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 2.4 kg, and the length of the wire is 1.37 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

Explanation / Answer

(a) v = (2gh) = (2 * 9.8m/s² * 1.37m) = 5.18m/s

(b) conserve momentum: 1.7kg * 5.18m/s + 0 = 1.6kg * u + 2.4kg * v
where u, v are the post-impact velocities of the ball, block respectively. So
8.806 = 1.7u + 2.4v

For an elastic, linear collision, we know that the relative velocity of approach = relative velocity of separation, or 5.18m/s = v - u
v = u + 5.18 plug that into the momentum equation
7.63 = 1.7u + 2.4(u + 5.18) = 4.1u + 12.432
u = -4.802 / 4.1 = -1.17 m/s (b)

Check: initial KE = ½ * 1.7kg * (5.18m/s)² =22.80 J
final KE = ½(1.7kg * (-0.1.17m/s)² + 2.4kg * (-0.1.7m/s + 5.18m/s)²) =
energy was conserved.

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