1D collision, varying masses but same initial velocity

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  • #1
mncyapntsi
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Homework Statement:
Trying to solve this problem to study for an exam next week!

There are two masses : 2m and m and they collide with v and -v. I need to find the final velocities (it is also a perfectly elastic collision).
Relevant Equations:
KEi = KEf gives (1/2)(m1)(vi1)^2 + (1/2)(m2)(vi2)^2 = (1/2)(m1)(vf1)^2 + (1/2)(m2)(vf2)^2
pi = pf gives (m1)(vi1) + (m2)(vi2) = (m1)(vf1) + (m2)(vf2)
m1 = 2m
m2 = m
I know I need to look at the conversation of momentum, as well as the conservation of kinetic energy. However I get stuck with my equations. Any help would be greatly appreciated! I've already got (don't know where I am going wrong):

(v)^2 + (1/2)(m)(v)^2 = (vf1)^2 + (1/2)(m)(vf2)^2
(3/2)v^2 = (vf1)^2 + (1/2)(vf2)^2

3mv = 2m(vf1) + m(vf2)
3v = (vf1) + (vf2)
(vf1) = 3v - vf2

Plugging into KE conservation equation gives me:
(3/2)v^2 = (3v - vf2)^2 + (1/2)(vf2)^2
(3/2)v^2 = 9v^2 -6v(vf2) + (vf2)^2 + (1/2)(vf2)^2
- 6v (vf2) = (7.5)v^2 + (3/2)(vf2)^2

And this is where I get stuck – I indeed have two variables, however since there is a v multiplied by vf2 I don't know how to continue. I can't spot my error anywhere... Would anyone have any advice? Thank you!
 

Answers and Replies

  • #2
kuruman
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3mv = 2m(vf1) + m(vf2)
How do you figure 3mv on the left? Read the problem. What are the initial velocities? The energy conservation equation and the momentum conservation equation should give you a system of two equations and two unknowns. Look at this

3mv = 2m(vf1) + m(vf2)
3v = (vf1) + (vf2)

I can see canceling the masses, but what happened to the factor of 2 in the top equation? If you don't make any careless mistakes like this, the solution should come out.
 
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  • #3
mncyapntsi
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How do you figure 3mv on the left? Read the problem. What are the initial velocities? The energy conservation equation and the momentum conservation equation should give you a system of two equations and two unknowns. Look at this

3mv = 2m(vf1) + m(vf2)
3v = (vf1) + (vf2)

I can see canceling the masses, but what happened to the factor of 2 in the top equation? If you don't make any careless mistakes like this, the solution should come out.
Thanks for catching that! So with that fixed I've got:

(v)^2 + (1/2)(m)(v)^2 = (vf1)^2 + (1/2)(m)(vf2)^2
(3/2)v^2 = (vf1)^2 + (1/2)(vf2)^2

3mv = 2m(vf1) + m(vf2)
3v = 2(vf1) + (vf2)
vf2 = 3v - 2(vf1)

Plugging into KE conservation equation gives me:
(3/2)v^2 = (vf1)^2 + (1/2)(3v - 2(vf1))^2
(3/2)v^2 = (vf1)^2 + (1/2)(9v^2 - 12v(vf1) + 4(vf1))
(3/2)v^2 = (vf1)^2 + (9/2)v^2 - 6v(vf1) + 2(vf1)
-6v(vf1) = (6/2)v^2 + (vf1)^2 + 2(vf1)

And this is a similar kind of problem as before... Is there some other mistake? I have factors of v(vf1), v^2, vf1^2, AND vf1 which doesn't seem right...
Thanks again!
 
  • #4
Frabjous
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Focus on the first comment of kuruman.
 
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  • #5
mncyapntsi
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Focus on the first comment of kuruman.
Ah yep! forgot about that:
I assumed that the initial momentum was 3mv because the momentum of the 2m mass is 2mv and the momentum of the m mass is mv, so the total initial momentum would be 2mv+mv = 3mv. Where am I going wrong?
Thanks!
 
  • #6
Frabjous
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Velocity is a vector so it has a direction.
 
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  • #7
kuruman
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##\dots## so the total initial momentum would be 2mv+mv = 3mv. Where am I going wrong?
There are two masses : 2m and m and they collide with v and -v.
And 2mv + m(-v) = 3mv, right?
 
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  • #8
mncyapntsi
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Velocity is a vector so it has a direction.
Ah! That's it...

(v)^2 + (1/2)(m)(v)^2 = (vf1)^2 + (1/2)(m)(vf2)^2
(3/2)v^2 = (vf1)^2 + (1/2)(vf2)^2

mv = 2m(vf1) + m(vf2)
v = 2(vf1) + (vf2)

Plugging into KE conservation equation gives me:
(3/2)v^2 = (vf1)^2 + (1/2)(vf2))^2
(3/2)(2(vf1) + (vf2))^2 = (vf1)^2 + (1/2)(vf2))^2
(3/2)(4(vf1)^2 + 8(vf1)(vf2) + (vf2)^2) = (vf1)^2 + (1/2)(vf2))^2
6(vf1)^2 + 12(vf1)(vf2) + (3/2)(vf2)^2 = (vf1)^2 + (1/2)(vf2))^2
5(vf1)^2 + 12(vf1)(vf2) = - (vf2))^2

I am so sorry – I run into the same problem... What am I doing wrong now ://
 
  • #9
Frabjous
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You know v so you do not want to substitute for it.
 
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  • #10
kuruman
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Yes, your goal is to find vf1 and vf2 in terms of v.
 
  • #11
mncyapntsi
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You know v so you do not want to substitute for it.
Oh my goodness - I think I overthought this problem and lost track of what I was effectively doing. Thank you for your patience and help! So if I substitute for vf2 instead...

(v)^2 + (1/2)(m)(v)^2 = (vf1)^2 + (1/2)(m)(vf2)^2
(3/2)v^2 = (vf1)^2 + (1/2)(vf2)^2

mv = 2m(vf1) + m(vf2)
v = 2(vf1) + (vf2)
(vf2) = v - 2(vf1)

Plugging into KE conservation equation gives me:
(3/2)v^2 = (vf1)^2 + (1/2)(v - 2(vf1)))^2
(3/2)v^2 = (vf1)^2 + (1/2)(v^2 - 4v(vf1) + 4(vf1)^2)
(3/2)v^2 = (vf1)^2 + (1/2)(v^2) - 2v(vf1) + 2(vf1)^2
v^2 = - 2v(vf1) + 3(vf1)^2
v^2 + 2v(vf1) = 3(vf1)^2
From here, how would I find vf1?
Sorry!
 
  • #12
Frabjous
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Quadratic equation.
 
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  • #13
Henryk
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your first equation is wrong. You are adding squared velocity term to kinetic energy term! ##v^2 + \frac 1 2 mv^2##
 
  • #14
mncyapntsi
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your first equation is wrong. You are adding squared velocity term to kinetic energy term! ##v^2 + \frac 1 2 mv^2##
Oh that's a typo (sorry I don't know proper Latex yet), I meant to cancel all the m's, but did so in the next step:
(v)^2 + (1/2)(v)^2 = (vf1)^2 + (1/2)(vf2)^2
(3/2)v^2 = (vf1)^2 + (1/2)(vf2)^2
 
  • #15
mncyapntsi
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Quadratic equation.
Right!
When I do that I get either (v1f) = v, or (v1f) = -(1/3)v. Does that second result sound reasonable?
 
  • #16
kuruman
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Right!
When I do that I get either (v1f) = v, or (v1f) = -(1/3)v. Does that second result sound reasonable?
It sounds reasonable, but only one of the two is the answer to this question for v1f. Which one and why? Also, you still have to find v2f.
 

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