Problem 7 on Motion in a Straight Line

Description:

Problem

A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one − tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.

Solution

Salient points about the problem −

• Ball’s motion will reverse its direction at every impact.

• Equations of motion will be applicable because the motion is happening under constant acceleration.

• We will analyze each impact and subsequent motion after it step-by-step.

• Sign convention chosen for this is,

  • Up is negative
  • Dowm is positive

Before First Impact

Ball drops from the height to hit the ground for the first impact.

Height = 90 m

Calculating velocity of first impact using, v² = u² + 2gH, (u = 0; S = 90),

v₁ = √2gH = 30√2 m/s = 42.42 m/s

Time elapsed before first impact using, H = ut + 1/2 gt², (u = 0; H = 90),

t = 4.24 s

Between First and Second Impact

Problem says, ball loses one − tenth of its speed upon every impact. This implies nine − tenth of the velocity is retained.

Therefore, magnitude of take-off velocity after first impact,

u₂ = 9/10 × v₁ = 9/10 × 30√2 = 27√2 m/s = 38.18 m/s

Time taken to reach the highest point using, v = u₂ + at, (v = 0)

0 = -27√2 + gt

t = 3.82 s

Total time before second impact, (Time taken to go up and then come down is simply twice the time taken to go up)

2 × t = 7.64 s

Between Second and Third Impact

Problem says, ball loses one − tenth of its speed upon every impact. This implies nine − tenth of the velocity is retained.

Magnitude of take-off velocity after second impact,

u₃ = 9/10 × u₂ = 9/10 × 27√2 = (24.3)√2 m/s = 34.36 m/s

Time taken to reach the highest point using, v = u₃ + at, (v = 0)

0 = -24.3√2 + gt

t = 3.44 s

Total time before third impact, (Time taken to go up and then come down is simply twice the time taken to go up)

2 × t = 6.88 s

The final Velocity – Time graph

The final Speed – Time graph