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

In summary, at the moment of collision, the spring system has not yet exerted any force on B or C. B's velocity therefore becomes v0, and C's velocity becomes v0/2. From this, the velocities of B and C relative to the point halfway between B and C can be calculated.
  • #1
jessij
2
0

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?

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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?
 
Last edited:
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  • #2
jessij said:
v0=a+b+cv
jessij said:
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 vb-vc will be compressing the spring.

Hope this helps.

Regards,
Buzz
 
  • #3
But the compression of the spring is not given ?!
 
  • #4
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
 
  • #5
jessij said:
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?
Buzz Bloom said:
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?
 
  • #6
drvrm said:
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
 
Last edited:
  • #7
Buzz Bloom said:
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.
 
  • #8
drvrm said:
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
 
  • Like
Likes drvrm
  • #9
jessij said:
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.
 

1. How do I find the velocity after a collision if one object is at rest?

If one object is at rest, the final velocity of the other object can be found using the formula v = (m1 * u1) / m2, where v is the final velocity, m1 is the mass of the moving object, u1 is the initial velocity of the moving object, and m2 is the mass of the stationary object.

2. What is the formula for finding the velocity after a collision if both objects are moving?

The formula for finding the velocity after a collision when both objects are moving is v = (m1 * u1 + m2 * u2) / (m1 + m2), where v is the final velocity, m1 and m2 are the masses of the two objects, and u1 and u2 are the initial velocities of the two objects, respectively.

3. Can the velocity after a collision be negative?

Yes, the velocity after a collision can be negative. This indicates that the object is moving in the opposite direction of its initial velocity.

4. How do I calculate the velocity after a collision in a two-dimensional scenario?

In a two-dimensional scenario, the final velocity can be calculated using vector addition. This means that the velocities in the x and y directions must be calculated separately and then combined using the Pythagorean theorem (v = √(vx^2 + vy^2)).

5. Is there a difference in calculating the velocity after a collision for elastic and inelastic collisions?

Yes, there is a difference in calculating the velocity after a collision for elastic and inelastic collisions. In elastic collisions, both kinetic energy and momentum are conserved, while in inelastic collisions, only momentum is conserved. This means that the final velocities will be different for the two types of collisions.

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