# Collision with 3 particles and inelastic string

1. Oct 14, 2012

### Perpendicular

The problem :

3 identical balls lie on a smooth horizontal surface, the 3rd ball moving with a velocity v while the other 2 are stationary and inter-connected via a taut inelastic string. Ball 3 strikes ball 2 normally making angle = 30 degrees with the string. Energy is conserved. We need to find post-impact velocities and impulsive tension.

My attempt :

By the usual principle that elastic collisions between identical bodies exchange velocities, ball 3 should become stationary while ball 2 acquires V and a part of it gets canceled by tension which then acts on ball 1 and the final result is such that both balls 1 and 2 move with equal final velocities.

However this doesn't match the options given. What exactly am I doing wrong ?

2. Oct 14, 2012

### Staff: Mentor

Do you have a sketch of the setup? Where is ball 1, relative to the collision of ball3/ball2?

Ball 1 and 2 could rotate around their common center of mass afterwards (together with a common movement), depending on the setup. In addition, ball 1 could be involved in the first collision, as the string is inelastic.

3. Oct 14, 2012

### Perpendicular

Crude ASCII art :

B1
....-
.....-
......-
.......-
........-
..........-
B3->.....B2

The angle B3B2B1 is 30 degrees, obviously. The string is represented by the "-"s.

4. Oct 15, 2012

### Staff: Mentor

Well in that case, B2 cannot move with the original velocity of B3 - it has to transmit a force to B1, therefore B2 will be slower.
You can set up an equation system with all conserved quantities as equations and solve that.

5. Oct 15, 2012

### Perpendicular

So conserving angular momentum, linear momentum, and energy should be enough, yeah ?

6. Oct 15, 2012

### Staff: Mentor

You need an additional equation to express that force in the string can act in the direction of the string only.
1 equation for angular momentum
2 equations for linear momentum
1 equation for energy
1 equation for the string
This leaves one degree of freedom, which corresponds to the impact point of B3. If they collide head-on, you can an additional equation for the direction of momentum transfer (here, it is identical with zero vertical velocity of B3 after the collision).