A 75-g ball strikes a wall horizontally at 12 m/s and rebounds in the opposite direction at 10 m/s. The ball’s change of momentum has magnitude

Question

BlackTom AI · Accepted Answer

The problem gives a 75-g ball (mass m = 0.075 kg) traveling horizontally toward a wall at 12 m/s and rebounding in the opposite direction at 10 m/s. To find the change in momentum, use Δp = p_final − p_initial.

- Before impact: p_initial = m × v_initial = 0.075 kg × (+12 m/s) = +0.90 kg·m/s.
- After impact: p_final = m × v_final = 0.075 kg × (−10 m/s) = −0.75 kg·m/s.

Now compute the change: Δp = (−0.75) − (+0.90) = −1.65 kg·m/s.

The magnitude of the change in momentum is |Δp| = 1.65 kg·m/s.

If we consider possible answer options (even though none are listed here), an option stating 1.65 kg·m/s would match the calculated magnitude. Any incorrect option would typically differ in sign, use an incorrect mass, or misapply the velocity values, leading to incorrect magnitude or sign for Δp.

In summary, the computed magnitude of the momentum change for this collision is 1.65 kg·m/s, derived directly from Δp = m(v_final − v_initial) with the given velocities.

SPHY001 Practice Quiz - Unit 08 | LA048

A 75-g ball strikes a wall horizontally at 12 m/s and rebounds in the opposite direction at 10 m/s. The ball’s change of momentum has magnitude

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