# A ball of mass m moving with velocity Vi strikes a vertical wall. The angle between the ball's initial velocity vector and the wall is θi. 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.

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

I think I need to find the x and y components and then put them together but I don't know how to find the components.

## To find the x and y components of the ball's initial velocity vector, we can use trigonometry. Let's denote the initial velocity vector as Vi.

The x-component of Vi can be found using the formula:

Vix = Vi * cos(θi)

The y-component of Vi can be found using the formula:

Viy = Vi * sin(θi)

After the ball collides with the wall, the x-component of velocity remains the same, while the y-component of velocity changes direction. Since the collision is elastic, the magnitude of the velocities before and after the collision will be the same.

Therefore, the x-component of the final velocity vector, Vfx, will be the same as the initial x-component: Vfx = Vix.

However, the y-component of the final velocity vector, Vfy, will have the same magnitude as the initial y-component, but in the opposite direction. Therefore, we can write:

Vfy = -Viy

Now we can find the final angle θf that the ball's velocity vector makes with the negative y-axis using trigonometry.

θf = arctan(Vfy/Vfx)

Substituting in the known values, the final equation becomes:

θf = arctan((-Viy)/Vix)

By substituting the expressions for Vix and Viy, we have:

θf = arctan((-Vi * sin(θi))/ (Vi * cos(θi)))

Simplifying further:

θf = arctan(-tan(θi))

Therefore, the final angle θf that the ball's velocity vector makes with the negative y-axis is given by θf = arctan(-tan(θi)).

## To find the final angle θf that the ball's velocity vector makes with the negative y-axis, we need to break down the ball's initial velocity into its x and y components, and then analyze the collision to determine the resulting velocities.

Let's denote the vertical component of the initial velocity as Viy and the horizontal component as Vix. We can calculate these components using trigonometry:

Viy = Vi * sin(θi)

Vix = Vi * cos(θi)

During the collision, the vertical component of the ball's velocity is reversed while the horizontal component remains unchanged. This is because the force exerted by the wall is parallel to the x-axis, meaning it only affects the motion of the ball in the x-direction. Since the collision is completely elastic, the magnitude of the ball's velocities in both directions should remain the same.

After the collision, the resulting velocity components can be denoted as Vfy and Vfx. Since the magnitude of the velocities does not change, we have:

Vfy = |Viy|

Vfx = |Vix|

The final angle θf that the ball's velocity vector makes with the negative y-axis can be determined using inverse trigonometric functions:

θf = atan(Vfy / Vfx)

By substituting the values we found earlier, the final angle θf can be calculated.