A steel ball of mass 0.790 kg is fastened to a cord that is 99.0 cm long and fix

Question

A steel ball of mass

0.790 kg is fastened to a cord that is 99.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal, as shown in the figure. At the bottom of its path, the ball strikes a 4.80 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.

A steel ball of mass 0.790 kg is fastened to a cord that is 99.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal, as shown in the figure. At the bottom of its path, the ball strikes a 4.80 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.

Explanation / Answer

Potential energy of the ball = mgh = .790 x 9.8 x 0.99 = 7.66 J

KE = PE

KE = .5 x m x v^2

.5 x .790 x v^2 = 7.66

v = 4.4 m/s

initial momentum = 4.4 x .790 = 3.5 kg m/s

ball has velocity v towards left and block u towards right

final momentum = 4.8u - 0.79v = 3.5

v = 6u - 4.43

Initial KE = 7.66

final KE = 0.395v^2 + 2.4u^2 = 7.99

0.395 [ 6u - 4.43]^2 + 2.4u^2 - 7.99 = 0

u = 1.27 m/s

v = 3.22 m/s

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