Force and Laws of Motion — Practice

Question 1

Define one newton.

Accepted Answer

Step 1 — From F = ma.
  Force = mass × acceleration.

Step 2 — Set both to 1 in SI.
  1 N = 1 kg × 1 m/s².

✦ Answer: 1 newton is the force that produces an acceleration of 1 m/s² in a body of mass 1 kg.

Question 2

What force is needed to accelerate a 2 kg object at 5 m/s²?

Accepted Answer

Step 1 — Use F = ma.
  F = 2 × 5 = 10 N.

✦ Answer: 10 N.

Question 3

Find the momentum of a 60 kg cyclist moving at 5 m/s.

Accepted Answer

Step 1 — p = mv = 60 × 5 = 300 kg·m/s.

✦ Answer: 300 kg·m/s.

Question 4

Why do passengers fall forward when a moving bus suddenly stops?

Accepted Answer

Step 1 — Identify the type of inertia.
  Inertia of MOTION — bodies in motion tend to stay in motion.

Step 2 — Apply to the situation.
  Before braking, both the bus and the passenger are moving forward at the same velocity.
  When the bus suddenly stops (large deceleration on the bus), passengers' bodies — having inertia of motion — tend to continue moving forward.
  The lower body (in contact with the seat) decelerates with the bus, but the upper body lurches forward.

✦ Answer: Passengers' bodies have inertia of motion — they continue moving forward when the bus suddenly stops (which decelerates only the bus, not the loose passengers). They jerk forward until the seat belt or other force stops them.

Question 5

When you walk on the ground, identify the action-reaction force pair.

Accepted Answer

Step 1 — When walking, your feet push the ground backward.
  This is the ACTION force (your foot on the ground, directed backward).

Step 2 — By Newton's 3rd law, the ground pushes back on your foot.
  This is the REACTION force (the ground on your foot, directed forward).
  The forward reaction force is what propels you forward.

✦ Answer: Action: your foot pushes ground backward. Reaction: ground pushes your foot forward. You move forward thanks to the reaction force.

Question 6

A car of mass 1000 kg accelerates from rest to 20 m/s in 10 s. Find the net force on it.

Accepted Answer

Step 1 — Find acceleration.
  a = (v − u)/t = (20 − 0)/10 = 2 m/s².

Step 2 — Apply F = ma.
  F = 1000 × 2 = 2000 N = 2 kN.

✦ Answer: Net force = 2000 N (2 kN), in the direction of motion.

Note: This is the NET force. The actual engine force is larger (counters air drag + friction), but the net is what gives the acceleration.

Question 7

A body of mass 5 kg moving with velocity 10 m/s is brought to rest in 2 s. Find (a) the impulse, (b) the average force.

Accepted Answer

Step 1 — Find change in momentum.
  Initial p = mv = 5 × 10 = 50 kg·m/s.
  Final p = 0.
  Change Δp = 0 − 50 = −50 kg·m/s.

Step 2 (a) — Impulse = Δp.
  Impulse = −50 kg·m/s = 50 N·s (magnitude).

Step 3 (b) — Average force.
  F = Δp/Δt = −50/2 = −25 N (magnitude 25 N, directed opposite to motion).

✦ Answer: (a) Impulse = 50 N·s (in the direction of stopping force); (b) Average force = 25 N (decelerating). The negative sign reflects that the force opposes the original motion.

Question 8

A gun of mass 4 kg fires a bullet of mass 20 g with a muzzle velocity of 300 m/s. Find the recoil velocity of the gun.

Accepted Answer

Step 1 — Conservation of momentum.
  Initial total momentum = 0 (both at rest).
  Final total momentum = m_bullet × v_bullet + m_gun × v_gun = 0.

Step 2 — Substitute.
  (0.020)(300) + (4)(v_gun) = 0
  6 + 4 × v_gun = 0
  v_gun = −1.5 m/s.

Step 3 — Interpret sign.
  Negative means gun recoils in the opposite direction to the bullet.

✦ Answer: Recoil velocity of the gun = 1.5 m/s, in the direction opposite to the bullet's motion.

This is why a heavy rifle is easier to fire: same momentum, more mass → smaller recoil velocity (less kick on your shoulder).

Question 9

A 50 kg boy running at 4 m/s jumps onto a 200 kg cart moving in the same direction at 1 m/s. Find their common velocity after he lands (assume he sticks to the cart).

Accepted Answer

Step 1 — Conservation of momentum (inelastic collision, two bodies become one).
  m_boy × v_boy + m_cart × v_cart = (m_boy + m_cart) × v_common

Step 2 — Substitute.
  50 × 4 + 200 × 1 = (50 + 200) × v
  200 + 200 = 250 × v
  400 = 250v
  v = 400/250 = 1.6 m/s.

✦ Answer: Common velocity = 1.6 m/s (in the direction of original motion).

Observation: the boy slows down (4 → 1.6), the cart speeds up (1 → 1.6). Total momentum 400 kg·m/s is unchanged.

Question 10

A book is at rest on a table. Identify the action-reaction pair acting between the book and the table.

Accepted Answer

Step 1 — Recognise: action and reaction are between TWO BODIES, not on the same body.

Step 2 — Identify the pair.
  Action: the book pushes DOWN on the table due to its weight (book on table).
  Reaction: the table pushes UP on the book (table on book) — this is the 'normal force'.

Step 3 — NOTE: gravity (the Earth pulling the book down) is NOT the action paired with the normal force. The reaction to gravity is the book pulling the Earth up (negligible but real, Newton's 3rd law).

✦ Answer: Action: book pushes the table downward (force = weight). Reaction: table pushes the book upward (normal force). These act on different bodies; that's why they don't 'cancel' — they're a Newton's 3rd-law pair.

The weight (mg) and the normal force happen to balance on the book — but balanced forces and action-reaction pairs are different concepts.

Question 11

A bus of mass 5000 kg moves with velocity 72 km/h. The driver applies brakes and the bus stops in 5 s. Find: (a) the deceleration, (b) the braking force.

Accepted Answer

Step 1 — Convert speed.
  72 km/h = 72 × 5/18 = 20 m/s.

Step 2 (a) — Deceleration.
  a = (v − u)/t = (0 − 20)/5 = −4 m/s².
  Magnitude 4 m/s² (deceleration).

Step 3 (b) — Braking force.
  F = ma = 5000 × (−4) = −20,000 N.
  Magnitude 20,000 N = 20 kN, directed opposite to motion.

✦ Answer: Deceleration = 4 m/s². Braking force = 20,000 N (20 kN).

Real-world calibration: heavy braking systems for buses produce 10–25 kN. The number is plausible for an emergency stop.

Question 12

Explain why a fast-moving cricket ball hurts more if caught with a stiff hand than if the hand is moved back during the catch.

Accepted Answer

Step 1 — Impulse = F × t = change in momentum.
  When a cricket ball is caught, Δp (change in ball's momentum) is fixed — it goes from high momentum to zero.

Step 2 — Same Δp, different time.
  Stiff hand: ball stops in a very short time t. Impulse formula → F × t = Δp → F = Δp/t (small t → large F → painful).
  Hand moved back: stopping time t is increased. Same Δp → smaller force F. Less pain.

Step 3 — Practical principle.
  Spreading impact over time reduces peak force. Same idea behind:
  • Air bags (slow stop in 50 ms instead of 5 ms).
  • Crumple zones in cars.
  • Padded gloves in boxing.
  • Springs in a parachute landing.

✦ Answer: For a given change in momentum (Δp), extending the contact time t reduces the average force (F = Δp/t). Moving the hand back stretches the stopping time → smaller force → less pain. Same principle as airbags and crumple zones.

Question 13

Two blocks of masses 3 kg and 2 kg are connected by a string and placed on a frictionless horizontal surface. A force of 10 N is applied to the 3 kg block, pulling both forward. Find (a) the acceleration of the system, (b) the tension in the string.

Accepted Answer

Step 1 — Analyse the system as a whole.
  Total mass = 3 + 2 = 5 kg.
  Net force = 10 N (only applied force, no friction).
  Acceleration: a = F/m = 10/5 = 2 m/s².

Step 2 — Find the tension by analysing just the 2 kg block.
  Only the string pulls the 2 kg block forward with tension T.
  T = m × a = 2 × 2 = 4 N.

Step 3 — Verify by analysing the 3 kg block.
  Forces on 3 kg block: +10 N applied, −T (string pulls back).
  Net force = 10 − T = 3 × 2 = 6 N (mass × acceleration).
  T = 10 − 6 = 4 N. ✓

✦ Answer: (a) Acceleration of system = 2 m/s²; (b) Tension in string = 4 N.

Question 14

An object of mass 1 kg moving at 10 m/s collides head-on with another stationary object of mass 4 kg. After collision, the 1 kg object moves backward at 2 m/s. Find the velocity of the 4 kg object after collision.

Accepted Answer

Step 1 — Pick a positive direction (rightward = positive).
  Before: m₁ = 1 kg at v₁ = +10 m/s; m₂ = 4 kg at v₂ = 0.
  After: 1 kg moves BACKWARD at 2 m/s → v₁' = −2 m/s. v₂' = ?

Step 2 — Conservation of momentum.
  m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
  1 × 10 + 4 × 0 = 1 × (−2) + 4 × v₂'
  10 = −2 + 4 × v₂'
  4 × v₂' = 12
  v₂' = 3 m/s.

Step 3 — Interpret.
  Positive sign → 4 kg object moves forward (in the original direction of the 1 kg).

✦ Answer: The 4 kg object moves forward at 3 m/s after the collision.

Sanity check on momentum: Before = +10 kg·m/s. After = 1×(−2) + 4×3 = −2 + 12 = +10 kg·m/s. ✓

Question 15

A rocket of mass 1000 kg is to be lifted vertically. To lift, the rocket must overcome gravity AND accelerate upward. If the upward acceleration is 5 m/s², find the thrust (force exerted by the engines). (g = 10 m/s²)

Accepted Answer

Step 1 — Forces on the rocket (vertical only).
  Weight (downward) = mg = 1000 × 10 = 10,000 N.
  Thrust (upward) = T.
  Net force = T − mg (upward positive).

Step 2 — Apply Newton's 2nd law.
  Net force = ma.
  T − mg = m × a.
  T = m(g + a) = 1000 × (10 + 5) = 15,000 N.

Step 3 — Interpret.
  Of the 15 kN thrust, 10 kN cancels gravity, and 5 kN provides the upward acceleration.

✦ Answer: Thrust required = 15,000 N (15 kN).

The rocket's thrust must always exceed its weight (mg) for any net upward motion. Just hovering needs T = mg; accelerating up needs T > mg.

Practice — Force and Laws of Motion

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