# 1D collision, varying masses but same initial velocity

• mncyapntsi
In summary: The velocity cannot be negative. So (vf1) = 0.In summary, the conversation discusses the conservation of momentum and kinetic energy. The equations provided show a system of two equations and two unknowns for the initial and final velocities. After fixing a few careless errors, the solution shows that one of the final velocities, (vf1), is equal to the initial velocity, v, while the other final velocity, (vf2), is equal to zero.
mncyapntsi
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!

mncyapntsi said:
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.

mncyapntsi and topsquark
kuruman said:
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!

Focus on the first comment of kuruman.

topsquark and mncyapntsi
Frabjous said:
Focus on the first comment of kuruman.
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!

Velocity is a vector so it has a direction.

topsquark and mncyapntsi
mncyapntsi said:
##\dots## so the total initial momentum would be 2mv+mv = 3mv. Where am I going wrong?
mncyapntsi said:
There are two masses : 2m and m and they collide with v and -v.
And 2mv + m(-v) = 3mv, right?

topsquark and mncyapntsi
Frabjous said:
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 ://

You know v so you do not want to substitute for it.

topsquark and mncyapntsi
Yes, your goal is to find vf1 and vf2 in terms of v.

mncyapntsi
Frabjous said:
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!

topsquark and mncyapntsi
your first equation is wrong. You are adding squared velocity term to kinetic energy term! ##v^2 + \frac 1 2 mv^2##

mncyapntsi
Henryk said:
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

Frabjous said:
Right!
When I do that I get either (v1f) = v, or (v1f) = -(1/3)v. Does that second result sound reasonable?

mncyapntsi said:
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.

topsquark

## 1. What is a 1D collision with varying masses but same initial velocity?

A 1D collision with varying masses but same initial velocity refers to a scenario where two objects with different masses collide in a straight line (1D) with the same initial velocity. This means that both objects are moving at the same speed before the collision occurs.

## 2. How is momentum conserved in a 1D collision with varying masses but same initial velocity?

In a 1D collision with varying masses but same initial velocity, momentum is conserved. This means that the total momentum of the two objects before the collision is equal to the total momentum after the collision. This is because in a closed system, the total momentum remains constant.

## 3. What is the equation for calculating the final velocities in a 1D collision with varying masses but same initial velocity?

The equation for calculating the final velocities in a 1D collision with varying masses but same initial velocity is:
vf1 = (m1 - m2) / (m1 + m2) * vi
vf2 = 2m1 / (m1 + m2) * vi
where vf1 and vf2 are the final velocities of the two objects, m1 and m2 are their masses, and vi is the initial velocity.

## 4. How does the mass of the objects affect the final velocities in a 1D collision with varying masses but same initial velocity?

The mass of the objects affects the final velocities in a 1D collision with varying masses but same initial velocity. The object with a larger mass will have a smaller final velocity, while the object with a smaller mass will have a larger final velocity. This is because the larger mass will experience less change in velocity compared to the smaller mass.

## 5. Can the final velocities in a 1D collision with varying masses but same initial velocity be negative?

Yes, the final velocities in a 1D collision with varying masses but same initial velocity can be negative. This occurs when one of the objects changes direction after the collision. The negative velocity indicates that the object is moving in the opposite direction compared to its initial velocity.

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