Elastic collision in one dimension

[munchy35]

Homework Statement

Block 1 of mass m1 slides along an x-axis on a frictionless floor with a speed of 2.40 m/s. Then it undergoes a one-dimensional elastic collision with stationary block 2 of mass m2 = 2.00m1. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 = 2.00m2.

what is the speed of block 3? the answer is 1.78.

i just can't figure it out.


more questions that relate to problem...are the speed, kinetic energy, and the momentum of block 3 greater than or less than, or th same as the initial values for block one?

the answers are less, less, greater...

but i don't need help with them because i haven't attempted them yet.

Homework Equations

v1f = m1 - m2 / m1 + m2 * v1i

v2f = 2m1 / m1 + m2 *v1i

The Attempt at a Solution

v1f = v2i = m1 - m2 / m1 + m2 * v1i
= m - 2m/ m + 2m *4
= -m/3m * 4
= -4/3

v2f = v3i = 2m1 / m1 + m2 * v1i
= 2m / m +2m *4
=2/3 *4
=8/3

v3f = m2 - m3 / m2 + m3 * v2i
= 2 - 4 / 2 + 4 * (-4/3)
= -2/6 (-4/3)
=-8/18

clearly this is wrong. i really did try a lot of different things. what am i doing wrong?

 

[rl.bhat]

In this problem collision is elastic. So you have to use the conservation of momentum and energy.
m(v1)i + 0 = m(v1)f + 2mv(2f) ...(1)
m(v1)i^2 + 0 = m(v1)f^2 + 2mv(2f)^2 ...(2)
Rewrite the first equation as
m(vi) - 2m(vf) = m(vf)...(3) Square both sides and subtract from equation (2) and solve for vf.
Repeat the same thing for the second collision.

 

[munchy35]

i'm confused. won't doing that give me the same equations that i gave in the known equations?

 

[rl.bhat]

Yes. You are right.
You have made mistake in v3f.. Formula also wrong. You have to use second formula.
In that v2f becomes v2i for the second collision.

 

[munchy35]

yes thank you! i see it now. i got the answer of 1.78 m/s. =)