Question

A ball of mass m moving with velocity v⃗ i strikes a vertical wall as shown in (Figure 1). The angle between the ball's initial velocity vector and the wall is θi as shown on the diagram, which depicts the situation as seen from above. The duration of the collision between the ball and the wall is Δt, and this collision is completely elastic. Friction is negligible, so the ball does not start spinning. In this idealized collision, the force exerted on the ball by the wall is parallel to the x axis.

Answer 1

Part A

What is the final angle θf that the ball’s velocity vector makes with the negative y axis?

Express your answer in terms of quantities given in the problem introduction.

Typically, when one object hits another and bounces off of it, the angle of deflection is the same as the initial angle. You probably know this intuitively based on the behavior of light – if you shine a beam of light at a mirror, it will reflect off the mirror at the same angle with which it hit the mirror. For physical objects, the behavior is the same.

θf = θi (answer)

Part B

What is the magnitude F of the average force exerted on the ball?

Express your answer in terms of variables given in the problem introduction and/or vix.

To solve this problem, we need to understand what type of collision we are dealing with. Since the problem tells us the collision is perfectly elastic, we know that the initial and final momentums will have the same magnitude. You might also notice that the y-direction of the ball isn’t changing – so we know that there will be no change in the y-component of the force.

We know that F = ma. But how do we get from force to momentum? Well, force is just the change in momentum. Have a look:

F = ma

a = Δv / Δt

so:

F = m * (Δv / Δt)

and:

ρ = mv

If we consider the change in momentum, we end up with:

Δρ = m(Δv)

so:

F = m * (Δv / Δt)

Δρ = m * (Δv)

F = m * Δρ / Δt

Since the ball is simply changing the direction of it’s x-movement, the change in momentum is just twice the initial momentum (in the x-direction):

Δρ = 2 * mv * sin(θ)

To get the force, just divide by (Δt):

F = 2mvisin(θf)/(Δt)

F = 2m*v_i sin(θf)/(Δt)   (ans)