Question 1

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 10g 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.

Accepted Answer

Solution:
(a) Since the drop of rain is falling with constant speed, acceleration is zero. Hence, net force = 0.
(b) Cork floating on water is at rest, so net force = 0.
(c) Kite held stationary means velocity is zero and not changing, so net force = 0.
(d) Car moving with constant velocity means acceleration is zero, so net force = 0.
(e) Electron moving at high speed in free space with no fields or objects experiences no force, so net force = 0. — According to Newton's first law, if velocity is constant (including zero), net force is zero. Hence, in all cases where motion is uniform or stationary, net force is zero.

Question 2

2 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?

Ignore air resistance.

Accepted Answer

Solution:
Mass, m = 0.05 kg
Acceleration due to gravity, g = 10 m/s²

(a) During upward motion, the only force acting is gravity downward.
Net force = mg downward = 0.05 × 10 = 0.5 N downward.

(b) During downward motion, gravity acts downward.
Net force = mg downward = 0.5 N downward.

(c) At the highest point, velocity is zero but acceleration due to gravity still acts downward.
Net force = mg downward = 0.5 N downward.

If thrown at 45°, the vertical component of motion is similar, so net force remains mg downward in all cases. — Gravity acts downward throughout the motion. Air resistance is ignored, so no other forces act. Hence net force magnitude is constant mg downward.

Question 3

3 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/s²,
(d) lying on the floor of a train which is accelerating with 1 m/s², the stone being at rest relative to the train.

Neglect air resistance throughout.

Accepted Answer

Solution:
Mass, m = 0.1 kg
Acceleration due to gravity, g = 10 m/s²

(a) Dropped from stationary train:
Net force = mg downward = 0.1 × 10 = 1 N downward.

(b) Dropped from train moving at constant velocity:
Velocity constant means no acceleration, so net force = mg downward = 1 N downward.

(c) Dropped from accelerating train:
Just after dropping, stone has horizontal velocity equal to train's velocity but no horizontal acceleration.
Net force = mg downward = 1 N downward.

(d) Stone lying on accelerating train floor at rest relative to train:
Horizontal acceleration a = 1 m/s²
Net force = horizontal force = m × a = 0.1 × 1 = 0.1 N in direction of train's acceleration.
Vertically, normal force balances mg, so net vertical force = 0.
Hence net force is 0.1 N horizontally in direction of acceleration. — Gravity acts vertically downward always. Horizontal acceleration affects only when stone is on accelerating train floor and at rest relative to train.

Question 4

4 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 - \frac{mv^2}{l},
(iii) T + \frac{mv^2}{l},
(iv) 0

T is the tension in the string. [Choose the correct alternative].

Accepted Answer

Answer: (i) T

Explanation:
The net force acting on the particle moving in a circle is the centripetal force directed towards the centre.
This force is provided by the tension T in the string.
Hence, net force = T.

Options (ii) and (iii) incorrectly add or subtract the centripetal force term from T.
Option (iv) zero force is incorrect as centripetal force is required for circular motion. — For uniform circular motion, net force = centripetal force = Tension T directed towards centre.

Question 5

5 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. How long does the body take to stop?

Accepted Answer

Given:
Mass, m = 20 kg
Initial velocity, u = 15 m/s
Retarding force, F = 50 N
Final velocity, v = 0 (body stops)

Using Newton's second law:
F = m a
=> a = F / m = 50 / 20 = 2.5 m/s² (retarding, so acceleration is -2.5 m/s²)

Using equation of motion:
v = u + a t
0 = 15 - 2.5 t
=> t = 15 / 2.5 = 6 s

Answer: The body takes 6 seconds to stop. — Retarding force causes negative acceleration. Using kinematic equation with final velocity zero gives stopping time.

Question 6

6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m/s to 3.5 m/s in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Accepted Answer

Given:
Mass, m = 3.0 kg
Initial velocity, u = 2.0 m/s
Final velocity, v = 3.5 m/s
Time, t = 25 s

Acceleration, a = (v - u) / t = (3.5 - 2.0) / 25 = 1.5 / 25 = 0.06 m/s²

Force, F = m a = 3.0 × 0.06 = 0.18 N

Direction: Since velocity increases, force is in the direction of motion. — Force calculated using Newton's second law with acceleration from velocity change over time.

Question 7

7 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.

Accepted Answer

Given:
Mass, m = 5 kg
Forces: F1 = 8 N, F2 = 6 N perpendicular

Resultant force, F = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 N

Acceleration, a = F / m = 10 / 5 = 2 m/s²

Direction: Angle 	heta with 8 N force,
	an 	heta = 6 / 8 = 0.75
	heta = 	an^{-1}(0.75) ≈ 37° from 8 N force. — Use Pythagoras theorem for resultant force and Newton's second law for acceleration magnitude and direction.

Question 8

8 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.

Accepted Answer

Given:
Initial speed, u = 36 km/h = 36 × (1000/3600) = 10 m/s
Final speed, v = 0
Time, t = 4 s
Mass of vehicle, m1 = 400 kg
Mass of driver, m2 = 65 kg
Total mass, m = 400 + 65 = 465 kg

Acceleration, a = (v - u) / t = (0 - 10) / 4 = -2.5 m/s²

Retarding force, F = m × a = 465 × (-2.5) = -1162.5 N

Magnitude of force = 1162.5 N opposite to direction of motion. — Calculate acceleration from velocity change and multiply by total mass to get retarding force.

Question 9

9 A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m/s². Calculate the initial thrust (force) of the blast.

Accepted Answer

Given:
Mass, m = 20,000 kg
Acceleration, a = 5.0 m/s²
Acceleration due to gravity, g = 10 m/s²

Total force required = Force to overcome gravity + force to accelerate upwards
= mg + ma = m(g + a)
= 20,000 × (10 + 5) = 20,000 × 15 = 300,000 N

Answer: Initial thrust = 300,000 N. — Thrust must balance weight and provide upward acceleration, so total force = m(g + a).

Question 10

10 A body of mass 0.40 kg moving initially with a constant speed of 10 m/s 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.

Accepted Answer

Given:
Mass, m = 0.40 kg
Initial velocity, u = 10 m/s north
Force, F = 8.0 N south
Time intervals: t = -5 s, 25 s, 100 s

Acceleration, a = F / m = 8.0 / 0.40 = 20 m/s² south (opposite to initial velocity)

At t = 0, position x = 0

For t < 0 (before force applied), motion uniform:
Position at t = -5 s:
x = u × t = 10 × (-5) = -50 m (50 m south of origin)

For t > 0, motion with acceleration a south:
Position at time t:
x = ut + (1/2) a t²
At t = 25 s:
x = 10 × 25 + 0.5 × (-20) × 25² = 250 - 0.5 × 20 × 625 = 250 - 6250 = -6000 m (6000 m south)

At t = 100 s:
x = 10 × 100 + 0.5 × (-20) × 10000 = 1000 - 100000 = -99000 m (99000 m south)

Answer:
Position at -5 s: 50 m south
Position at 25 s: 6000 m south
Position at 100 s: 99000 m south. — Before force application, motion uniform. After force application, acceleration opposite to initial velocity causes deceleration and then motion southward.

Question 11

11 A truck starts from rest and accelerates uniformly at 2.0 m/s². 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.)

Accepted Answer

Given:
Acceleration of truck, a = 2.0 m/s²
At t = 10 s, stone dropped from height h = 6 m
Find velocity and acceleration of stone at t = 11 s

(a) Velocity of truck at t=10 s:
v = u + at = 0 + 2 × 10 = 20 m/s (horizontal)

When stone is dropped, initial velocity of stone = velocity of truck = 20 m/s horizontally

After 1 s (t=11 s), vertical velocity due to gravity:
v_y = g × t = 10 × 1 = 10 m/s downward

Horizontal velocity remains 20 m/s (no horizontal acceleration)

Total velocity magnitude:
v = \sqrt{(20)^2 + (10)^2} = \sqrt{400 + 100} = \sqrt{500} ≈ 22.36 m/s

Direction: angle 	heta = 	an^{-1}(10/20) = 	an^{-1}(0.5) ≈ 26.6° below horizontal

(b) Acceleration of stone:
Only gravity acts vertically downward, so acceleration = 10 m/s² downward

Horizontal acceleration = 0 (no force)

Answer:
(a) Velocity ≈ 22.36 m/s at 26.6° below horizontal
(b) Acceleration = 10 m/s² downward. — Stone retains horizontal velocity of truck at drop, gains vertical velocity due to gravity. Acceleration is only due to gravity vertically.

Question 12

12 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 1 m/s. 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.

Accepted Answer

Given:
Mass m = 0.1 kg (not needed for trajectory)
Length of string l = 2 m
Speed at mean position v = 1 m/s

(a) If string is cut at extreme position:
At extreme position, bob momentarily at rest (speed zero).
When string is cut, bob has zero velocity and starts free fall vertically downward under gravity.
Trajectory: straight vertical downward fall.

(b) If string is cut at mean position:
At mean position, bob has maximum speed (1 m/s) tangent to circular path.
When string is cut, bob moves in projectile motion with initial velocity tangent to circle.
Trajectory: parabolic path under gravity in direction of velocity at mean position. — Cutting string removes centripetal force. At rest position, bob falls vertically. At mean position, bob moves tangentially with projectile motion.

Question 13

13 A man of mass 70 kg stands on a weighing scale in a lift which is moving

(a) upwards with a uniform speed of 10 m/s,
(b) downwards with a uniform acceleration of 5 m/s²,
(c) upwards with a uniform acceleration of 5 m/s².

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?

Accepted Answer

Given:
Mass m = 70 kg
Acceleration due to gravity g = 10 m/s²

Weight W = mg = 70 × 10 = 700 N

(a) Lift moving upwards with uniform speed:
Acceleration a = 0
Apparent weight = mg = 700 N

(b) Lift moving downwards with acceleration 5 m/s²:
Apparent weight = m(g - a) = 70 × (10 - 5) = 70 × 5 = 350 N

(c) Lift moving upwards with acceleration 5 m/s²:
Apparent weight = m(g + a) = 70 × (10 + 5) = 70 × 15 = 1050 N

(d) Lift falling freely under gravity:
Acceleration a = g = 10 m/s² downward
Apparent weight = m(g - g) = 0 N

Reading on scale = 0 (weightlessness). — Apparent weight changes with acceleration of lift. Scale reads normal force which changes accordingly.

Question 14

14 Figure 4.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only).

Accepted Answer

Given:
Mass m = 4 kg

From position-time graph (Fig. 4.16):

(a) For t < 0: Position is constant (horizontal line), velocity = 0, acceleration = 0
Force F = m × a = 0

For 0 < t < 4 s: Position changes linearly with time, constant slope implies constant velocity, acceleration = 0
Force F = 0

For t > 4 s: Position constant again, velocity = 0, acceleration = 0
Force F = 0

(b) Impulse at t = 0 and t = 4 s:
Impulse = change in momentum = m × change in velocity
At t=0: Velocity changes from 0 to some value v (from graph slope)
Impulse = m × v
At t=4 s: Velocity changes from v to 0
Impulse = m × (-v) = -m v

Exact numerical values depend on slope of graph at intervals, which is velocity.

Hence, force is zero except at instants of velocity change (impulse), and impulses are equal in magnitude and opposite in direction at t=0 and t=4 s. — Force is related to acceleration (change in velocity). Impulse is change in momentum at velocity discontinuities.

Question 15

15 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 \mathbf{F} = 600 \mathbf{N} is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?

Accepted Answer

Given:
Masses: m_A = 10 kg, m_B = 20 kg
Force F = 600 N

Case (i): Force applied to A (10 kg)

Total mass = 10 + 20 = 30 kg
Acceleration, a = F / total mass = 600 / 30 = 20 m/s²

Tension T in string pulls B:
Force on B = m_B × a = 20 × 20 = 400 N

Case (ii): Force applied to B (20 kg)

Acceleration, a = F / total mass = 600 / 30 = 20 m/s²

Tension T pulls A:
Force on A = m_A × a = 10 × 20 = 200 N

Answer:
(i) Tension = 400 N
(ii) Tension = 200 N — Acceleration same in both cases; tension equals force needed to accelerate the other mass.

Motion in a Plane

Motion in a Plane — Study Notes

4.1 Introduction

4.2 Aristotle's fallacy

4.3 The law of inertia

Practice Questions — Motion in a Plane

All 7 Chapters in Physics Part-I