A 2 kg steel ball strikes a wall with a speed

of 13.9 m/s at an angle of 54.5
◦ with the
normal to the wall. It bounces off with the
same speed and angle, as shown.
If the ball is in contact with the wall for
0.277 s, what is the magnitude of the average
force exerted on the ball by the wall?
Answer in units of N.

1 answer

the component of velocity normal to the wall is reversed on impact, the other component stays the same.
So the change of momentum is dependent on this part of velocity.

changemomentum=2*mass*cos54.5*13.9
the 2 in front is because the initial normal momentum is reversed.

Impulse=changemomentum
Force*time=2*2*13.9cos54.5
solve for force.