Conservation of energy and momentum?

[leena19]

Homework Statement

2 boys of identical masses are standing on 2 identical trolleys A and B(facing each other) which are at rest on a frictionless horizontal surface.The boy on trolley A then throws a ball of mass m horizontally with velocity V with respect to the Earth and the boy o trolley B catches it.If the mass of a trolley with a boy is M,the respective final velocities of trolleys A and B are

1) -mV/M and -mV/M+m
2)-mV/M+m and mV/M+m
3) -mV/M and mV/M+m
4) -mV/M-m and mV/M+m
5) -V and V

Homework Equations

The Attempt at a Solution

momentum before = after
mV = (M+m ) VB - MVA

conservation of energy
1/2mV² = 1/2 MV_A² + 1/2(M+m)V_B²
and VA = \sqrt{[mV^{2}-(M+m)VB^{2}/M]}
which just looks so wrong,i'm too afraid to continue.

Hope someone can help.

Thanx

 

[tiny-tim]

… conservation of energy …

Hi leena19!

First, energy is not conserved in collisions unless the exam question somehow tells you it is!

In this case, the question actually tells you that these are perfectly inelastic collisions (because both masses either start or finish with the same velocity), and so energy certainly isn't conserved.

momentum before = after
mV = (M+m ) VB - MVA

Yes, but that's one equation for all three masses.

You have two collisions here, so do one equation for each collision separately!

 

[leena19]

Thanx!
So momentum before throwing ball = momentum just after throwing the ball ?
0 = mV - MV_A
mV = MV_A
V_A= mV/M ?

Momentum before catching the ball= momentum after catching it ?
mV - MV_A = MV_B+ mV_B
0 = (M+m)V_B
V_B=0 which is obviously wrong
though I'm not sure where I'm going wrong :(

EDIT:
The answer is (3),but I still don't know how to get it .

 

[leena19]

I get the speeds ,but the directions are still wrong?
mV = (m+M)V_B
V_B=mV/(m+M) ?

 

[cepheid]

Thanx!
So momentum before throwing ball = momentum just after throwing the ball ?
0 = mV - MV_A

This equation seems to assume that the ball is thrown in the positive direction, and that therefore the trolley recoils in the negative direction, and the negative sign is put in explicitly. However, if you do this, it means that your V_A has a different meaning from the V_A in the answer key. Yours is the magnitude of the velocity of trolley A only, with no information about the direction (this is contained in the minus sign). Theanswer key assumes V_A indicates both the magnitude and the direction of the velocity (i.e. its sign is intrinsic). You can do this by NOT assuming anything about the direction (sign) of V_A and merely writing:

sum of final momenta = 0

mV + MV_A = 0

In this way, the fact that V_A is negative shows up naturally in the math, in the final answer.

 

[leena19]

OK.
Thank you very much!