EK 1001 physics #408

[2010premed]

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if m2 is 3 times greater than m2, what is the velocity of m2 after the collision? (Note: Assume a completely elastic collision)
A: 1/2 v

in the diagram of the picture, bob m1 is swung from a height h and allowed to collide with bob m2, which is at rest. The bobs are the same size, but have different masses. The velocity of m1 just before striking m2 is v

 

[Isoprop]

I'm guessing you meant m2 is 3 times greater than m1. There's 2 ways to formally solve this problem. And then there's the MCAT way to solve this problem.

Method 1: solve systems of equations (i use u for initial velocities and v for final)
since u2 = 0 and m2 = 3*m1
m1*u1 + 0 = m1*v1 + (3*m1)*v2 (momentum is conserved)
1/2*m1*u1^2 + 0 = 1/2*m1*v1^2 + 1/2*m2*v2^2 (KE is conserved)

and then do some messy algebra to get to the right answer

Method 2: memorize the solutions to elastic collisions
since u2 = 0 and m2 = 3*m1\

v2 = [u2*(m2-m1) + 2*m1*u1]/[m1 + m2]
= (0 + 2*m1*u1)/(4*m1)
= 2*u2/4
= 1/2*u2

Method 3: MCAT method aka use process of elimination
We know that the kinetic energy of the second mass cannot be greater than the kinetic energy of the system before the collision (work-energy theorem). So we figure out the max KE mass 2 can have.

Total KE bofore collision = Max KE of mass 2 after collision =
= 1/2*m1*u1^2

KE of mass 2 after collision = 1/2*m2*v2^2 = 1/2*(3*m1)*v2^2

Thus...

1/2*(3*m1)*v2^2 < 1/2*m1*u1^2
3*v2^2 < u1^2
v2 < sqrroot(1/3)*u1

So the final velocity must be between 0 and sqrroot(1/3)*u1. sqrroot of (1/3) is a little more than 0.5. So eliminate the rest, guess, and move on.

 

[2010premed]

Thanks for the long explanation.
I'm following the math for the MCAT way, but I'm wondering how'd you know that the final KE of the second mass cannot be more than the initial KE of the system?
since this is an elastic collision, the formula is m1v1 = m1v1 + m2v2 and
1/2 m1v1^2 = 1/2m1fv1f^2 + 1/2 m2v2f^2
so I would think that the combined final KE cannot exceed the initial KE, why is it that just the second mass's ke cannot exceed the initial? Thanks =)

 

[Isoprop]

Thanks for the long explanation.
I'm following the math for the MCAT way, but I'm wondering how'd you know that the final KE of the second mass cannot be more than the initial KE of the system?
since this is an elastic collision, the formula is m1v1 = m1v1 + m2v2 and
1/2 m1v1^2 = 1/2m1fv1f^2 + 1/2 m2v2f^2
so I would think that the combined final KE cannot exceed the initial KE, why is it that just the second mass's ke cannot exceed the initial? Thanks =)

Both are true. The combined KE of the system after the collision is equal to the initial KE of mass 1 before the collision (because it equals the KE of the entire system before the collision). But we also know that the final KE of only mass 2 must be less than the total final KE. So, the KE of mass 2 must also be less than the initial total KE.

Total Kinetic Energy Initial = Total Kinetic Energy Final
KE1(initial) + KE2(initial) = KE1(final) + KE2(final)

We know that KE2(initial) = 0, and KE1(final) must be greater or equal to 0.

So KE1(initial) >= KE2(final)


Thinking of it in another way: the most KE mass 2 can gain is the most work mass 1 exerts on it, which is mass 1's KE.

 

[2010premed]

thanks!

 

[ShySpliceosome]

Sorry to bump an old thread, but this question is driving me crazy...it still doesn't make sense to me, even after reading this whole thread! I initially tried to use momentum conservation to answer it, and ended up getting 1/3v as my answer. Then I used KE conservation, and ended up getting 1/2v, the correct answer. I am really confused when it is appropriate to use momentum conservation vs. KE conservation to solve a problem 😕

 

[ShySpliceosome]

**BUMP**

Anyone? When is it appropriate to use momentum conservation vs. KE conservation to solve a collision problem?

 

[bostoncollegepremed]

here's the way I see it.

MCAT way:

elastic collision means that momentum and KE are conserved so:
since m1<m2 by a factor of 3 I assume that it will be moving back after it bounces with m2 (aka it will have a -v) that means:

1. KE(initial) = KE(final)
1/2(m1)(v1i^2) = 1/2(m2)(v2f^2) + 1/2(m1)(-v1f^2) ... substitute the m1=m and m2=3m
1/2(m)(v1i^2) = 1/2(3m)(v2f^2) + 1/2(m)(-v1f^2) ...eliminate 1/2 and m from both sides
(v1i^2) = 3(v2f^2) + (-v1f^2) (we can see that v2f HAS TO BE less v1i so choice A is gone - if we kept going with the math the way others have done above you would find that basically v2f < sqrt[(v1i^2)/3] ~ equal to .58 but since theres no time i jump to part 2)

2. Momentum (initial)
(m1)(v1i) = (m2)(v2f) + (m1)(-v1f)
same as (m1)(v1i) = (m2)(v2f) - (m1)(v1f)
or (m1)(v1i) + (m1)(v1f)= (m2)(v2f) (so we see that m2 has to have a greater momentum that what m1 started with - now substitute the m1=m and m2=3m
(m1)(v1i) + (m1)(v1f)= (3m2)(v2f) (from this we see that choices C and D simply don't give you a big enough momentum so the only choice is B)

ELIMINATION IS YOUR FRIEND.