LAWS OF MOTION-Exercise

Q1. 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.

Solution

Q2. A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and 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° with the horizontal direction?

Solution

A pebble of mass 0.05 kg experiences only the gravitational force mg downward during free flight, regardless of motion direction. The net force magnitude is always 0.49 N vertically downward by Newton's second law.

Upward Motion

During upward motion, velocity points upward but acceleration remains \( \vec{g} = 9.8 \, \hat{j}_{\text{down}} \). Net force equals weight: \( F_{\text{net}} = mg = 0.05 \times 9.8 = 0.49 \, \text{N} \) directed vertically downward. The pebble decelerates as it rises.

Downward Motion

When falling downward, both velocity and acceleration point downward. The net force stays \( 0.49 \, \text{N} \) vertically downward, causing acceleration \( g \). Gravity alone governs the motion in both ascent and descent phases.

Highest Point

At the highest point, velocity becomes instantaneously zero but acceleration due to gravity persists at \( 9.8 \, \text{m s}^{-2} \) downward. Thus net force magnitude remains \( 0.49 \, \text{N} \) vertically downward, consistent with non-zero acceleration despite zero speed.

Angled Projection

When thrown at 45° to horizontal, the vertical motion component mirrors the vertical case exactly. Horizontal motion has zero net force (no air resistance), but vertical net force stays \( mg = 0.49 \, \text{N} \) downward throughout, including at peak where vertical velocity vanishes.

Q3. 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.

Solution

The stone of mass 0.05 kg experiences gravitational force mg downward in free fall cases, while on the train floor it experiences normal reaction balancing effective weight. Net force depends on the reference frame and motion state per Newton's laws from NCERT Class XI Chapter 4. [memory:6][memory:7]

Stationary Train Drop

Just after dropping from a stationary train, the stone is in free fall with acceleration \( g = 9.8 \, \text{m s}^{-2} \) downward. Net force magnitude is \( F_{net} = mg = 0.1 \times 9.8 = 0.98 \, \text{N} \) vertically downward.

Constant Velocity Train

A train at constant 36 km/h (10 m/s) is an inertial frame identical to stationary case. Stone acquires horizontal velocity instantly but vertical motion remains free fall. Net force stays \( 0.98 \, \text{N} \) vertically downward, independent of horizontal constant velocity.

Accelerating Train Drop

From accelerating train (\( a = 1 \, \text{m s}^{-2} \)), stone has initial horizontal velocity matching train's instantaneous speed but no horizontal acceleration after release. Only vertical gravitational force acts, so net force remains \( 0.98 \, \text{N} \) vertically downward.

Stone on Accelerating Floor

Lying at rest relative to accelerating train (\( a = 1 \, \text{m s}^{-2} \)), stone accelerates horizontally with train. Floor provides normal force \( N = mg = 0.98 \, \text{N} \) upward and friction \( f = ma = 0.1 \times 1 = 0.1 \, \text{N} \) horizontally. Net force magnitude \( F_{net} = \sqrt{(mg)^2 + (ma)^2} = \sqrt{0.98^2 + 0.1^2} \approx 0.985 \, \text{N} \) at angle \( \tan^{-1}(a/g) \approx 5.8^\circ \) from vertical toward acceleration direction.

Q4. 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 :

1. \(T\)
2. \(T-\dfrac{mv^2}{l}\)
3. \(T+\dfrac{mv^2}{l}\)
4. 0

Solution

The particle moves in uniform circular motion on a smooth table, constrained by string tension T providing centripetal force.

Net Force Analysis

On a smooth horizontal table, the only horizontal force is tension T directed toward the center. This tension supplies the required centripetal force \( \frac{mv^2}{l} \). Thus net force toward center equals T.

Correct Option

The net force on the particle (directed towards the centre) is (i) T. Tension balances exactly with centripetal requirement \( T = \frac{mv^2}{l} \), so net force magnitude is T.

Option Elimination

(ii) and (iii) imply additional forces, but smooth table has no friction. (iv) zero net force would mean straight-line motion, contradicting circular path. Only tension acts centrally.

Q5. A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s-1. How long does the body take to stop ?

Solution

A constant retarding force of 50 N acts opposite to the direction of motion of a body of mass 20 kg moving initially with a speed of 15 m s\(^{-1}\). The force produces a uniform deceleration that eventually brings the body to rest.

Given:
Retarding force, \( F = 50 \, \text{N} \)
Mass of the body, \( m = 20 \, \text{kg} \)
Initial velocity, \( u = 15 \, \text{m s}^{-1} \)
Final velocity for the body to stop, \( v = 0 \, \text{m s}^{-1} \).

Using Newton's second law of motion, the magnitude of the acceleration is obtained from \[\begin{aligned} F &= ma \\\\ \Rightarrow a &= \dfrac{F}{m}\\\\ &= \dfrac{50}{20}\\\\ &= \dfrac{5}{2} \, \text{m s}^{-2} \end{aligned}\] Since the force is retarding, this acceleration actually acts opposite to the direction of motion, i.e. it is a deceleration.

The time taken by the body to come to rest can be found using the first equation of motion \[ v = u - at. \] Substituting the known values, \[\begin{aligned} 0 &= 15 - \dfrac{5}{2} t\\\\ \Rightarrow \dfrac{5}{2} t &= 15\\\\ \Rightarrow t &= \dfrac{15 \times 2}{5}\\\\ &= 6 \, \text{s}. \end{aligned}\]

Therefore, the body takes \( 6 \, \text{s} \) to come to a complete stop under the action of the constant retarding force.

Q6. A constant force acting on a body of mass 3.0 kg changes its speed from \(\mathrm{2.0\ m\ s^{-1}}\) to \(\mathrm{3.5\ m\ s^{-1}}\) in \(\mathrm{25\ s}\). The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force ?

Solution

A constant force acts on a body of mass \( 3.0 \, \text{kg} \) and changes its speed from \( 2.0 \, \text{m s}^{-1} \) to \( 3.5 \, \text{m s}^{-1} \) in \( 25 \, \text{s} \), without changing the direction of motion. The force therefore produces a uniform acceleration along the direction of motion.

Given:
Mass of the body, \( m = 3.0 \, \text{kg} \)
Initial velocity, \( u = 2.0 \, \text{m s}^{-1} \)
Final velocity, \( v = 3.5 \, \text{m s}^{-1} \)
Time interval, \( t = 25 \, \text{s} \).

Using the first equation of motion to find the acceleration, \[ \begin{aligned} v &= u + a t \\ 3.5 &= 2.0 + a \times 25 \\ 3.5 - 2.0 &= 25 a \\ 1.5 &= 25 a \\ a &= \dfrac{1.5}{25} \\ &= \dfrac{3}{50} \, \text{m s}^{-2}. \end{aligned} \] Thus the body has a small uniform acceleration of \( \dfrac{3}{50} \, \text{m s}^{-2} \) in the direction of its motion.

By Newton's second law, the force is \[ \begin{aligned} F &= m a \\ &= 3 \times \dfrac{3}{50} \\ &= \dfrac{9}{50} \, \text{N} \\ &= 0.18 \, \text{N}. \end{aligned} \]

Therefore, the magnitude of the force is \( 0.18 \, \text{N} \), and since the speed is increasing while the direction of motion remains unchanged, the force acts along the direction of motion of the body.

Q7. 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.

Solution

A body of mass \( 5 \, \text{kg} \) is acted upon by two mutually perpendicular forces of magnitudes \( 8 \, \text{N} \) and \( 6 \, \text{N} \). To find the acceleration, the resultant of these two forces is first obtained using vector addition. [web:85][web:88]

Let the two forces be \( F_{1} = 8 \, \text{N} \) and \( F_{2} = 6 \, \text{N} \), and the mass be \( m = 5 \, \text{kg} \). Since the forces act at right angles to each other, the magnitude of the resultant force \( F_{R} \) is \[ \begin{aligned} F_{R} &= \sqrt{F_{1}^{2} + F_{2}^{2}} \\ &= \sqrt{8^{2} + 6^{2}} \\ &= \sqrt{64 + 36} \\ &= \sqrt{100} \\ &= 10 \, \text{N} \end{aligned} \]

Using Newton's second law, the magnitude of the acceleration is \[ \begin{aligned} F_{R} &= m a \\ a &= \dfrac{F_{R}}{m} \\ &= \dfrac{10}{5} \\ &= 2 \, \text{m s}^{-2} \end{aligned} \]

The direction of the acceleration is the same as that of the resultant force. If the \( 8 \, \text{N} \) force is taken along the x-axis and the \( 6 \, \text{N} \) force along the y-axis, then the angle \( \theta \) which the resultant makes with the \( 8 \, \text{N} \) force is \[ \begin{aligned} \theta &= \tan^{-1} \left( \dfrac{F_{2}}{F_{1}} \right) \\ &= \tan^{-1} \left( \dfrac{6}{8} \right) \\ &= \tan^{-1} \left( \dfrac{3}{4} \right) \\ &\approx 37^{\circ} \end{aligned} \]

Hence, the body has an acceleration of magnitude \( 2 \, \text{m s}^{-2} \), directed at an angle of about \( 37^{\circ} \) to the \( 8 \, \text{N} \) force, towards the \( 6 \, \text{N} \) force.

Q8. 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 brings 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.

Solution

The driver of a three-wheeler is moving with an initial speed of \( 36 \, \text{km h}^{-1} \). This speed is converted into SI units as follows: \[ \begin{aligned} u &= 36 \times \dfrac{5}{18} \\ &= 10 \, \text{m s}^{-1}. \end{aligned} \] The vehicle is brought to rest in \( 4.0 \, \text{s} \), so the motion is uniformly retarded.

The time taken to stop the vehicle is \( t = 4 \, \text{s} \). The mass of the three-wheeler is \( 400 \, \text{kg} \) and the mass of the driver is \( 65 \, \text{kg} \), so the total mass of the system is \[ \begin{aligned} m &= 400 + 65 \\ &= 465 \, \text{kg}. \end{aligned} \] The final velocity after braking is \( v = 0 \, \text{m s}^{-1} \).

Using the first equation of motion to find the acceleration: \[ \begin{aligned} v &= u + a t \\ 0 &= 10 + a \times 4 \\ 0 &= 10 + 4a \\ 4a &= -10 \\ a &= -\dfrac{10}{4} \\ &= -2.5 \, \text{m s}^{-2} \end{aligned} \] The negative sign shows that the acceleration (retardation) is opposite to the direction of motion. [web:110]

The average retarding force on the vehicle is obtained from Newton's second law: \[ \begin{aligned} F &= m a \\ &= 465 \times (-2.5) \\ &= -1162.5 \, \text{N} \end{aligned} \]

Thus, the magnitude of the average retarding force is \( 1.16 \times 10^{3} \, \text{N} \) (approximately), and it acts opposite to the direction of motion of the three-wheeler.

Q9. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of \(\mathrm{5.0\ m\ s^{-2}}\). Calculate the initial thrust (force) of the blast.

Solution

The rocket has a lift-off mass of \( 20{,}000 \, \text{kg} \) and is blasted vertically upwards with an initial acceleration of \( 5.0 \, \text{m s}^{-2} \). The thrust must both overcome the weight of the rocket and provide this upward acceleration.

Let the initial thrust be \( F \). Taking upward as positive, the net force on the rocket is \[ \begin{aligned} F - mg &= ma \\ F &= ma + mg \\ &= m(a + g) \end{aligned} \]

Substituting the given values \( m = 20{,}000 \, \text{kg} \), \( a = 5 \, \text{m s}^{-2} \) and \( g \approx 10 \, \text{m s}^{-2} \), \[ \begin{aligned} F &= 20{,}000 \left( 5 + 10 \right) \\ &= 20{,}000 \times 15 \\ &= 3.0 \times 10^{5} \, \text{N} \end{aligned} \]

Therefore, the initial thrust of the blast is \( 3.0 \times 10^{5} \, \text{N} \) directed vertically upwards.

Q10. A body of mass 0.40 kg moving initially with a constant speed of \(\mathrm{10\ m\ s^{-1}}\) to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.

Solution

A body of mass \( 0.40 \, \text{kg} \) moves initially with a constant speed of \( 10 \, \text{m s}^{-1} \) towards the north. At time \( t = 0 \), a constant force of \( 8.0 \, \text{N} \) directed towards the south begins to act on it and continues for \( 30 \, \text{s} \). The position at \( t = 0 \) is taken as \( x = 0 \).

The acceleration produced by the constant force is \[ \begin{aligned} F &= 8.0 \, \text{N} \\ m &= 0.40 \, \text{kg} \\ a &= \dfrac{F}{m} \\ &= \dfrac{8.0}{0.40} \\ &= \dfrac{80}{4} \\ &= 20 \, \text{m s}^{-2} \end{aligned} \] Since the force is towards the south and the initial velocity is towards the north, taking north as positive, the acceleration is actually \( a = -20 \, \text{m s}^{-2} \).

At \( t = -5 \, \text{s} \), the force has not yet started acting, so the body moves with uniform velocity \( 10 \, \text{m s}^{-1} \) to the north. With \( x(0) = 0 \) and constant velocity, the position is \[ \begin{aligned} x(-5) &= u t \\ &= 10 \times (-5) \\ &= -50 \, \text{m}. \end{aligned} \] Thus, at \( t = -5 \, \text{s} \), the body was \( 50 \, \text{m} \) to the south of the origin.

For \( 0 \le t \le 30 \, \text{s} \), the motion is uniformly accelerated (actually retarded) with \( u = 10 \, \text{m s}^{-1} \) and \( a = -20 \, \text{m s}^{-2} \). The position at \( t = 25 \, \text{s} \) is \[ \begin{aligned} x(25) &= u t + \dfrac{1}{2} a t^{2} \\ &= 10 \times 25 + \dfrac{1}{2} \times (-20) \times 25^{2} \\ &= 250 - 10 \times 625 \\ &= 250 - 6250 \\ &= -6000 \, \text{m} \\ &= -6 \, \text{km} \end{aligned} \] So at \( t = 25 \, \text{s} \), the body is \( 6 \, \text{km} \) to the south of the origin.

For \( t = 100 \, \text{s} \), the motion must be split into two intervals. For the first \( 30 \, \text{s} \), the retarding force acts; after \( 30 \, \text{s} \), the problem states that the body continues with constant speed for the remaining \( 70 \, \text{s} \).

Displacement in the first \( 30 \, \text{s} \) is \[ \begin{aligned} S_{1} &= u t + \dfrac{1}{2} a t^{2} \\ &= 10 \times 30 + \dfrac{1}{2} \times (-20) \times 30^{2} \\ &= 300 - 10 \times 900 \\ &= 300 - 9000 \\ &= -8700 \, \text{m} \end{aligned} \]

The velocity at \( t = 30 \, \text{s} \) is \[ \begin{aligned} v &= u + a t \\ &= 10 + (-20) \times 30 \\ &= 10 - 600 \\ &= -590 \, \text{m s}^{-1} \end{aligned} \] The negative sign shows that by \( t = 30 \, \text{s} \) the body is moving towards the south with speed \( 590 \, \text{m s}^{-1} \).

For the remaining \( 70 \, \text{s} \) (from \( t = 30 \, \text{s} \) to \( t = 100 \, \text{s} \)), the body moves with this constant velocity. The displacement in this interval is \[ \begin{aligned} S_{2} &= v t \\ &= (-590) \times 70 \\ &= -41{,}300 \, \text{m} \end{aligned} \]

The total displacement from \( t = 0 \) to \( t = 100 \, \text{s} \) is \[ \begin{aligned} S &= S_{1} + S_{2} \\ &= -8700 - 41{,}300 \\ &= -50{,}000 \, \text{m} \\ &= -50 \, \text{km} \end{aligned} \]

Therefore, the predicted positions are: \( x(-5 \, \text{s}) = -50 \, \text{m} \), \( x(25 \, \text{s}) = -6000 \, \text{m} \), and \( x(100 \, \text{s}) = -50{,}000 \, \text{m} \), all measured along the north–south line with negative sign indicating positions to the south of the origin.

Q11. A truck starts from rest and accelerates uniformly at \(\mathrm{2.0\ m\ s^{-2}}\). At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the
(a) velocity, and
(b) acceleration of the stone at t = 11 s ? (Neglect air resistance.)

Solution

The truck starts from rest and moves with a uniform acceleration of \( 2.0 \, \text{m s}^{-2} \). The stone is dropped at \( t = 10 \, \text{s} \) from a height of \( 6 \, \text{m} \) above the ground. At the instant of dropping, the stone has the same horizontal velocity as the truck and zero initial vertical velocity.

The horizontal velocity of the truck (and hence of the stone at the instant it is dropped) at \( t = 10 \, \text{s} \) is \[ \begin{aligned} u_{x} &= 0, \\ a_{x} &= 2.0 \, \text{m s}^{-2}, \\ t &= 10 \, \text{s} \\\\ v_{x} &= u_{x} + a_{x} t \\ &= 0 + 2.0 \times 10 \\ &= 20 \, \text{m s}^{-1} \end{aligned} \]

After \( t = 10 \, \text{s} \), the stone is in free fall. For the vertical motion, taking upward as positive, the initial vertical velocity at the instant of dropping is \[ \begin{aligned} u_{y} &= 0, \\ a_{y} &= -g \approx -10 \, \text{m s}^{-2}, \\ \Delta t &= 11 - 10 = 1 \, \text{s} \\\\ v_{y} &= u_{y} + a_{y} \Delta t \\ &= 0 + (-10) \times 1 \\ &= -10 \, \text{m s}^{-1} \end{aligned} \] Thus, at \( t = 11 \, \text{s} \), the vertical component of velocity is \( 10 \, \text{m s}^{-1} \) downward.

The resultant speed of the stone at \( t = 11 \, \text{s} \) is obtained by combining the horizontal and vertical components: \[ \begin{aligned} v &= \sqrt{v_{x}^{2} + v_{y}^{2}} \\ &= \sqrt{20^{2} + 10^{2}} \\ &= \sqrt{400 + 100} \\ &= \sqrt{500} \\ &= 10 \sqrt{5} \\ &\approx 22.4 \, \text{m s}^{-1} \end{aligned} \] The velocity is directed forward (along the truck’s direction of motion) and downward.

Once the stone is released, the only force acting on it (neglecting air resistance) is gravity. Therefore, its acceleration at \( t = 11 \, \text{s} \) is \[ \begin{aligned} \vec{a} &= -g \, \hat{j} \\ &\approx -10 \, \text{m s}^{-2} \quad \text{(vertically downward)}. \end{aligned} \]

Q12. A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is \(\mathrm{1\ m\ s^{-1}}\). What is the trajectory of the bob if the string is cut when the bob is
(a) at one of its extreme positions,
(b) at its mean position.

Solution

When the string is cut, the bob simply moves under gravity, so its later path depends only on its velocity at that instant.

Extreme position

At an extreme position of oscillation the bob is momentarily at rest, so its instantaneous velocity is zero.

​ Once the string is cut there, the only force on the bob is its weight, and it therefore falls straight vertically downward along a line.

Mean position

At the mean position the bob has maximum speed\(\mathrm{ 1m\ s^{−1}}\), directed horizontally along the tangent to the arc.

​ If the string is cut at this instant, the bob is a horizontally projected body under gravity and follows a parabolic trajectory (projectile motion).

Q13. A man of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of \(\mathrm{10\ m\ s^{-1}}\),
(b) downwards with a uniform acceleration of \(\mathrm{5\ m\ s^{-2}}\),
(c) upwards with a uniform acceleration of \(\mathrm{5\ m\ s^{-2}}\). What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ?

Solution

Let the mass of the man be \( m = 70 \, \text{kg} \) and take \( g \approx 10 \, \text{m s}^{-2} \). The weighing scale reads the normal reaction \( N \) exerted by the floor on the man, which is interpreted as the apparent weight.

Case (a): Lift moving upwards with uniform speed \( 10 \, \text{m s}^{-1} \). Since the speed is uniform, the acceleration is \[ \begin{aligned} a &= 0, \\ N &= mg \\ &= 70 \times 10 \\ &= 700 \, \text{N} \end{aligned} \] Thus the scale reading is equivalent to \( 70 \, \text{kg} \).

Case (b): Lift moving downwards with uniform acceleration \( 5 \, \text{m s}^{-2} \). Taking upward as positive, the acceleration of the man is \( a = -5 \, \text{m s}^{-2} \). The equation of motion along the vertical is \[ \begin{aligned} N - mg &= m a \\ N - 70 \times 10 &= 70 \times (-5) \\ N - 700 &= -350 \\ N &= 700 - 350 \\ &= 350 \, \text{N} \end{aligned} \] The scale therefore reads an apparent weight corresponding to \( 35 \, \text{kg} \).

Case (c): Lift moving upwards with uniform acceleration \( 5 \, \text{m s}^{-2} \). Now \( a = +5 \, \text{m s}^{-2} \) (upward), and the equation of motion is \[ \begin{aligned} N - mg &= m a \\ N - 70 \times 10 &= 70 \times 5 \\ N - 700 &= 350 \\ N &= 700 + 350 \\ &= 1050 \, \text{N} \end{aligned} \] The scale reading corresponds to an apparent weight of \( 105 \, \text{kg} \).

Case (d): Lift in free fall when the mechanism fails. In this situation the lift, the man, and the scale all accelerate downward with acceleration \( g \). Using \( a = -g \) (downward), \[ \begin{aligned} N - mg &= m a \\ N - mg &= m(-g) \\ N - mg &= -mg \\ N &= 0 \end{aligned} \] Hence, the scale reading is zero; the man experiences weightlessness.

Q15. Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?

Solution

Two blocks of masses \( m_{1} = 10 \, \text{kg} \) and \( m_{2} = 20 \, \text{kg} \) are placed on a smooth horizontal surface and connected by a light string. A horizontal force \( F = 600 \, \text{N} \) acts along the string, first on block A (mass \( 10 \, \text{kg} \)) and then on block B (mass \( 20 \, \text{kg} \)). Since the surface is smooth and the string light, both blocks have the same acceleration.

The total mass of the system is \[ \begin{aligned} m_{1} + m_{2} &= 10 + 20 \\ &= 30 \, \text{kg}, \end{aligned} \] so the common acceleration is \[ \begin{aligned} a &= \dfrac{F}{m_{1} + m_{2}} \\ &= \dfrac{600}{30} \\ &= 20 \, \text{m s}^{-2} \end{aligned} \]

When A (the 10 kg block) is pulled by the force \( F \), the free-body diagram of block A shows the applied force \( F \) forward and the tension \( T \) in the string backward. Applying Newton's second law to block A, \[ \begin{aligned} F - T &= m_{1} a \\ 600 - T &= 10 \times 20 \\ 600 - T &= 200 \\ T &= 600 - 200 \\ &= 400 \, \text{N} \end{aligned} \] Thus, when A is pulled, the tension in the string is \( 400 \, \text{N} \).

When B (the 20 kg block) is pulled instead, the applied force \( F \) acts on block B, and the tension \( T \) now pulls block A. Considering block B alone, the forces along the direction of motion are \( F \) forward and tension \( T \) backward. Newton's second law for block B gives \[ \begin{aligned} F - T &= m_{2} a \\ 600 - T &= 20 \times 20 \\ 600 - T &= 400 \\ T &= 600 - 400 \\ &= 200 \, \text{N} \end{aligned} \]

Therefore, the tension in the string is \( 400 \, \text{N} \) when the 10 kg block (A) is pulled by the 600 N force, and \( 200 \, \text{N} \) when the 20 kg block (B) is pulled.

Q16. Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Solution

Two masses of \( 8 \, \text{kg} \) and \( 12 \, \text{kg} \) are connected by a light, inextensible string passing over a frictionless pulley. When the system is released, the heavier mass \( 12 \, \text{kg} \) moves down and the lighter mass \( 8 \, \text{kg} \) moves up with the same magnitude of acceleration \( a \), and the tension in the string is \( T \).

For the \( 8 \, \text{kg} \) mass (moving upward), taking upward as positive, \[ \begin{aligned} T - 8 \times 10 &= 8 a \\ T - 80 &= 8 a \\ T &= 8 a + 80 \end{aligned} \]

For the \( 12 \, \text{kg} \) mass (moving downward), taking downward as positive, \[ \begin{aligned} 12 \times 10 - T &= 12 a \\ 120 - T &= 12 a \\ T &= 120 - 12 a \end{aligned} \]

Equating the two expressions for \( T \), \[ \begin{aligned} 8 a + 80 &= 120 - 12 a \\ 8 a + 12 a &= 120 - 80 \\ 20 a &= 40 \\ a &= 2 \, \text{m s}^{-2} \end{aligned} \]

Substituting this value of \( a \) into \( T = 8 a + 80 \), \[ \begin{aligned} T &= 8 \times 2 + 80 \\ &= 16 + 80 \\ &= 96 \, \text{N} \end{aligned} \]

Thus, when the masses are released, the acceleration of each mass is \( 2 \, \text{m s}^{-2} \), and the tension in the string is \( 96 \, \text{N} \).

Q17. A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

Solution

Consider a nucleus of mass \( m \) at rest in the laboratory frame. Since it is at rest, its initial linear momentum is \[ \vec{p}_{\text{initial}} = m \vec{0} = \vec{0}. \]

Suppose the nucleus disintegrates into two smaller nuclei of masses \( m_{1} \) and \( m_{2} \), which move along the same straight line with velocities \( \vec{v}_{1} \) and \( \vec{v}_{2} \), respectively. By the law of conservation of linear momentum, the total momentum after disintegration must equal the initial momentum: \[ m_{1} \vec{v}_{1} + m_{2} \vec{v}_{2} = \vec{0} \]

Rearranging this relation gives \[\begin{aligned} m_{2} \vec{v}_{2} &= -\,m_{1} \vec{v}_{1} \\ \Rightarrow \vec{v}_{2} &= -\,\dfrac{m_{1}}{m_{2}} \, \vec{v}_{1} \end{aligned}\] The negative sign shows that the velocities \( \vec{v}_{1} \) and \( \vec{v}_{2} \) are directed opposite to each other along the same line.

Therefore, if a nucleus initially at rest disintegrates into two smaller nuclei, the two products must move in opposite directions so that their momenta cancel and the total linear momentum remains zero.

Q18. Two billiard balls each of mass 0.05 kg moving in opposite directions with speed \(\mathrm{6\ m\ s^{-1}}\) collide and rebound with the same speed. What is the impulse imparted to each ball due to the other ?

Solution

Each billiard ball has mass \( m = 0.05 \, \text{kg} \) and an initial speed of \( 6 \, \text{m s}^{-1} \), the two balls moving directly towards each other. The impulse imparted to a ball is equal to the change in its momentum.

For one ball, take its initial velocity as \( u = +6 \, \text{m s}^{-1} \). After the collision it rebounds with the same speed in the opposite direction, so its final velocity is \( v = -6 \, \text{m s}^{-1} \). The change in momentum of this ball is \[ \begin{aligned} \Delta p &= m v - m u \\ &= m (v - u) \\ &= m \bigl( -6 - 6 \bigr) \\ &= m (-12) \end{aligned} \]

Substituting \( m = 0.05 \, \text{kg} \), \[ \begin{aligned} \Delta p &= -12 \times 0.05 \\ &= -0.6 \, \text{kg m s}^{-1} \end{aligned} \] The impulse \( J \) on the ball (equal to its change in momentum) therefore has magnitude \[ J = |\Delta p| = 0.6 \, \text{kg m s}^{-1}, \] directed opposite to the ball’s initial motion.

By Newton’s third law, the other ball receives an impulse of the same magnitude \( 0.6 \, \text{kg m s}^{-1} \) but in the opposite direction, so the impulse imparted to each ball due to the other is \( 0.6 \, \text{kg m s}^{-1} \) in opposite directions.

Q19. A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is \(\mathrm{80\ m\ s^{-1}}\), what is the recoil speed of the gun ?

Solution

Let the mass of the shell be \( m_{s} = 0.020 \, \text{kg} \) and the mass of the gun be \( m_{g} = 100 \, \text{kg} \). The shell leaves the gun with muzzle speed \( v_{s} = 80 \, \text{m s}^{-1} \). Before firing, the gun–shell system is at rest, so its total initial momentum is zero.

Taking the forward direction of the shell as positive, the initial momentum is \[ \begin{aligned} p_{\text{initial}} &= 0 \end{aligned} \] After firing, let the recoil velocity of the gun be \( v_{g} \) (which will be negative, since it is backward), then by conservation of linear momentum: \[ \begin{aligned} m_{s} v_{s} + m_{g} v_{g} &= 0 \\ 0.020 \times 80 + 100 \, v_{g} &= 0 \end{aligned} \]

Solving for \( v_{g} \), \[ \begin{aligned} 100 \, v_{g} &= -0.020 \times 80 \\ v_{g} &= -\dfrac{0.020 \times 80}{100} \\ &= -0.016 \, \text{m s}^{-1} \end{aligned} \]

Thus, the gun recoils with speed \( 0.016 \, \text{m s}^{-1} \) in the direction opposite to the motion of the shell.

Q20. A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball ? (Mass of the ball is 0.15 kg.)

Solution

The ball has mass \( m = 0.15 \, \text{kg} \) and initial speed \( 54 \, \text{km h}^{-1} \). Converting this speed into SI units, \[ \begin{aligned} v &= 54 \times \dfrac{5}{18} \\ &= 15 \, \text{m s}^{-1} \end{aligned} \] The batsman deflects the ball through an angle of \( 45^{\circ} \) without changing its speed, so the initial and final velocity vectors have the same magnitude \( 15 \, \text{m s}^{-1} \) and are inclined at \( 45^{\circ} \) to each other.

Let the initial velocity be along the positive \( x \)-axis with magnitude \( v_{1} = 15 \, \text{m s}^{-1} \). After deflection, the velocity of the ball makes an angle \( 45^{\circ} \) with the initial direction but has the same magnitude \( v_{2} = 15 \, \text{m s}^{-1} \). The magnitude of the change in velocity is obtained using the vector relation for two equal vectors separated by angle \( \theta = 45^{\circ} \): \[ \begin{aligned} |\Delta \vec{v}|^{2} &= v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} \cos \theta \\ &= 15^{2} + 15^{2} - 2 \times 15 \times 15 \cos 45^{\circ} \\ &= 450 - 450 \cos 45^{\circ} \\ &= 450 \left( 1 - \dfrac{\sqrt{2}}{2} \right) \end{aligned} \]

Taking the square root, \[ \begin{aligned} |\Delta \vec{v}| &= \sqrt{450 \left( 1 - \dfrac{\sqrt{2}}{2} \right)} \\ &\approx 11.5 \, \text{m s}^{-1} \end{aligned} \] The impulse imparted to the ball is the change in its momentum: \[ \begin{aligned} J &= m \, |\Delta \vec{v}| \\ &= 0.15 \times 11.5 \\ &\approx 1.7 \, \text{kg m s}^{-1} \end{aligned} \]

Therefore, the magnitude of the impulse imparted to the ball is approximately \( 1.7 \, \text{kg m s}^{-1} \), directed along the change in its velocity from the initial to the final direction.

Q21. A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N ?

Solution

Q22. If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :
(a) the stone moves radially outwards,
(b) the stone flies off tangentially from the instant the string breaks,
(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?

Solution

The correct description is option (b): the stone flies off tangentially from the instant the string breaks.

At every point in uniform circular motion, the stone’s instantaneous velocity is directed along the tangent to the circular path, while the string tension provides the centripetal force towards the centre.

When the speed becomes too high and the string breaks, the centripetal force suddenly drops to zero but the stone still has its tangential velocity at that instant, so, by Newton’s first law, it continues in a straight line along that tangent.

Q23. Explain why
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawn mower than to push it,
(d) a cricketer moves his hands backwards while holding a catch.

Solution

(a) A horse pulls a cart by pushing the ground backward with its hooves; the ground then provides an equal and opposite reaction force that pushes the horse (and cart) forward.

In empty space there is no ground to provide this reaction, so no external forward force acts on the horse–cart system and it cannot accelerate or run.

(b) In a speeding bus, both bus and passengers share the same forward velocity.

When the bus stops suddenly, the lower parts of passengers’ bodies in contact with the bus are brought to rest, but their upper bodies, due to inertia of motion, continue moving forward, so they are thrown forward relative to the bus.

(c) When a lawn mower (or roller) is pushed, the applied force usually has a downward vertical component, which increases the normal reaction and hence the frictional force opposing motion.

When it is pulled, the vertical component of the pulling force acts upward, reducing the normal reaction and friction, so a smaller horizontal force is enough, making pulling easier than pushing.

(d) A fast-moving cricket ball has large momentum, and stopping it in a very short time would require a very large force on the hands.

By moving the hands backward while catching, the player increases the time over which the ball’s momentum is reduced to zero, decreasing the average force \((\text{impulse} =F\Delta t\)) and reducing the chance of injury.