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**[SOLVED] Linear momentum and impulse problem**

## Homework Statement

A 3.0-kg steel ball strikes a massive wall with a speed of 10 m/s at an angle of 60 degrees with the surface. It bounces off with the same speed and angle (see Figure 9.30). If the ball is in contact with the wall for 0.20 s, what is the average force exerted on the ball by the wall?

[Description of accompanying diagram (Figure 9.30) ...]

There are x and y axes, with the y-axis representing the wall. A ball is shown coming in at an angle of 210 degrees relative to the positive x-axis (or 60 degrees to the left of the negative y-axis), and leaving at an angle of 150 degrees relative to the positive x-axis (or 60 degrees to the left of the positive y-axis).

## Homework Equations

[tex]\overrightarrow{I} = \int (from t_i to t_f) \overrightarrow{F}dt = \Delta\overrightarrow{p} = \overline{F} \Delta t[/tex]

Note: [tex]\overrightarrow{I} = impulse; \overrightarrow{F} = force; \Delta\overrightarrow{p} = change in momentum; \overline{F} = time-averaged force; and \Delta t = change in time [/tex]

## The Attempt at a Solution

The answer in the back of the book says

*260 N to the left of the diagram*.

If I reverse engineer the problem, I find that 260 N of force corresponds to 52 N of impulse. Then ...

[tex]\overline{F} = \frac{\overrightarrow{I}}{\Delta t} = \frac{52 N}{0.20 s} = 260 N[/tex]

But I can't for the life of me figure out where the impulse comes from. Since the ball enters and exits at the same angle and speed, it looks like the impulse should be zero!

Please help. Thank you.

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