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Step-by-Step Solution
Step 1: Identify the Relevant Concepts
The problem involves a ball striking a wall and reflecting with the same speed but reversed normal (perpendicular) velocity component. The key physical concept here is the change in momentum along the direction perpendicular to the wall, and the average force is calculated using the impulse-momentum theorem:
$ \text{Average Force} = \dfrac{\Delta p}{\Delta t} $
Step 2: Determine the Velocity Component Normal to the Wall
Given:
Mass of the ball, $ m = 0.5 \text{ kg}$.
Speed of the ball, $ v = 12 \text{ m/s} $.
Angle of incidence (with respect to the plane of the wall), $ \theta = 30^\circ $.
Contact time with the wall, $ \Delta t = 0.25 \text{ s} $.
If the ball’s velocity makes an angle of $30^\circ$ with the plane of the wall, then the angle with the normal to the wall is $60^\circ$. The component of the velocity perpendicular to the wall is:
$ v_\perp = v \cos(60^\circ) = 12 \times \dfrac{1}{2} = 6 \text{ m/s} $
Step 3: Calculate the Change in Momentum
Before collision, the perpendicular velocity component of the ball is $+6 \text{ m/s}$ (taking “toward the wall” as positive). After collision, it is $-6 \text{ m/s}$, because the ball bounces back with the same speed but in the opposite normal direction. Hence, the change in velocity in the perpendicular direction is:
$ \Delta v_\perp = (-6) - (+6) = -12 \text{ m/s} $
(Magnitude of change is $12 \text{ m/s}$ but the sign indicates the reversal of direction.)
The change in momentum in the normal direction is:
$ \Delta p = m \Delta v_\perp = 0.5 \times (-12) = -6 \text{ kg·m/s} $
(Its magnitude is $6 \text{ kg·m/s}$; the negative sign shows that momentum has reversed direction as well.)
Step 4: Compute the Average Force
By the impulse-momentum relation,
$ \text{Average Force} = \dfrac{\Delta p}{\Delta t} = \dfrac{6}{0.25} = 24 \text{ N} $
Step 5: Final Answer
Thus, the average force acting on the wall is $24 \text{ N}$.
Reference Image (Given)
