A thin uniform rod of mass M and is suspended vertically from fixed e...
1. The momentum of the (ball + rod) system just before and after the collision relative to the ground frame will be conserved.
Explanation: The collision between the ball and rod is an internal force. There are no external horizontal forces acting on the system, so the horizontal component of the momentum will be conserved. Before the collision, the momentum of the system is mv0 in the horizontal direction, and after the collision, the momentum will remain the same. The vertical component of the momentum is zero as the rod is initially at rest and the ball is moving horizontally. So, the total momentum of the system will be conserved.
3. The angular momentum of the (ball + rod) system about the fixed point A just before and after the collision will be conserved.
Explanation: There are no external torques acting on the system about the fixed point A. The only forces acting on the system are gravity and the tension at the fixed point A, both of which pass through point A and hence do not produce any torque about point A. Therefore, angular momentum about point A will be conserved. Just before the collision, the angular momentum is m(v0)(l), and after the collision, the angular momentum will remain the same.
A thin uniform rod of mass M and is suspended vertically from fixed e...
Explanation:
Conservation of Momentum:
- In this collision, momentum of the ball-rod system just before and after collision relative to ground frame will be conserved.
- Before collision, the momentum of the system is given as mv0 downwards.
- After collision, the ball and rod move together as one body with a common velocity v downwards.
- Therefore, the momentum of the system after collision is (M+m)v downwards.
- As momentum is conserved, we can say that mv0 = (M+m)v, which gives us the common velocity of the ball and rod after collision.
Non-Conservation of Kinetic Energy:
- Kinetic energy of the ball-rod system just before and after collision relative to ground frame will not be conserved.
- Before collision, the kinetic energy of the system is given by (1/2)mv0^2.
- After collision, the ball and rod move together as one body with a common velocity v.
- Therefore, the kinetic energy of the system after collision is given by (1/2)(M+m)v^2.
- As v is less than v0, we can say that the kinetic energy after collision is less than the kinetic energy before collision.
- Therefore, kinetic energy is not conserved in this collision.
Conservation of Angular Momentum:
- Angular momentum of the ball-rod system about the fixed point A just before and after collision will be conserved.
- Before collision, the ball has no angular momentum about point A as it is moving along the line joining A and B.
- After collision, the ball and rod move together as one body with a common velocity v.
- As they rotate about point A, they acquire angular momentum.
- However, the angular momentum of the system about point A remains constant as there are no external torques acting on the system.
- Therefore, angular momentum about point A is conserved.
Non-Conservation of Angular Momentum about C.M.:
- Angular momentum of the ball-rod system about the C.M. of the rod just before and after collision will not be conserved.
- Before collision, the C.M. of the rod is at rest and has no angular momentum.
- After collision, the ball and rod move together as one body with a common velocity v.
- As they rotate about the C.M. of the rod, they acquire angular momentum.
- However, the angular momentum of the system about the C.M. of the rod is not conserved as the rod rotates about its own axis and its C.M. moves downwards.
- Therefore, angular momentum about the C.M. of the rod is not conserved.
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