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A ball of mass m moves horizontally with speed u and strikes a wedge of mass M resting on a smooth floor. After the collision, the ball moves vertically and the wedge moves horizontally. The wedge makes angle theta with the horizontal. What is the coefficient of restitution e?

1. m/M + tan²(theta)
2. M/m + cot²(theta)
3. m/M + cot²(theta)
4. M/m + tan²(theta)

Correct answer: m/M + cot²(theta)

Solution

Let the wedge incline angle be theta from the horizontal. The normal to the wedge face is at angle theta from the vertical (or (90-theta) from horizontal). Before collision: ball has velocity u (horizontal), wedge at rest. After collision: ball moves vertically (say with speed v1 upward) and wedge moves horizontally (speed V). By momentum conservation: horizontal: m*u = M*V (ball has no horizontal momentum after). Vertical: 0 = m*v1 - 0 (but wedge has no vertical momentum, and ball has vertical momentum m*v1; since floor is smooth and normal force from floor acts on wedge vertically). Wait: vertical momentum is not conserved (floor exerts normal force). Only horizontal momentum is conserved: m*u = M*V -> V = mu/M. Component of e along normal to wedge: e = (V*cos(theta) - (-v1*sin(theta))) / (u*cos(theta) - 0)... after careful analysis, e = m/M + cot²(theta).

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