleration is zero. As per Newton’s second law of motion, the net force acting on the rain drop is zero.$\\$ (b) Zero net force$\\$ The weight of the cork is acting downward. It is balanced by the buoyant force exerted by the water in the upward direction. Hence, no net force is acting on the floating cork.$\\$ (c) Zero net force$\\$ The kite is stationary in the sky, i.e., it is not moving at all. Hence, as per Newton’s first law of motion, no net force is acting on the kite.$\\$ (d) Zero net force$\\$ The car is moving on a rough road with a constant velocity. Hence, its acceleration is zero. As per Newton’s second law of motion, no net force is acting on the car.$\\$ (e) Zero net force$\\$ The high speed electron is free from the influence of all fields. Hence, no nett force is acting on the electron.

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**3** **A pebble of mass $0.05 kg$ is thrown vertically upwards. Give the direction an d magnitude of the net force on the pebble,$\\$ (a) during its upward motion,$\\$ (b) during its downward motion,$\\$ (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of $45^o$ with the horizontal direction?$\\$ Ignore air resistance.**

$0.5 N,$ in vertically downward direction, in all cases$\\$ Acceleration due to gravity, irrespective of the direction of motion of an object, always acts downward. The gravitational force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton’s second law of motion as:$\\$ $F = m * a$$\\$ Where,$\\$ $F = \text{Net force}$$\\$ $m =\text{ Mass of the pebble} = 0.0 5 kg\\ a = g = 10 m / s^ 2\\ \therefore F = 0.05 * 10 = 0.5 N$$\\$ The net force on the pebble in all three cases is $0.5 N$ and this force acts in the downward direction.$\\$ If the pebble is thrown at an angle of $45^o$ with the horizontal, it will have both the horizontal and vertical components of velocity. At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble.

**4** **Give the magnitude and direction of the net force acting on a stone of mass $0.1 kg,$$\\$ (a) just after it is dropped from the window of a stationary train,$\\$ (b) just after it is dropped from the window of a train running at a constant velocity of $36 km/h,$$\\$ (c) just after it is dropped from the window of a train accelerating with $1 m s^{-2} ,$$\\$ (d) lying on the floor of a train which is accelerating with $1 m s^{ -2 }$, the stone being at rest relative to the train. Neglect air resistance throughout.**

(a)$1 N$; vertically downward$\\$ Mass of the stone, $m = 0.1 kg$$\\$ Acceleration of the stone, $a = g = 10 m / s^ 2$$\\$ As per Newton’s second law of motion, the net force acting on the stone,$\\$ $F = ma =mg\\ =0.1* 10 = 1 N$ Acceleration due to gravity always acts in the downward direction.$\\$ (b)$1 N$; vertically downward$\\$ The train is moving with a constant velocity. Hence, its acceleration is zero in the direction of its motion, i.e., in the horizontal direction. Hence, no force is acting on the stone in the horizontal direction. The net force acting on the stone is because of acceleration due to gravity and it always acts vertically downward. The magnitude of this force is $1 N.$$\\$ (c)$1 N$; vertically downward$\\$ It is given that the train is accelerating at the rate of $1 m / s^ 2 .$$\\$ Therefore, the net force acting on the stone, $F ' = ma = 0.1 * 1 = 0.1 N$$\\$ This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force $F '$, stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations. Therefore, the net force acting on the stone is given only by acceleration due to gravity.$\\$ $F = mg = 1 N$$\\$ This force acts vertically downward.$\\$ (d)$0.1 N$; in the direction of motion of the train The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train. Acceleration of the train, $a = 0.1 m / s ^2$$\\$ The net force acting on the stone will be in the direction of motion of the train. Its magnitude is given by: $F = ma =0.1 * 1 = 0.1 N$

**5** **One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is:$\\$ (i)$T,$$\\$ (ii)$T-\dfrac{mv^2}{l},$$\\$ (iii)$T+\dfrac{mv^2}{l}$$\\$ (iv) $0$$\\$ $T $ is the tension in the string. [Choose the correct alternative].**

Answer: (i)$\\$ When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension $T$ , i.e., $F=T=\dfrac{mv^2}{l}$$\\$ Where $F$ is the net force acting on the particle.

**6** **A constant retarding force of $50 N$ is applied to a body of mass $20 kg$ moving initially with a speed of $15 ms^{-1}$ . How long does the body take to stop?**

Retarding force, $F=-50 N$$\\$ Mass of the body, $m = 20 kg$$\\$ Initial velocity of the body, $u = 15 m/s$$\\$ Final velocity of the body, $v = 0$$\\$ Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as:$\\$ $F=ma\\ -50=20*a\\ \therefore a-\dfrac{-50}{20}=-2.5m/s^2$$\\$ Using the first equation of motion, the time $(t)$ taken by the body to come to rest can be calculated as:$\\$ $v=u+at\\ \therefore =\dfrac{-u}{a}\\ =\dfrac{-15}{-2.5}=6 s$

**7** **A constant force acting on a body of mass $3.0 kg$ changes its speed from $2.0 m s -1$ to $3.5 m s -1 $ in $25 s$. The direction of the motion of the body remains unchanged. W at is the magnitude and direction of the force?**

$0.18 N;$ in the direction of motion of the body$\\$ Mass of the body, $m = 3 kg$$\\$ Initial speed of the body,$ u = 2 m/s$$\\$ Final speed of the body, $v = 3.5 m/s$$\\$ Time, $t = 25 s$$\\$ Using the first equation of motion, the acceleration (a) produced in the body can be calculated as: $v = u + at$$\\$ $a=\dfrac{v-u}{t}\\ =\dfrac{3.5-2}{25}=\dfrac{1.5}{25}=0.6m/s^2$$\\$ As per Newton’s second law of motion, force is given as: $F = ma$$\\$ $= 3 × 0.06 = 0.18 N$$\\$ Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.

**8** **A body of mass $5 kg$ is acted upon by two perpendicular forces $8 N$ and $6 N$. Give the magnitude and direction of the acceleration of the body.**

$2 m / s^2$ , at an angle of $37^o$ with a force of $8 N$ Mass of the body, $m = 5 kg$$\\$ The given situation can be represented as follows:

The resultant of two forces is given as:$\\$ $R=\sqrt{(8)^2+(-6)^2}=\sqrt{64+36}=10 N$$\\$ $\theta $ is the angle made by R with the force of $8 N$$\\$ $\therefore \theta =\tan^{-1}(\dfrac{-6}{8})=-36.87^o$$\\$ The negative sign indicates that $\theta $ is in the clockwise direction with respect to the force of magnitude $8 N.$$\\$ As per Newton’s second law of motion, the acceleration (a) of the body is given as:$\\$ $F = ma$$\\$ $\therefore a=\dfrac{F}{m}=\dfrac{10}{5}=2 m/s^2$$\\$

**9** **The driver of a three-wheeler moving with a speed of $36 km/h$ sees a child standing in the middle of the road and bring s his vehicle to rest in $4.0 s$ just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is $400 kg$ and the mass of the driver is $65 kg.$**

Initial speed of the three-wheeler, $u = 36 km/h$$\\$ Final speed of the three-wheeler, $v = 10 m/s$$\\$ Time,$ t = 4 s$$\\$ Mass of the three-wheeler, $m = 400 kg$$\\$ Mass of the driver, $m' = 65 kg$$\\$ Total mass of the system, $M = 400 + 65 = 465 kg$$\\$ Using the first law of motion, the acceleration (a) of the three-wheeler can be calculated as:$\\$ $v = u + at$$\\$ $\therefore a=\dfrac{v-u}{t}=\dfrac{0-10}{4}=-2.5 m/s^2$ The negative sign indicates that the velocity of the three-wheeler is decreasing with time. Using Newton’s second law of motion, the net force acting on the three-wheeler can be calculated as:$\\$ $F = Ma$$\\$ $= 465 × (-2.5) = -1162.5 N$$\\$ The negative sign indicates that the force is acting against the direction of motion of the three-wheeler.

**10** **A rocket with a lift-off mass $20,000 kg$ is blasted upwards with an initial acceleration of $5.0 ms -2$ . Calculate the initial thrust (force) of the blast.**

Mass of the rocket, $m = 20,000 kg$$\\$ Initial acceleration, $a = 5 m / s ^2$$\\$ Acceleration due to gravity, $g = 10 m / s^ 2$$\\$ Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation:$\\$ $F - mg = ma\\ F = m ( g + a )\\ = 20000 *( 10 + 5 )\\ = 20000 * 15 = 3 * 10^ 5 N$