Samacheer Kalvi 10th Science Chapter 1 Numerical Problems — Laws of Motion with Solutions

Numerical Problems — Chapter 1: Laws of Motion

Important Formulas:

Problem 1. A force of 30 N acts on a body of mass 5 kg. Find the acceleration produced.

• Force, F = 30 N
• Mass, m = 5 kg
• Acceleration, a = ?

By Newton's Second Law: F = ma

a = F/m = 30/5 = 6 m/s²

Problem 2. A body of mass 4 kg is moving with a velocity of 8 m/s. Find its momentum.

• Mass, m = 4 kg
• Velocity, v = 8 m/s
• Momentum, p = ?

p = mv = 4 × 8 = 32 kg·m/s

Problem 3. A force of 10 N acts on a body for 3 seconds. Find the impulse. If the mass of the body is 2 kg and it was initially at rest, find the final velocity.

• Force, F = 10 N
• Time, t = 3 s
• Mass, m = 2 kg
• Initial velocity, u = 0 m/s

Step 1: Find Impulse

Impulse = F × t = 10 × 3 = 30 N·s

Step 2: Find final velocity

Impulse = change in momentum = mv − mu

30 = 2 × v − 2 × 0

30 = 2v

v = 30/2 = 15 m/s

Problem 4. A cricket ball of mass 0.15 kg is moving with a velocity of 12 m/s. A batsman hits it and the ball moves in the opposite direction with a velocity of 20 m/s. If the contact time is 0.01 s, find: (a) The change in momentum of the ball (b) The average force exerted by the batsman

• Mass, m = 0.15 kg
• Initial velocity, u = 12 m/s (before hit, take as positive)
• Final velocity, v = −20 m/s (opposite direction, so negative)
• Time of contact, t = 0.01 s

Step 1: Change in momentum

Δp = mv − mu = m(v − u) = 0.15 × (−20 − 12) = 0.15 × (−32) = −4.8 kg·m/s

Magnitude of change in momentum = 4.8 kg·m/s

Step 2: Average force

F = Δp / t = 4.8 / 0.01 = 480 N

• Change in momentum = 4.8 kg·m/s (direction reversed)
• Average force = 480 N

Problem 5. A body of mass 2 kg is moving with velocity 4 m/s. A net force acts on it for 5 seconds and the velocity changes to 14 m/s. Find the net force.

• Mass, m = 2 kg
• Initial velocity, u = 4 m/s
• Final velocity, v = 14 m/s
• Time, t = 5 s
• Force, F = ?

Step 1: Find acceleration

a = (v − u) / t = (14 − 4) / 5 = 10 / 5 = 2 m/s²

Step 2: Find force

F = ma = 2 × 2 = 4 N

Problem 6. A gun of mass 3 kg fires a bullet of mass 30 g with a velocity of 400 m/s. Find the recoil velocity of the gun.

• Mass of gun, M = 3 kg
• Mass of bullet, m = 30 g = 0.030 kg
• Velocity of bullet, v = 400 m/s
• Recoil velocity of gun, V = ?

Before firing: Both at rest → Total momentum = 0

By conservation of momentum: MV + mv = 0

MV = −mv

V = −(mv) / M = −(0.030 × 400) / 3

V = −12 / 3 = −4 m/s

The negative sign means the gun moves opposite to the bullet.

Problem 7. An object of mass 5 kg is moving with a speed of 10 m/s in a circular path of radius 4 m. Find the centripetal force acting on it.

• Mass, m = 5 kg
• Speed, v = 10 m/s
• Radius, r = 4 m
• Centripetal force, F = ?

F = mv²/r = (5 × 10²) / 4 = (5 × 100) / 4 = 500 / 4 = 125 N

Problem 8. A force of 50 N acts on a body and produces an acceleration of 5 m/s². Find the mass of the body. Also, find what force is needed to produce an acceleration of 8 m/s² in the same body.

Given (Part 1):

• Force, F = 50 N
• Acceleration, a = 5 m/s²
• Mass, m = ?

Problem 9. Two objects of masses 2 kg and 4 kg are moving in opposite directions with velocities 6 m/s and 3 m/s respectively. After collision, they move together. Find their common velocity after collision.

• m₁ = 2 kg, u₁ = +6 m/s (positive direction)
• m₂ = 4 kg, u₂ = −3 m/s (opposite direction, negative)
• After collision: they move together with velocity v

By conservation of momentum: m₁u₁ + m₂u₂ = (m₁ + m₂)v

(2 × 6) + (4 × −3) = (2 + 4) × v

12 + (−12) = 6v

0 = 6v

v = 0 m/s

Problem 10. A vehicle of mass 1200 kg is moving with a velocity of 20 m/s. Find: (a) The momentum of the vehicle (b) The braking force required to stop the vehicle in 8 seconds

• Mass, m = 1200 kg
• Initial velocity, u = 20 m/s
• Final velocity, v = 0 (stopped)
• Time, t = 8 s

Part (a): Momentum

p = mu = 1200 × 20 = 24,000 kg·m/s

Part (b): Braking force

a = (v − u) / t = (0 − 20) / 8 = −20/8 = −2.5 m/s²

F = ma = 1200 × 2.5 = 3,000 N (decelerating force)

• Momentum = 24,000 kg·m/s
• Braking force = 3,000 N

Quick Formula Reference

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