A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall.The ball strikes the wall horizontally with a speed of $25 \mathrm {~m} / \mathrm { s } \text {, }$ and it bounces back with this same speed.The ball is in contact with the wall $0.050 \mathrm {~s} .$ What is the magnitude of the average force exerted on the wall by the ball?
A)800 N
B)400 N
C)55,000 N
D)27,500 N
E)13,750 N

Question

Quizplus · Accepted Answer

The average force exerted on the wall by the ball can be calculated using the change in momentum of the ball divided by the time it is in contact with the wall. The change in momentum is given by the difference in the final and initial momentum of the ball. Since the ball bounces back with the same speed, its velocity changes direction but not magnitude, resulting in a total change in velocity of $2 \times 25 \mathrm{~m/s} = 50 \mathrm{~m/s}$. The change in momentum ($\Delta p$) is mass ($m$) times change in velocity ($\Delta v$), so $\Delta p = 0.80 \mathrm{~kg} \times 50 \mathrm{~m/s} = 40 \mathrm{~kg \cdot m/s}$. The average force ($F$) is then $\Delta p / \Delta t = 40 \mathrm{~kg \cdot m/s} / 0.050 \mathrm{~s} = 800 \mathrm{~N}$.

A Girl of Mass 55 Kg Throws a Ball of Mass