A ball falls straight down onto a wedge that is sitting on frictionless ice. The

Question

A ball falls straight down onto a wedge that is sitting on frictionless ice. The ball has a mass of 2 kg, and the wedge has a mass of 4 kg. The ball is moving a speed of v = 4 m/s when it strikes the wedge, which is initially at rest (see the figure). Assuming that the collision in instantaneous and perfectly elastic. What is the velocity of the wedge after the collision in m/s? -2 squareroot 6/3 -4 squareroot 6/5 -2 squareroot 6 -2 squareroot 6/5 -3 squareroot 6/2 What is the velocity of the ball after the collision in m/s? 7 squareroot 6/3 squareroot 6 5 squareroot 6/3 4 squareroot 6/3 2 squareroot 6/3

Explanation / Answer

As far as I can see, the ball will bounce off in an initially horizontal direction because of the 45° angle and because the collision is instantaneous (so the wedge doesn't move while ball bounces off it).

Suppose that immediately after the collision, the wedge moves with velocity w and the ball moves with initial velocity u.

Note that w is in the -x direction and u is in the +x direction.

From conservation of momentum in the x direction:
2u + 4w = 0 => u = -2w (w will be negative)

From conservation of energy (as it's an elastic collision):
½(2)(4)² = ½(2)u² + ½(4)w² => 16 = u² + 2w² => 16 = 4w² + 2w² = 6w²

=> w = (26)/3  (w will be negative)

u = (46)/3

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