Elastic collision heavy particle problem

Do you think he should repeat all that here?In summary, the conversation discusses a problem involving a heavy particle colliding elastically with a very light particle initially at rest. It is shown that if the mass of the heavy particle is significantly greater than that of the light particle, the velocity of the heavy particle remains practically unchanged while the light particle acquires a velocity that is twice that of the heavy particle's initial velocity. The conversation also delves into the concept of conservation of momentum and kinetic energy in elastic collisions, and provides hints and equations to help solve the problem.
  • #1
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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  • #2
What about the kinetic energy of the system? What do you know about the kinetic energy in elastic collisions?
 
  • #3
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
 
  • #4
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.
 
  • #5
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.
 
  • #6
You have an error just before you form your final equation. Check it again.
 
  • #7
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.
 
  • #8
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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  • #9
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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1. What is an elastic collision in a heavy particle problem?

An elastic collision is a type of collision where both kinetic energy and momentum are conserved. In a heavy particle problem, it refers to a collision between two particles where one is significantly larger and more massive than the other.

2. How is momentum conserved in an elastic collision between heavy particles?

In an elastic collision between heavy particles, momentum is conserved because the total momentum of the system before and after the collision remains the same. This means that the sum of the momenta of the two particles before the collision is equal to the sum of the momenta after the collision.

3. What is the difference between an elastic and inelastic collision in a heavy particle problem?

The main difference between an elastic and inelastic collision in a heavy particle problem is that in an elastic collision, both kinetic energy and momentum are conserved, while in an inelastic collision, only momentum is conserved. In an inelastic collision, some of the kinetic energy is converted into other forms of energy, such as heat or sound.

4. How do you calculate the final velocities of heavy particles in an elastic collision?

To calculate the final velocities of heavy particles in an elastic collision, you can use the conservation of momentum and kinetic energy equations. These equations take into account the masses and initial velocities of the particles, as well as the angle of collision.

5. What factors can affect the outcome of an elastic collision between heavy particles?

The outcome of an elastic collision between heavy particles can be affected by factors such as the masses and initial velocities of the particles, the angle of collision, and any external forces acting on the system. The elasticity of the particles and the nature of the interaction between them can also play a role in the outcome of the collision.

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