Consider the situation shown in above the figure. Two blocks of masses m1 and m2 are moving on the same straight line on a frictionless horizontal table. The block m2, which is ahead of m1, is going with a speed V2 smaller than the speed V1 of m1. A spring is attached to the rear end of m2. Since V1> V2, the block m1 will touch the rear of the spring at some instant, say t. Then onwards, the velocity of the left end of the spring will be equal to the velocity of m1 (as they are in contact). The velocity of the right end of the spring will be same as that of m2 (as they are in contact). Since m1 moves faster than m2, the length of the spring will decrease. The spring will be compressed. As it is compressed, it pushes back both the blocks with forces k.x where x is the compression and k, the spring constant. This force is in the direction of the velocity of m2, hence m2 will accelerate. However, this is opposite to the velocity of m1 and so m1 will
decelerate. The velocity of the front block A (which was slower initially) will gradually increase, and the velocity of the rear block B (which was faster initially) will gradually decrease. The spring will continue to become more and more compressed as long as the rear block B is faster than the front block A. There will be an instant t1+ Δt when the two blocks will have equal velocities. At this instant, both the ends of the spring will move with the same velocity and no further compression will take place. This corresponds to the maximum compression of the spring. Thus, “the spring-compression is maximum when the two blocks attain equal velocities“.
As the spring is compressed it tries to regain its shape in a way pushes the two block so block A is accelerated and B is decelerated the elastic potential energy stored in spring is converted back into kinetic energy of the block. In the whole process as there is no external force acting on the system the initial and final momentum is conserved. And inial and final kinetic energy is same as thee is no loss in the spring elastic energy it is wholly converted to the kinetic energy.
The same model can be used for the collision of rigid bodies if there is no plastic deformation then the collision is elastic and kinetic energy before and after the collision remains the same.
COEFFICIENT OF RESTITUTION (e)
It is a ratio of the velocity of approach of two colliding bodies to the velocity of separation.
- PERFECTLY ELASTIC COLLISION E=1
- INELASTIC COLLISION 0<E<1
- PLASTIC COLLISIOM E=0
- SUPERELASTIC COLLISION E>1(THOUGH IT SEEMS IT VIOLATES ENERGY PRINCIPLE BY IT IS IN ACCORDANCE WITH EINSTEIN’S THEORY OF RELATIVITY)
POINTS TO BE REMEMBERED
- IN PERFECTLY ELASTIC COLLISION THE KINETIC ENERGY AFTER AND BEFORE THE COLLISION IS SAME BUT IT IS INCORRECT TO SAY THAT IT IS CONSTANT .
- IN COLLISION BETWEEN TWO BODIES AS SYSTEM IN ABSENCE OF EXTERNAL FORCE ,MOMENTUM IS CONSERVED.
- IN CASE OF INCLINED COLLISION ALWAYS WRITE THE EQATION OF MOMENTUM ALONG THE LINE PERPENDICULAR TO THE LINE OF IMPACT AS THE NORMAL FORCE, DURING COLLISION ACTS AS AN EXTERNAL FORCE.
- AND WRITE THE EQUATION OF COEFFICIENT OF RESTITUTION ALONG THE COMMON NORMAL i.e. ALONG THE LINE OF IMPACT
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