# Elastic collision in one dimension

#### munchy35

1. 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.

2. Homework Equations

v1f = m1 - m2 / m1 + m2 * v1i

v2f = 2m1 / m1 + m2 *v1i

3. 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?

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#### rl.bhat

Homework Helper
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

Homework Helper
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. =)

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