A particle moving horizontally hits a stationary
ring then about point of collision :
(a) angular momentum of the ring is conserved
(b) angular momentum of the particle is conserved
(c) angular momentum of particle-ring system is conserved
(d) all of these
Answers
Answer:
Given: A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end A of the rod with a velocity v0 in a direction perpendicular to AB. The collision is elastic. After the collision the particle comes to rest
Solution:
Let C be centre of mass of the rod of mass M and length L. Consider the rod and the particle together as a system. Let v be velocity of C and ω be angular velocity of the rod just after collision. The linear momentum of the system just before and just after the collision is,
pi=mv0,pf=Mv
where pi,f are the initial and final linear momentum of the system.
There is no external force on the system in x direction. Hence, linear momentum in x-direction is conserved i.e., pi=pf, which gives,
Mv=mv0⟹v0=mMv........(i).
The angular momentum of the system about C just before and just after the collision is,
Li=mv02L,Lf=ωIc=Mω12L2.
There is no external torque on the system about C. Hence, angular momentum of the system about C is conserved i.e., Li=Lf, which gives,
mv0=Mω6L........(ii).
The kinetic energy before and after the collision is,
Ki=21mv02,Kf=21
Answer:
A particle moving horizontally hits a stationary
ring then about point of collision :
(a) angular momentum of the ring is conserved
(b) angular momentum of the particle is conserved
(c) angular momentum of particle-ring system is conserved ✔️✔️
(d) all of these