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, 1.6 kg and 2.4 kg, and the length of the wire is 1.29 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

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, 1.6 kg and 2.4 kg, and the length of the wire is 1.29 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

Explanation / Answer

The ball starts with potential energy only and ends with kinetic energy only.
PE = m*g*h
KE = (1/2)m*v^2

KE = PE
(1.2)m*v*2 = m*g*h
v = SQRT[2*g*h]

h = 1.29 m
g = 9.81 m/s^2

v = SQRT[2*9.81*1.29] = 5.0305 m/s

So the velocity of the ball before impact is 5.03 m/s

For the collision we must have momentum and energy conservation.
u = velocity of ball after collision
w = velocity of block after collision
m = mass of ball = 1.6 kg
M = mass of block = 2.4 kg

Momentum before collision equals the momentum after. m*v = u*m + w*M
Kinetic energy before equal kinetic energy after: (1/2)m*v^2 = (1/2)m*u^2 + (1/2)M*w^2

From momentum: v = u + (M/m)w which gives u = v - (M/m)w
From energy: v^2 = u^2 + (M/m)w^2

v^2 = [v - (M/m)w]^2 + (M/m)w^2
v^2 = v^2 - 2*v*w(M/m) + (M/m)^2w^2 + (M/m)w^2
0 = -2*v*w(M/m) + (M/m)^2w^2 + (M/m)w^2
0 = -2*v + (M/m)w + w = -2*v + (1 + (M/m)]w
w = 2*v/[1 + (M/m)]
w = 4..024 m/s and u = v - (M/m)w = -1.006

Check momentum:
Pbefore = 1.6*5.0305 = 8.0488
Pafter = 1.6*(-1.006) + 2.4*4.024 = 8.048
So the sum of the momenum after the collision equals the momentum before.

Check energy.
KEbefore = (1/2)1.6*(5.0305)^2 = 20.244
KE after = (1/2)*1.6*(-1.006)^2 + (1/2)*2.4*(4.024)^2 = 20.240
So the KE before and after are equal.

So the velocity of the ball after impact is -1.006 m/s and the velocity of the block after impact is 4.024 m/s.

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