Motion and force

NUST NET (Engineering) · Physics · Mechanics. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

For motion in a straight line with constant acceleration, four equations link u, v, a, s and t. Pick the one that leaves out the quantity you do not know or need. On a velocity-time graph the gradient is the acceleration and the area underneath is the displacement.

Newton's second law says force is the rate of change of momentum, so a force acting for a time gives an impulse equal to the change in momentum. Total momentum is conserved in every collision; kinetic energy is conserved only in an elastic one.

A projectile is two motions at once: constant velocity horizontally (u cos θ) and free fall vertically (starting at u sin θ).

Rules to remember

Common mistake

Saying a projectile has zero velocity at its highest point. Only the vertical part is zero there; the speed is still u cos θ and the acceleration is still g downwards.

Worked example

A stone is thrown horizontally at 15 m s−1 from the top of a cliff 20 m high. Taking g = 10 m s−2, how far from the foot of the cliff does it land?

  1. Vertically it starts from rest: h = ½gt2, so t2 = 2h/g = (2 × 20)/10 = 4 and t = 2 s.
  2. Horizontally the velocity stays at 15 m s−1.
  3. Distance = 15 × 2 = 30 m.

Answer: 30 m

Practice questions

Try each one, then open the answer.

1. A 1 kg ball moving at 4 m s−1 makes a head-on elastic collision with a 3 kg ball at rest. Just after the collision the 1 kg ball

  1. A
    stops
  2. B
    moves forward at 2 m s−1
  3. C
    moves back at 2 m s−1
  4. D
    moves back at 4 m s−1
Show answer

Answer: C. v1 = (m1 − m2)u/(m1 + m2) = (1 − 3) × 4/4 = −2 m s−1, so it bounces back at 2 m s−1. The 3 kg ball moves forward at 2m1u/(m1 + m2) = 2 m s−1. A ball stops only when the masses are equal.

2. A car starts from rest and its speed rises uniformly to 12 m s−1 in 4 s. It then keeps this speed for another 6 s. The total distance covered is

  1. A
    72 m
  2. B
    96 m
  3. C
    120 m
  4. D
    48 m
Show answer

Answer: B. Distance = area under the v–t graph = ½ × 4 × 12 + 12 × 6 = 24 + 72 = 96 m. 120 m treats the whole 10 s as if the car were at 12 m s−1 throughout.

3. A projectile is launched at 25 m s−1 at an angle θ above the horizontal, where sin θ = 0.8. Taking g = 10 m s−2, it reaches its highest point after

  1. A
    2.0 s
  2. B
    4.0 s
  3. C
    1.5 s
  4. D
    2.5 s
Show answer

Answer: A. At the top the vertical velocity is zero: t = v sin θ/g = 25 × 0.8/10 = 2.0 s. 4.0 s is the whole time of flight, and 1.5 s uses cos θ = 0.6.

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