Saved Bookmarks
| 1. |
Two smooth spheres. A and B. having equal radii, lie on a horizontal table. A is of mass m and B is of mass 3m. The spheres are projected towards each other with velocity vector 5hati+2hatj and 2hati-hatj, respectively, and when they collide the lige joining their centres is parallel to the vector hati. If the coefficient of restitution between A and B is 1//3, find the velocities after impact and the loss in kinetic energy caused by the collision. Find also the magnitude of the impulses that act at the instant of impact. |
Answer» Solution :The velocity component of `A` and `B` perpendicular to the line connecting centres are unchaned by the impact. It is because no FORCE acts on `A` and `B` in a direction perpendicular to the line connecting the centre  Applying conservation of inear momentum nd the law of restitution we have in `x` direction `5m+(3m)(2)="mu"+3MV`............i `and 1/3(5-2)=v-u`.........ii Solving these EQUATIONS, we have `u=2` and `v=3` The velocities of `A` and`B` after the impact are `2hati+2hatj` and `3hati-hatj` respectively. Before the kinetic ENERGY of `A` is `1/2m(5^(2)+2^(2))=29m//2` and of `B` is `1/2(3m)(2^(2)+1^(2))=15m//2` After the impact the kinetic energy of `A` is `1/2m(2^(2)+2^(2))=4m` and that of `B` is `1/2(3m)(3^(2)+1^(2))=15m` Therefore, the loss in `KE` at impact is `29/2m+15/2m-4m-15m=3m` To find the value of `J` we CONSIDER the change in momemtum along `hati` for one sphere only. For sphere `B, J=3m(3-2)` or `J=3m,` |
|