A ball is attached to one end of a wire, the other end being fastened to the cei

Question

A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.7 kg and 2.50 kg, and the length of the wire is 1.11 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

Explanation / Answer

First, use conservation of energy to find the velocity at the bottom before collision.
mgh = 1/2mv2

v = 2*g*h = 2*9.8*1.11m = 4.66m/s

For the second part, you need to use conservation of momentum and energy

m1v1i + m2v2i = m1v1f + m2v2f

The block is stopped to begin with so the second part of the equation goes away but the final velocities are both unknown.

v2f = m1(v1i + v1f)/m2

Plug that into conservation of energy

1/2m1v12 = 1/2m1v1f2 + 1/2mv2f2

solve for v1f and plug into conservation of momentum equation

and you get v1f = m1-m2/m1 + m2 * v1i = -0.888m/s the ball

v2f = 2m1/m1 + m2 * v1i = the block = 3.77m/s

Hope that helps

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