Answer to Question #141198 in Mechanics | Relativity for qwetdfg

Question

A 3.00 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of = 60.08 with the surface. It bounces off at the same speed and angle. If the ball is in contact with the wall for 0.200 s, what is the average force exerted by the wall on the ball?

Answer 1

We can find the average force exerted by the wall on the ball from the impulse-momentum change equation:

$$F_{avg}\Delta t=\Delta p=m(v_f-v_i),$$

here, $F_{avg}$ is the average force exerted by the wall on the ball, $\Delta t = 0.2\ s$ is the time at which the ball is in contact with the wall, $\Delta p$ is the change in momentum, $m=3.0\ kg$ is the mass of the ball, $v_i$ is the initial velocity of the ball and $v_f$ is the final velocity of the ball.

Let's write the change in momentum in projections on axis $x$ and $y$:

$$\Delta p_x=m(-vsin\theta-vsin\theta),$$ $$\Delta p_x=3.0\ kg\cdot(-10\ \dfrac{m}{s}\cdot sin60^{\circ}-10\ \dfrac{m}{s}\cdot sin60^{\circ})=-52\ \dfrac{kgm}{s},$$ $$\Delta p_y=m(vcos\theta-vcos\theta),$$ $$\Delta p_y=3.0\ kg\cdot(10\ \dfrac{m}{s}\cdot cos60^{\circ}-10\ \dfrac{m}{s}\cdot cos60^{\circ})=0.$$

Finally, we can find the average force exerted by the wall on the ball:

$$F_{avg,x}=\dfrac{\Delta p_x}{\Delta t}=\dfrac{-52\ \dfrac{kgm}{s}}{0.2\ s}=-260\ N.$$

The sign minus means that the average force exerted by the wall on the ball directed in the negative $x$-direction.

Answer:

$F_{avg,x}=260\ N,$ in the negative $x$-direction.