Originally Posted by **qweiop90**
the below diagram shows the initial ppostion of two bodies A n B respectively of mass (a) and mass ( b) . A is released from rest and it undergoes elastic collision with B.
1) if M(a) = M (b) and the vertical displacement for B after collision is L/2, show that the speed of A is zero instantly after the collision with B and find the value of cos (alpha).
2) if M(b) > M(a), describe the motion of A subsequent to the collision with B.
3) if M(a) > M ( b), what is the ratio of angular speed of A after collision with b/ angular speed of A before collision with B |

Sorry if I have answered it wrongly.

Finding $\displaystyle cos \alpha$, I guess it is 2.5/3 which is 5/6 after simplification.

As after collision, the vertical displacement of B is L/2. Constructing a simple triangle, we would see the hypotenue which always equals to 3L and the adjacent side equatls( 3L-0.5L)=2.5L

Sorry that I don't know the way to prove velocity is zero.

For question 2, due to conservation of momentum, if the momentum of b is greater than that of a, then momentum of a after collision should be negative keeping momentum conserved. Then a should swing backwards.

For question 3, is there something related to mgh= mv^2/2 ?Or maybe related to SHM. I 'm not quite sure.Sorry.

By the way, does the notation M(a)>M(b) means the momentum of a is greater than that of b after collision?