A thin uniform bar lies on a frictionless horizonta surface and is free to move in any way on th surface. Its mass is 0.16 kg and length `sqrt3`meters. Two particless, each of mass 0.08 kg, are moving on the same surface and towards the bar in a direction perpendicular to the bar, one with a velocity of 10 m/s, and other with 6m/s as shown in fig. The first particle strikes the bar at point A and the other at point B. Points A and B are at a distane of 0.5m from the centre of the bar. The particles strike the bar at the same instant of time and stick to the bar on collision. Calculate the loss of the kinetic energy of the system in the above collision process.

Answer

B

Explanation

Initial Kinetic Energy
`=(1)/(2)m_1 v_1^2 +(1)/(2)m_2 v_2^2 + (1)/(2)MV^2`
`=(1)/(2) 0.08xx10^2 +(1)/(2) 0.08xx6^2 _ 0 = 5.44 J …(i)`

Applying law of conservation of linear momentum during
collision
`m_1 xx v_1 + m_2xxv_2 = (M + m_1 +m_2) V_c`
where `v_c` is the velocity on it after collision
`0.08xx10+0.08xx6 = (0.16 + 0.08 + 0.08)v_c`
`rArr V_c = 4m/s`
`:.` Translational kinetic energy after collision
`=(1)/(2)(M + m_1+m_2)V_c^2 = 2.56J ...(ii)`
Applying conservation of angular momentum of the bar
and two particle system about the centre of the bar.
Since extrnal torque is zero, the inital angular momentum
is equal to final angular momentum.
Initial angular momentum
`=m_1v_1xx x - m_2 v_2x`
`=0.08xx10x0.5 - 0.08xx6xx0.5`
`=0.4 - 0.24 = 0.16kgm^2s^(-1)` (in clockwise direction)`
Final angular momentum `= I omega`
`=[(Ml^2)/(12) + m_1x^2+m_2x^2]omega`
`=[((0.16)(sqrt3)^2/(12 + 0.08xx(0.5)^2 + (0.08)(0.5)^2]omega`
`=0.08 omega`
`:. 0.08 omega = 0.16 rArr omega =2rad/s ...(iii)`
The rotational kinetic energy
=Translational K.E. + Rotational K.E.
= 2.56+0.16 = 2.72 J
The change in K.E. = Initial K.E.- Final K.E.
=5.44 - 2.72 = 2.72J