1. **State the problem:** A 3 kg steel ball hits a wall at 10 m/s at a 60° angle to the surface and bounces off with the same speed and angle. The contact time is 0.2 s. We need to find the average force exerted by the wall on the ball. 2. **Understand the physics:** The ball's velocity component perpendicular to the wall changes direction, while the parallel component remains the same. The force is related to the change in momentum perpendicular to the wall. 3. **Resolve velocity into components:** - Velocity perpendicular to the wall: $v_\perp = 10 \times \cos 60^\circ = 10 \times 0.5 = 5$ m/s - Velocity parallel to the wall: $v_\parallel = 10 \times \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ m/s 4. **Calculate change in perpendicular velocity:** - Initial perpendicular velocity towards the wall: $+5$ m/s - Final perpendicular velocity away from the wall: $-5$ m/s - Change in velocity: $\Delta v_\perp = -5 - (+5) = -10$ m/s 5. **Calculate change in momentum perpendicular to the wall:** $$\Delta p = m \times \Delta v_\perp = 3 \times (-10) = -30 \text{ kg m/s}$$ 6. **Calculate average force:** $$F = \frac{\Delta p}{\Delta t} = \frac{-30}{0.2} = -150 \text{ N}$$ The negative sign indicates direction opposite to initial velocity. 7. **Interpretation:** The average force exerted by the wall on the ball is 150 N directed normal to the wall. **Note:** The problem states 260 N normal to the wall, which suggests considering the total force magnitude including the parallel component or a different approach, but based on momentum change normal to the wall, the force is 150 N. Final answer: **150 N normal to the wall**.