Question

What is the final angle θf that the ball's velocity vector makes with the negative y-axis? Express your answer in terms of some or all of the variables m, vᵢ, θᵢ, δt.

a) θᵢ + (1/2)δt
b) θᵢ - (1/2)δt
c) θᵢ + vᵢδt
d) θᵢ - vᵢδt

Answer 1

The final angle θf that the ball's velocity vector makes with the negative y-axis is θᵢ + vᵢδt.

Therefore, the correct answer is:

`\[ \boxed{\text{c) } \theta_i + v_i \delta t} \]`

Step-by-step explanation:

To determine the final angle
`\( \theta_f \)` that the ball's velocity vector makes with the negative y-axis, we need to consider the change in angle
`(\( \Delta \theta \))` during the time interval
`\( \delta t \).`

The change in angle
`(\( \Delta \theta \))` is given by the angular displacement formula:

`\[ \Delta \theta = \omega \cdot \delta t \]`

where
`\( \omega \)` is the angular velocity.

For circular motion, the angular velocity
`\( \omega \)` is related to the linear velocity
`\( v \)` by the formula:

`\[ \omega = (v)/(r) \]`

In this case, the radius
`\( r \)` is not given, but we can express
`\( \omega \)` in terms of
`\( v \)` and
`\( \delta t \):`

`\[ \omega = (\Delta \theta)/(\delta t) \]`

Now, substitute the expression for
`\( \omega \)` in terms of
`\( v \)` and
`\( \delta t \)` into the angular displacement formula:

`\[ \Delta \theta = (v)/(\delta t) \cdot \delta t \]`

The
`\( \delta t \)` cancels out, and we are left with:

`\[ \Delta \theta = v \]`

This means that the change in angle is equal to the initial velocity
`\( v_i \).`

Now, the final angle
`\( \theta_f \)` is given by:

`\[ \theta_f = \theta_i + \Delta \theta \]`

Substitute
`\( \Delta \theta = v_i \)` into the equation:

`\[ \theta_f = \theta_i + v_i \]`

Therefore, the correct answer is:

`\[ \boxed{\text{c) } \theta_i + v_i \delta t} \]`