Question

A 5.50 −kg−kg ball is dropped from a height of 10.0 mm above one end of a uniform bar that pivots at its center. The bar has mass 9.50 kgkg and is 6.80 mm in length. At the other end of the bar sits another 5.30 −kg−kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision.

How high will the other ball go after the collision?

Answer 1

Given,

Mass of ball,mb= 5.50 kg

Height from which ball drop ,H= 10×10^-3 m

Mass of bar,ml= 9.50 kg

Length of bar,L= 6.80×10^-3 m

Mass at other end of bar,m0= 5.30 kg

If we consider the 3 masses (2 balls and bar) as a system ,we can isolate the system and apply principle of conservation of angular momentum as there are no extra unbalance force is acting on this system.

Angular momentum for ball dropped before collision (M1)

M1= mb×Vb×L/2

Vb is the speed of ball.This can be determined by using principle of conservation of energy

∆P.E= ∆K.E

mb.g.H= 1/2 mb.Vb^2

Vb= √2gH=√2×9.8×10×10^-3

=0.443 m/s

Angular momentum of system before collision is

M1= mb.Vb.L/2=8.284×10^-3 Kg m^2/s

After collision momentum transferred to other ball

Ie, M2= mo× Vo×(L/2)

From principle of conservation of angular momentum

M1=M2

Therefore,V0=M1/(mo.L/2)

=8.284×10^-3/0.01802

= 0.4597 m/s

The speed of other ball after collision is Vo and max.height can be determined by principle of conservation of energy

∆P.E=∆K.E

mo.g.h= 1/2 moVo^2

Therefore,h= Vo^2/2g=0.010 m= 10.78 m