Elastic collision heavy particle problem

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Homework Help Overview

The discussion revolves around an elastic collision problem involving a heavy particle (mass m1) colliding with a very light particle (mass m2) that is initially at rest. The goal is to demonstrate that if m1 is significantly greater than m2, the velocity of m1 remains nearly unchanged, while m2 acquires a velocity of v2' = 2v1.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the conservation of momentum and kinetic energy principles in elastic collisions. Some express confusion over the manipulation of equations and the implications of mass ratios on the velocities post-collision.

Discussion Status

There are multiple hints and suggestions provided regarding the use of conservation laws to analyze the collision. Participants are actively questioning assumptions and clarifying their understanding of the relationships between the variables involved.

Contextual Notes

Some participants note errors in their calculations and express confusion about the implications of the mass ratio on the final velocities. There is an emphasis on the need to correctly apply the conservation equations without making incorrect substitutions.

Abu

Homework Statement


Suppose a heavy particle (mass m1) has an elastic head-on collision with a very light particle of mass m2 initially at rest. Show that if m1>>m2, the velocity of the projectile (m1) is practically unchanged, whereas the target particle (m2) acquires a velocity v2' = 2v1

Homework Equations


m1v1+ m2v2 = m1v1'+m2v2'
v1-v2 = v2'-v1'

The Attempt at a Solution


When I first tried to solve it I thought I had it correct.
Basically I did something really simple like this:
v1-v2 = v2'-v1'
v1 - 0 = v2'-v1'
v1 = v2'-v1'

Then here I just went to rearrange v1' with v1, forgetting that they are not the same values. Then I got the 2v1 = v2' but I did it incorrectly, obviously. I don't know where to go from here though... Thank you.
 
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What about the kinetic energy of the system? What do you know about the kinetic energy in elastic collisions?
 
DoItForYourself said:
What about the kinetic energy of the system? What do you know about the kinetic energy in elastic collisions?
Oh right... kinetic energy in elastic collisions is conserved. I know the equation of it, if I type it though it will look really messy. Here it is though:

1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2
 
Hint 1 : Initially use the momentum conservation principle to show that the velocity of 1 is practically unchanged.

Hint 2 : Take v1' from the first equation and put it in the kinetic energy conservation equation.
 
DoItForYourself said:
Hint 1 : Initially use the momentum conservation principle to show that the velocity of 1 is practically unchanged.

Hint 2 : Take v1' from the first equation and put it in the kinetic energy conservation equation.
Okay so what I did was solve for the velocity of the large mass after the collision. What I found was that the velocity of the large mass after the collision is it's velocity before the collision minus the momentum of the small mass after the collision. It looked like this
m1v1 +m2v2 = m1v1' + m2v2'
m1v1 + 0 = m1v1' + m2v2'
v1' = v1 - m2v2'

Does this prove the first hint? Because won't the momentum of the smaller mass after the collision still be a lot smaller than the speed of the larger mass before the collision? I feel like I didn't get it right.
 
You have an error just before you form your final equation. Check it again.
 
Abu said:

Homework Statement


Suppose a heavy particle (mass m1) has an elastic head-on collision with a very light particle of mass m2 initially at rest. Show that if m1>>m2, the velocity of the projectile (m1) is practically unchanged, whereas the target particle (m2) acquires a velocity v2' = 2v1

Homework Equations


m1v1+ m2v2 = m1v1'+m2v2'
v1-v2 = v2'-v1'

The Attempt at a Solution


When I first tried to solve it I thought I had it correct.
Basically I did something really simple like this:
v1-v2 = v2'-v1'
v1 - 0 = v2'-v1'
v1 = v2'-v1'

Then here I just went to rearrange v1' with v1, forgetting that they are not the same values. Then I got the 2v1 = v2' but I did it incorrectly, obviously. I don't know where to go from here though... Thank you.
You can not simply replace v1' with v1. Use also the first equation m1v1+ m2v2 = m1v1'+m2v2' --->m1v1=m1v1'+m2v2' to get v1' and v2'.
After that, you can see what happens when m1>>m2.
 
DoItForYourself said:
What about the kinetic energy of the system? What do you know about the kinetic energy in elastic collisions?
Abu got the second equation in the OP by using conservation both of momentum and kinetic energy.
 
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ehild said:
You can not simply replace v1' with v1. Use also the first equation m1v1+ m2v2 = m1v1'+m2v2' --->m1v1=m1v1'+m2v2' to get v1' and v2'.
After that, you can see what happens when m1>>m2.
I'm getting confused, I'm really sorry.

So i did what you said and I got
v1' = m1v1 - m2v2'/m1
and
v2' = m1v1 - m1v1'/m2

But I don't see how that can tell me what happens when m1>>m2.
 
  • #10
ehild said:
Abu got the second equation in the OP by using conservation both of momentum and kinetic energy.

Oh yes, you are right!

But still, I think he must show how he reached that point (that he started with the kinetic energy conservation law).
 
  • #11
Abu,

please check your equations again. You did an error when dividing by m1 and m2.
 
  • #12
Abu said:
I'm getting confused, I'm really sorry.

So i did what you said and I got
v1' = m1v1 - m2v2'/m1
and
v2' = m1v1 - m1v1'/m2

But I don't see how that can tell me what happens when m1>>m2.
Use the other equation, too, that you got in the OP : v1 = v2'-v1', together with the equation m1v1 = m1v1'+m2v2'.
v1 is given. Write v1' and v2' in terms of v1 and the masses.
 
  • #13
DoItForYourself said:
Oh yes, you are right!

But still, I think he must show how he reached that point (that he started with the kinetic energy conservation law).
You are right, but certainly he was taught how to get that equation in case of elastic collision.
 
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