Question

What did I do wrong on this question?

A 0.2-kg ball is bounced against a wall. It hits the wall with a speed of 20 m/s at a 35 degree angle and rebounds elastically at a 35 degree angle. What is the magnitude of the change in momentum of the ball in the horizontal direction?

I found v1 =-20

V2= 20

-20(cos35)=-16.4

20(cos35)=16.4

I then took 16.4-(-16.4) and multiplied that by 0.2, and got 6.56, however, I was told the answer was between 4.5 and 4.6. What did I do wrong?

Answer 1

Change in momentum in horizontal direction will be

P = m*Vi - m*Vf

Your whole process is correct, but mistake was that you need to take horizontal component of velocity, which will be

Vx = 20*sin 35 deg

Since angle 35 deg is with the wall, So vertical component of velocity will be Vy = V*cos 35 deg

and horizontal component will be Vx = V*sin 35 deg

generally given angles are with x-axis, since ground is assumed to be x-axis, but in this case angle is with the wall, so wall is x-axis and ground is vertical axis. And horizontal direction is ground, So we need to take vertical component of velocity for horizontal direction.

I hope I was able to clear your confusion, if you still have any query, please comment below.

Solution is:

dP = m*dV = 0.2*(20*sin 35 deg - (-20*sin 35 deg))

dP = 4.58

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