Classical mech. - inelastic collision

[areverseay]

Homework Statement

A particle A of mass 2m travelling with speed u undergoes an inelastic collision with a
stationary particle B of mass 10m. After the collision, A travels with a speed vA at 90 degrees to its original direction of motion and B travels with speed vB at an angle θ = arcsin(3/5) to the original direction of motion of A. Express vA and vB in terms of u; what fraction of kinetic energy is lost as a result of the collision?

Relevant Equations

p = mv
KE = (p^2)/2m

vA = 3u/4 and vB = u/4, and 1/8 KE is lost. I can't get to these answers however: for the first part, I got to u = vA + 3vB using conservation of momentum, and the fact that particle B is at an angle, hence I would think its momentum should be 10mvBsin(arcsin(3/5)). Doing the same for A with angle sin90 gives 2mu = 2mvA + 6mvB, hence the equation I first listed, but I don't know how to isolate vA and vB in terms of u. For the second part, even with vA and vB in terms of u, I can't get to 1/8 lost: using KE = (p^2)/2m and summing the KE for particles A and B, I get (54mu^2)/80.

 

[haruspex]

2mu = 2mvA + 6mvB

How do you figure that? u and vA are at right angles.

 

[haruspex]

sorry for 3 questions

Per forum guidelines, please separate into three threads,

 

[areverseay]

Per forum guidelines, please separate into three threads,

Thanks for this, I've made this post just the first question.

In retrospect it doesn't really make much sense, since as you say u and vA are perpendicular, but I'm not sure how else to get anything for vA - if I were to use cos90, the particle would be stationary, which isn't true, and I don't know how else to get an expression.

 

[haruspex]

Thanks for this, I've made this post just the first question.

In retrospect it doesn't really make much sense, since as you say u and vA are perpendicular, but I'm not sure how else to get anything for vA - if I were to use cos90, the particle would be stationary, which isn't true, and I don't know how else to get an expression.

No, it would not be stationary, it just wouldn’t have a velocity component in its original direction. That is perfectly possible in general.
The component of vB in that direction is the cos of the angle, not the sine.

 

[erobz]

Thanks for this, I've made this post just the first question.

In retrospect it doesn't really make much sense, since as you say u and vA are perpendicular, but I'm not sure how else to get anything for vA - if I were to use cos90, the particle would be stationary, which isn't true, and I don't know how else to get an expression.

You must write conservation of momentum for each velocity component before and immediately after collision. It applies to both horizontal and vertical directions. The mass $2m$ leaves the impact with only a vertical momentum component (w.r.t. to the coordinates I chose), which balances the vertical momentum component of the $10m$ mass leaving the impact. Apply this concept to the horizontal direction too. You can get the velocities of each mass in terms of $u$, which can then be used to determine the energy loss.

 

[areverseay]

You must write conservation of momentum for each velocity component before and immediately after collision. It applies to both horizontal and vertical directions. The mass $2m$ leaves the impact with only a vertical momentum component (w.r.t. to the coordinates I chose), which balances the vertical momentum component of the $10m$ mass leaving the impact. Apply this concept to the horizontal direction too. You can get the velocities of each mass in terms of $u$, which can then be used to determine the energy loss.

Both of these replies are very helpful, but I'm still stuck on the horizontal component. So, the vertical component of the momentum of B is the same as the original momentum of A, since there is no vertical component in the new momentum of A, as cos90 = 0 / they are perpendicular. But how do you get the horizontal component? Since there is none in the original momentum, I guess the horizontal component of both new A and B sums to 0, but how do I link that back to u?

 

[haruspex]

the horizontal component of both new A and B sums to 0

Yes

how do I link that back to u?

Write the equations and that may become apparent.

 

[erobz]

Both of these replies are very helpful, but I'm still stuck on the horizontal component. So, the vertical component of the momentum of B is the same as the original momentum of A,

No.

Please write conservation of momentum for the vertical direction, $p_{before} ~\hat{\text{j}} = p_{after} ~\hat{\text{j}}$

What do you get for this expression? Is there any momentum in the vertical direction before the collision? What are the vertical components of each masses momentum after the collision?

 

[haruspex]

No.

Are you making an assumption about which is vertical and which is horizontal in the OP's view?

 

[erobz]

I don't know. Where is the diagram that labels these vectors? We are talking about vertical momentum. I take that to mean up and down on the screen, and left and right as horizontal. I said before that is with respect to the coordinates I chose. Of course the OP's problem could be oriented in anyway.

To the OP: orient it however you want, just give an indication as to the directions you chose? I diagram would be helpful in either case here. I'm not trying to confuse you, but we better first agree on what is up/down, left/right ...

 

[areverseay]

I don't know. Where is the diagram that labels these vectors? We are talking about vertical momentum. I take that to mean up and down on the screen, and left and right as horizontal. I said before that is with respect to the coordinates I chose. Of course the OP's problem could be oriented in anyway.

To the OP: orient it however you want, just give an indication as to the directions you chose? I diagram would be helpful in either case here.

Yes, I think we had different views of which orientation everything was. But, I've figured out the correct answers for all of the parts, so it's fine. Thanks for both of your help, explicitly writing the equations for conservation of momentum did indeed help a lot!

 

[erobz]

Yes, I think we had different views of which orientation everything was. But, I've figured out the correct answers for all of the parts, so it's fine. Thanks for both of your help, explicitly writing the equations for conservation of momentum did indeed help a lot!

That's good news...time for bed!