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, 1.8 kg and 2.5 kg, and the length of the wire is 1.21 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision. I found a, but not b...

Explanation / Answer

You sort of do need the diagram. The critical piece of information is where is the block. I'm going to assume that the block is at the place where the ball started. I'm assuming that the ball started where if it was left alone would be straight down. Part a ===== Begin by realizing this can be done with energy equations. The beginning potential energy = the ending kinetic energy when there is no potential left. Begin It's all potential. That means PE = mgh End It's all kinetic. That means KE = 1/2 m v^2 So PE = KE mgh = 1/2 mv^2 The m's cancel out and you get v = sqr(2gh) g = 9.81 m/s^2 h = 1.34 m v = sqr(2*9.81*1.34) v = 5.13 m/s Part b ===== Part b is a momentum question. momentum before = momentum after That missing piece of information is back to bite us. I can set up part of the problem, but not all. m1*v1 + m2*v2 = m1*v3 + m2*v4 m1 = 1.7 v1 = 5.13 m2 = 2.5 v2 = 0 v3 = ????!!! v4 = Here's the problem. I don't know how to get v4.

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