Posted by Seth on .
A 2 kg steel ball strikes a wall with a speed of 8.15 m/s at an angle of 54.4◦ with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. If the ball is in contact with the wall for 0.243 s, what is the magnitude of the average force exerted on the ball by the wall?
Answer in units of N
Divide the momentum change magnitude by 0.243 seconds.
I would need to see the figure to know what the momentum change is. It could bounce off in the same direction that it came (backwards), or in a 54.4 degree direction on the other side of the normal.