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 the formula for momentum

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

Step2: Define initial and final velocities

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

Step3: Calculate initial and final momentum

Initial momentum \( p_i = m v_i = 2 \times 5 = 10 \, \text{kg·m/s} \).
Final momentum \( p_f = m v_f = 2 \times (-2.5) = -5 \, \text{kg·m/s} \).

Step4: Calculate momentum change

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