Question

A ball weighing 10g hits a hard surface vertically with a speed of 5 ms^-1 and rebounds with the same speed. The ball remains in contact with the surface for 0.01 s. The average force exerted by the surface on ball is.

• A. 100N
• B. 10N
• C. 1 N
• D. 0.1N

Answer 1

Answer: (B)

10N

Answer 2

Impulse = Ft = change in momentum = mv – (-mv)

= 2mv = 2 × 0.01 × 5 = 0.1

∴ F = $\frac{0.1}{0.01}$ = 10 N