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 the problem, we will calculate the magnitude of the impulse imparted to the ball when it hits the ground and rebounds. The relevant concepts include impulse and momentum. Impulse is defined as the change in momentum of the object. 

1. The initial velocity (\(v_i\)) of the ball just before it strikes the ground can be calculated using the equation for velocity under gravity: \(v_i = \sqrt{2gh}\), where \(g = 10 \, \text{m/s}^2\) and \(h = 10 \, \text{m}\).

\(v_i = \sqrt{2 \times 10 \times 10} = \sqrt{200} \approx 14.14 \, \text{m/s}\)

1. When the ball rebounds to the same height, its final velocity (\(v_f\)) at the point of impact upon rebounding will be equal in magnitude but opposite in direction to \(v_i\). Thus, \(v_f = -v_i \approx -14.14 \, \text{m/s}\).
2. Impulse (\(J\)) is given by the change in momentum, which can be expressed as: \(J = m(v_f - v_i)\), where \(m = 0.15 \, \text{kg}\).

\(J = 0.15 \times (-14.14 - 14.14) = 0.15 \times (-28.28) \approx -4.242 \, \text{kg m/s}\)

1. We take the magnitude of impulse as \(|J| \approx 4.24 \, \text{kg m/s}\)which, when rounded, is approximately \(4.2 \, \text{kg m/s}\).

Therefore, the magnitude of the impulse imparted to the ball is approximately 4.2 kg m/s, which corresponds with option 3.

Answer 2

The correct answer is option (C): 4.2 kg m/s

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 can be determined by-