Elastic collision between two balls with different masses

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Homework Help Overview

The discussion revolves around an elastic collision between two balls of different masses, where one ball is initially moving and the other is stationary. Participants are examining the conservation of momentum and energy in this context, leading to differing conclusions about the mass ratio of the two balls.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the application of conservation laws, noting discrepancies between momentum and energy calculations. Questions arise regarding the nature of the collision and the definitions of elastic versus inelastic collisions.

Discussion Status

There is an active exploration of the definitions and implications of elastic collisions, with some participants suggesting that the problem may be incorrectly framed. Guidance is offered regarding the distinction between elastic and inelastic collisions, particularly in relation to kinetic energy conservation.

Contextual Notes

Participants note that the problem states the balls collide elastically, yet the resulting behavior suggests a different type of collision. The discussion includes considerations of the coefficient of restitution and the implications of the balls moving together post-collision.

kshitij
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Homework Statement
A ball of mass m1 is moving with velocity 3v. It collides head on elastically with a stationary ball of mass m2 . The velocity of both the balls become v after collision. Then the value of the ratio m2/m1 is
Relevant Equations
Conservation of momentum, Conservation of energy
Since in an elastic collision, both momentum and energy is conserved,
P(initial)=P(final)
m1(3v)=m1v+m2v
m2/m1=2
Which was the given answer but if we use conservation of energy,
K.E(initial)=K.E(final)
1/2*m1*(3v)^2=1/2*m2*v^2+1/2*m1*v^2
m2/m1=8
Why do we get two different answers and why doesn't conservation of energy gives us the right answer here?
 
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kshitij said:
Homework Statement:: A ball of mass m1 is moving with velocity 3v. It collides head on elastically with a stationary ball of mass m2 . The velocity of both the balls become v after collision. Then the value of the ratio m2/m1 is
Relevant Equations:: Conservation of momentum, Conservation of energy

Since in an elastic collision, both momentum and energy is conserved,
P(initial)=P(final)
m1(3v)=m1v+m2v
m2/m1=2
Which was the given answer but if we use conservation of energy,
K.E(initial)=K.E(final)
1/2*m1*(3v)^2=1/2*m2*v^2+1/2*m1*v^2
m2/m1=8
Why do we get two different answers and why doesn't conservation of energy gives us the right answer here?
Why do you think energy is conserved in this collision?
 
PeroK said:
Why do you think energy is conserved in this collision?
A perfectly elastic collision is defined as one in which there is no loss in kinetic energy in the collision. That's what I've been taught.
 
kshitij said:
A perfectly elastic collision is defined as one in which there is no loss in kinetic energy in the collision. That's what I've been taught.
Why do you think this is a perfectly elastic collision?
 
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PeroK said:
Why do you think this is a perfectly elastic collision?
The question states that they collide elastically but if you're saying that they didn't include the word perfectly then I understand.
 
kshitij said:
The question states that they collide elastically but if you're saying that they didn't include the word perfectly then I understand.
Sorry, I missed that. The question is wrong.

Another way to describe an elastic collision is that the "separation speed" is conserved. I.e. the speed at which the objects are moving apart after the collision is the same as the speed at which they are moving together before the collision. With a bit of algebra you can show that is the same as conservation of kinetic energy.

In this case, as the balls stick together after the collision, it cannot be elastic. In fact, when they stick together is is called a totally inelastic collision. And, again with a bit of algebra, you can show that this represents the maximum possible energy loss.

In short, you cannot have conservation of energy in this scenario.

You always have conservation of momentum. So, in this case the solution is simply to use that.
 
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PeroK said:
Sorry, I missed that. The question is wrong.

Another way to describe an elastic collision is that the "separation speed" is conserved. I.e. the speed at which the objects are moving apart after the collision is the same as the speed at which they are moving together before the collision. With a bit of algebra you can show that is the same as conservation of kinetic energy.

In this case, as the balls stick together after the collision, it cannot be elastic. In fact, when they stick together is is called a totally inelastic collision. And, again with a bit of algebra, you can show that this represents the maximum possible energy loss.

In short, you cannot have conservation of energy in this scenario.

You always have conservation of momentum. So, in this case the solution is simply to use that.
I thought the catch was that every collision is elastic but only in perfectly elastic collisions coefficient of restitution(e) is exactly equal to 1 and in perfectly inelastic collision, e=0.

But as you said that they move together with the same velocity after collision so here e=0 so they should write perfectly inelastic instead of elastic in the question but anyway thanks for helping me.
 
kshitij said:
I thought the catch was that every collision is elastic but only in perfectly elastic collisions coefficient of restitution(e) is exactly equal to 1 and in perfectly inelastic collision, e=0.

But as you said that they move together with the same velocity after collision so here e=0 so they should write perfectly inelastic instead of elastic in the question but anyway thanks for helping me.
Elastic is unambigously the same as perfectly elastic: no loss of KE.

Inelastic has more variety, in that there is a range of KE than may be lost, up to the maximum amount where the objects stick together, which is called perfectly inelastic.

https://en.wikipedia.org/wiki/Elastic_collision

https://en.wikipedia.org/wiki/Inelastic_collision
 
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"The velocity of both the balls become v after collision"
It occurred to me that they might mean (or perhaps even wrote) speed rather than velocity. But it doesn't help, merely creating the extra possible solution to the momentum equation m2 =4m1.
 

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