Question

A 0.4-kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 40 m/s and rebounds up at 20 m/s. Find the impulse on the ball during the collision.

Answer 1

The impulse experienced during the collision is calculated as the change in momentum. For the 0.4-kg ball, the impulse is found to be 24 kg⋅m/s upward.

Step-by-step explanation:

The impulse experienced by the ball during the collision can be found by applying the impulse-momentum theorem, which states that impulse is equal to the change in momentum of an object. The change in momentum (Δp) is calculated by subtracting the initial momentum from the final momentum. Since momentum is the product of mass and velocity (p = mv), we can calculate the initial and final momentum using the velocities before and after the rebound.

The initial momentum (before hitting the sidewalk) is:

• p_initial = m * v_initial
• p_initial = 0.4 kg * (-40 m/s) = -16 kg⋅m/s (negative sign indicates the downward direction)

The final momentum (after rebounding) is:

• p_final = m * v_final
• p_final = 0.4 kg * 20 m/s = 8 kg⋅m/s

The impulse (J) provided to the ball is calculated as:

• J = Δp = p_final - p_initial
• J = 8 kg⋅m/s - (-16 kg⋅m/s) = 8 kg⋅m/s + 16 kg⋅m/s
• J = 24 kg⋅m/s

The impulse on the ball during the collision is 24 kg⋅m/s upward.