# Question regarding elastic collisions and the conservation of momentum / energy [closed]

I've been working on this question for a few hours and I cannot quite convince myself that my answer is correct. I've read the relevant section in my textbook, but I guess the lightbulb just isn't clicking on. The question (which I'll post the full text of below), asks for final velocity of a puck in three body collision in which two of the bodies are initially at rest. I was thinking that, since I only need final velocity, I could just use the conservation of linear momentum in the relevant plane (since this equation has a term for v) and solve it out for v. Other solution I've seen use both linear momentum AND conservation of energy equations. Which way is correct? See picture for question and attempted solution. Any help is greatly appreciated.

Question: Three identical round pucks A, B, and C are placed on frictionless surface as shown. The puck A is shot with initial velocity V0 along the symmetry line. The pucks B and C are initially touching each other. The pucks undergo elastic collision. This question asks that I find the final velocity of Puck A.

EDIT: My attempt is included in the image below. The problem I'm currently having is that I'm not sure if I can treat Pucks A & B as a single object for the purposes of this solution.

• Balls $B$ and $C$ go off along the lines joining their centres to the centre of ball $A$ and thus you can find the angles by using geometry. – Farcher Jul 24 '16 at 21:00
• But I'm asked for the final velocity of puck A. – PhysicsStudent Jul 24 '16 at 21:02
• So use conservation of momentum and conservation of kinetic energy which will give you the equations you need to find the final speeds of balls $B$ and $C$ (which will be the same by symmetry) and ball $A$. – Farcher Jul 24 '16 at 21:05
• I've been kind of confused about how those two work together. When I solve KE conservation equation, which variable am I solving it for? v final for A, B, C on the right side? – PhysicsStudent Jul 24 '16 at 21:10
• In your latest solution attempt, take a very close look at your very first equation. – Inquisitive Jul 24 '16 at 22:32

You will agree that the velocity of A depends on the elasticity of the collision. Now, momentum is conserved regardless of the elasticity of the collision. Therefore, you cannot expect momentum conservation alone to determine the outgoing velocity of A.

In detail, the total outgoing momentum depends on two unknowns, i.e. A's velocity and B and C's common velocity. You need an extra equation to uniquely determine your solution, and that is what energy conservation does.

• OH! Thank you! That makes sense. In other words, when momentum is conserved, if I were to only use conservation of momentum equation I would get no change? So I must also apply conservation of energy? – PhysicsStudent Jul 24 '16 at 20:07
• By using conservation of momentum alone you would end up with a relation between $v^A_x$ and $v^{BC}_x$. Just try it. – Arek' Fu Jul 24 '16 at 20:09
• Do I need to use the conservation of momentum equation at all? – PhysicsStudent Jul 24 '16 at 20:09
• Yes, of course you do. Ask yourself how many unknown variables you have in the final state; you will need that many equations to solve the problem. – Arek' Fu Jul 24 '16 at 20:10
• If I'm understanding this: I have three objects (A, B, C). I want to find out the final velocity of A. Since momentum is conserved I can use it to find a value (v final) for A, and B,C (theirs will be the same, right?). I then plug those values, v, into the conservation of energy equation and solve for final velocity of puck A? – PhysicsStudent Jul 24 '16 at 20:19

Assume all masses are the same and we have a perfectly symmetrical collision. It seems a logical assumption doesn't it? Then, try to understand the following conservation of energy equation. Use it to solve for $v_{af}$:

$$\frac{1}{2}mv_{ai}^2 = \frac{1}{2}mv_{af}^2 + \frac{1}{2}mv_{b}^2 +\frac{1}{2}mv_{c}^2$$

Once you've done that, go back and solve the problem using conservation of momentum. That's trickier because you have to remember that you can have conservation of total momentum, you can have conservation of momentum in the $x$-direction, and you can have conservation of momentum in the $y$-direction. You will also have to handle the angles in order to solve it with conservation of momentum. Good luck!

• I'm thinking to just conserve x momentum. I'll post my updated attempt in an edi tnow. Thanks for your help! – PhysicsStudent Jul 24 '16 at 22:09