How to find the velocity after a collision if one or more....

[jessij]

Homework Statement

A body with mass of m kg (A) is sliding horizontally in a frictionless surface at the velocity,V₀.

In front of it are two bodies (B and C) , each with a mass of m kg attached with a spring whose force constant is k.

If the collision between A and B is elastic, how to calculate the velocities of A,B and C after the collision?

I took a,b and c respectively as velocities of A,B and C after the collision

Homework Equations

Using C.O.L.M

mv₀=ma+mb+mc

v₀=a+b+c

Using C.O.M.E

1/2mv₀²=1/2ma²+1/2mb²+1/2mc²

What should i do next?

 

[Buzz Bloom]

v₀=a+b+cv

1/2mv₀²=1/2ma²+12mb²+1/2mc²

Hi jess:

I think the first thing you need to do is to correct the typos.
Next, you need to think about the energy in the spring. The velocity difference v_b-v_c will be compressing the spring.

Hope this helps.

Regards,
Buzz

 

[jessij]

But the compression of the spring is not given ?!

 

[Buzz Bloom]

Hi Jess:

I think it will help you to choose an initial distance d₀ between B and C. When A hits B there will be instantaneous transfer of energy and momentum to B, and and this will cause B to start to move away from A. B's motion towards C will compress the spring. Since the spring has no mass specified for it, the change in the center of gravity of the spring has no energy or momentum associated with it. However, as the spring is compressed, it will absorb energy, and the spring will exert a force on B and C. This force will act to move C away from B. The center of gravity of B and C, located half way between B and C, will have the same momentum as if B and C were attached directly to each other without the spring.

Now the next part is a guess, since I haven't done the math. The distance between B and C will oscillate as the spring is compressed, and subsequently expands and contracts, etc. This part of the problem is the same as if there was no A, but B has an initial velocity relative to C which is initially stationary.

Regards,
Buzz

 

[drvrm]

I took a,b and c respectively as velocities of A,B and C after the collision

Homework Equations

Using C.O.L.M

mv0=ma+mb+mc

v0=a+b+cv

Using C.O.M.E

1/2mv02=1/2ma2+12mb2+1/2mc2

What should i do next?

instantaneous transfer of energy and momentum to B, and and this will cause B to start to move away from A. B's motion towards C will compress the spring. Since the spring has no mass specified for it, the change in the center of gravity of the spring has no energy or momentum associated with it. However, as the spring is compressed, it will absorb energy, and the spring will exert a force on B and C. This force will act to move C away from B. The center of gravity of B and C, located half way between B and C, will have the same momentum as if B and C were attached directly to each other without the spring.

is there any relation ship between the velocities of masses B and C as they are attached through a spring.?
Can one treat spring forces as internal forces between the two bodies-
and the motion seen as external force applied by momentum change of A.
then the conservation of energy and momentum leads to a solution- this is my guess as going in detail about internal motion will lead to a complication!
However if B and c would be joined together by elastic mechanism so m is hitting a mass of 2m and both B and c should move with same speed-is it a possible picture?

 

[Buzz Bloom]

However if B and c would be joined together by elastic mechanism so m is hitting a mass of 2m and both B and c should move with same speed-is it a possible picture?

In the alternate scenario I introduced in which there is no spring, and B and C are therefore considered to have a single 2m mass, then B and C would move with the same speed. However, I think the following is a more likely scenario. At the moment of the collision, the spring has not yet become compressed, and therefore it does not yet exert any force on B or C. Therefore , my guess is that the system would initially behave as if A hit B with no C present. In that case A's velocity would become zero, and B's velocity would become v₀.

ADDED
Therefore at the moment of collision, the B and C system would have a momentum of mv₀. Therefore the center of mass of the B C system would have a velocity of v₀/2. From that the initial velocities of B and C relative to the point halfway between B and C can be calculated.

Now think about how the subsequent dynamics of this system relative to the midpoint between B and C.

Regards,
Buzz

 

[drvrm]

In the alternate scenario I introduced in which there is no spring, and B and C are therefore considered to have a single 2m mass, then B and C would move with the same speed.

i do not think that will be proper as if you calculate the impulse of the force produced by the impact of A with B there will be reaction and C participates in the reaction Force by A on B -should be equal to Force by the two on A in opposite direction - if it is elastic combination then the internal forces communicated by B on C and by C on B are equal and opposite.
Even if one considers a mass of 2m the layers of the masses are elastically connected together and gives the force of action and reaction called internal forces.

 

[Buzz Bloom]

i do not think that will be proper as if you calculate the impulse of the force produced by the impact of A with B there will be reaction and C participates in the reaction Force by A on B -should be equal to Force by the two on A in opposite direction - if it is elastic combination then the internal forces communicated by B on C and by C on B are equal and opposite.

Hi drvrm:

I apologize that I am unable to parse this sentence. When I discussed the scenario with no spring, I did not intend that to mean B a nd C had an elastic connection as if they remained separate balls just touching. I intended for B and C to be physically bound together.

Regards,
Buzz

 

[drvrm]

If the collision between A and B is elastic, how to calculate the velocities of A,B and C after the collision?

well on the above proposition you can get numbers for velocities and let us see those values.