Collision + observation

Classical Mechanics Level 5

A particle of mass $m$ strikes with a rod of mass $m$ at its one of end as shown in diagram with $e = 0$ , without sticking to the rod. If we observe the rod and particle from the point $A$ on the rod, then:

$(A)$ Angular momentum of rod and particle will be conserved just before and just after collision.

$(B)$ Linear momentum of rod and particle will be conserved.

$(C)$ Observed motion of rod after collision will be pure rotation.

$(D)$ Observed motion of particle will be straight line after collision.

B,D

A,B,D

C

A,C

A,B,C,D

A,B

1 solution

Nishant Rai
May 15, 2015

Please explain how observed motion will be pure rotation.

Ashish R Nair - 5 years, 2 months ago

i was sure abt c but b confused me, nice q buddy :)

A Former Brilliant Member - 4 years, 1 month ago

But the particle will go in pure rotatory motion about its IAOR and that is the CM of the body in this case. @Nishant Rai

Utkarsh Tiwari - 5 years, 10 months ago

But their is nither net external torque nor any other force on rod particle system then why angular and linear momentum is not conserved.

Akshay Sharma - 5 years, 5 months ago

@Akshay Sharma

$e=0$ denotes the velocity of separation will be $0 m/s$ after the collision. But the question is also specifying that the body does not sticks to the rod after collision. This is only possible when the smaller body stays at the point $B$ and the rod goes into rotational motion about its COM. Frame of A is non-inertial.

Nishant Rai - 5 years, 5 months ago

Oh yeah... Thanks for clearing my doubts.

Akshay Sharma - 5 years, 5 months ago

But isn't e = (relative speed of departure)/(relative speed of approach) of the ball and point B. So doesn't the ball move in a straight line? I am not getting this point....

Arunava Das - 3 years, 2 months ago

Collision + observation

1 solution

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