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 2.5 kg, and the length of the wire is 1.18 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, 1.7 kg and 2.5 kg, and the length of the wire is 1.18 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

Explanation / Answer

a) USing energy conservation,

ball comes down 1.18 m , so its P.E. get converts into K.E.

so mu^2 /2 = mgh

u = sqrt(2gh) = sqrt(2*9.81*1.18) = 4.81 m/s ........Ans

b) After collision velocity of ball is vB in opposite direction and velocity of block is vBl .

for elastic collision,

velocity of approach = velocity of sepration

u = (vB + vBL)

vBL = 4.81 - vB

Using momentum conservation for this collision,

1.7 x 4.81 + 2.5 x 0 = 1.7 x (-vB) + 2.5 x vBL

2.5vBL - 1.7vB = 8.18

putting the values of vBL in this equation,

2.5 ( 4.81 - vB )   - 1.7vB = 8.18

12.025 - 2.5vB - 1.7vB = 8.18

vB = 0.915 m/s .............Ans

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