Question

A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g=10 m/s2) nearly

• A. 1.4 kg m/s
• B. 0 kg m/s
• C. 4.2 kg m/s
• D. 2.1 kg m/s

Answer 1

The Correct Option is C

To solve this problem, we need to determine the impulse imparted to the ball when it strikes the ground and then rebounds to the same height.

1. First, let's understand the concept of impulse. Impulse is defined as the change in momentum, and is given by:

\(I = \Delta p = m(v_{\text{final}} - v_{\text{initial}})\)

1. Calculate the initial velocity just before hitting the ground. Since the ball is dropped from rest, we use the equation of motion:

\(v^2 = u^2 + 2gh\)

Here, \(u = 0\) (initial velocity), \(g = 10 \, \text{m/s}^2\), and \(h = 10 \, \text{m}\).

Substituting the values:

\(v^2 = 0 + 2 \times 10 \times 10 = 200\)

Thus, the velocity just before impact:

\(v = \sqrt{200} = 14.14 \, \text{m/s}\)

1. Determine the final velocity just after the ball rebounds to the same height. Since it rebounds to the same height, its speed upon leaving the ground is the same as its speed right before impact, but in the opposite direction:

\(v_{\text{rebound}} = 14.14 \, \text{m/s} \, \text{(upward)}\)

1. Calculate the change in velocity (\(\Delta v\)):

\(\Delta v = v_{\text{rebound}} - (-v_{\text{impact}}) = 14.14 - (-14.14) = 14.14 + 14.14 = 28.28 \, \text{m/s}\)

1. Now calculate the impulse imparted to the ball:

\(I = m \times \Delta v = 0.15 \times 28.28 = 4.242 \, \text{kg m/s}\)

Rounding to one decimal place, the impulse is approximately \(4.2 \, \text{kg m/s}\).

Thus, the correct answer is 4.2 kg m/s.