Question

A ball of mass 500 g strikes a fixed wall with a speed of 10 m/s. After striking, it returns back with the same speed. If the ball was in contact with the wall for 0.01 s, find the magnitude of the force applied on the ball by the wall.

• A. 500 N
• B. 1000 N
• C. 2000 N
• D. 1500 N

Answer 1

The correct option is B 1000 N

The initial momentum is given as mass times the initial velocity.
⇒ p${}_i$ = m x v${}_i$ = 0.5 kg x 10 m/s = 5 kg m/s

After the collision with the wall, the velocity of the ball becomes opposite to the initial velocity.

⇒ Final velocity, v${}_f$ = -10 m/s

Here, the negative sign implies the opposite direction.

The final momentum is given as mass times the final velocity. ⇒ p${}_f$ = m x v${}_f$ = 0.5 kg x (-10 m/s) = -5 kg m/s

Change in momentum = p${}_f$ - p${}_i$
Change in momentum = -5 - 5 = -10 kg m/s

⇒ Magnitude of change in momentum = 10 kg m/s

From Newton’s second law, the force applied is equal to the rate of change of momentum.
F = $\frac{p_f-p_i}{t}$
F = $\frac{-10}{0.01}$
= -1000 N

Here, the negative sign implies that the force applied on the ball is in the direction opposite to its initial velocity.