If a steel ball dropped from a height of 4.00 m rebounds to its original height, what is the impulse imparted to the ball during a 2.00 × 10⁻³ s collision?

To determine the impulse imparted to the ball, we first need to consider how impulse relates to momentum and the change in velocity of the ball during the impact. Impulse can be calculated using the formula:

[ I = \Delta p = m(v_f - v_i) ]

where ( I ) is impulse, ( m ) is the mass of the ball, ( v_f ) is the final velocity (just after the rebound), and ( v_i ) is the initial velocity (just before impact).

Initially, when the ball is dropped from a height of 4.00 m, it accelerates due to gravity, reaching a speed just before it strikes the ground. The final velocity right before impact can be calculated using the equation for an object in free fall:

[ v_i = \sqrt{2gh} ]

where ( g ) is the acceleration due to gravity (approximately 9.81 m/s²) and ( h ) is the height (4.00 m).

Calculating this gives:

[ v_i = \sqrt{2 \times 9.81 , \text{m/s}^2 \times 4.00 , \text{m}}