Question

1. a 2 kg ball hits a surface with a speed of 5 m/s and bounces back with a speed of 2.5 m/s. determine the momentum change of the ball.

Explanation:

Step1: Recall momentum formula

Momentum \( p = mv \), where \( m \) is mass and \( v \) is velocity. Momentum change \( \Delta p = p_f - p_i \).

Step2: Define initial and final velocities

Let initial velocity \( v_i = 5 \, \text{m/s} \) (towards the surface, positive direction), final velocity \( v_f = -2.5 \, \text{m/s} \) (bounces back, negative direction). Mass \( m = 2 \, \text{kg} \).

Step3: Calculate initial momentum

\( p_i = m v_i = 2 \times 5 = 10 \, \text{kg·m/s} \)

Step4: Calculate final momentum

\( p_f = m v_f = 2 \times (-2.5) = -5 \, \text{kg·m/s} \)

Step5: Calculate momentum change

\( \Delta p = p_f - p_i = -5 - 10 = -15 \, \text{kg·m/s} \). The negative sign indicates direction, magnitude of change is \( 15 \, \text{kg·m/s} \).

Answer:

The momentum change of the ball is \(\boldsymbol{-15 \, \text{kg·m/s}}\) (or a magnitude of \(15 \, \text{kg·m/s}\) in the direction opposite to the initial velocity).