Question

Elastic Collision in One Dimension HW?

Block 1, of mass m_1, moves across a frictionless surface with speed u_i. It collides elastically with block 2, of mass m_2, which is at rest (v_i=0). After the collision, block 1 moves with speed u_f, while block 2 moves with speed v_f. Assume that m_1 > m_2, so that after the collision, the two objects move off in the direction of the first object before the collision.

What is the final speed u_f of block 1?

Express u_f in terms of m_1, m_2, and u_i

What is the final speed v_f of block 2?

Express v_f in terms of m_1, m_2, and u_i.

Answers

I don't like the nomenclature you gave, so I will make up my own.

In fact, my strategy is necessary such that you DO NOT USE u FOR LABORATORY REFERENCE FRAME velocities.

Nomenclature:

m: mass

v: laboratory reference frame velocity

u: center of mass reference frame velocity

1: block 1

2: block 2

i: initial

f: final

Since nothing is special about the LRF (other than that we use it to take measurements) we progress to the COMRF. First find the center of mass velocity in LRF:

vcm = (m1*v1i + m2*v2i)/(m1 + m2)

Init velocities in COMRF:

u1i = v1i - vcm

u2i = v2i - vcm

Interesting trick: to avoid the big quadratic formula, simply invert the velocities in the center of mass reference frame. This is the only way to have a collision occur, and conserve both energy and momentum.

u1f = - u1i

u2f = -u2i

Translate back to LRF:

v1f = u1f + vcm

v2f = u2f + vcm

Hence:

v1f = 2*vcm - v1i

v2f = 2*vcm - v2i

Substitute vcm:

v1f = 2*(m1*v1i + m2*v2i)/(m1 + m2) - v1i

v2f = 2*(m1*v1i + m2*v2i)/(m1 + m2) - v2i

Simplify for special case of v2i=0:

v1f = v1i*(m1 - m2)/(m1 + m2)

v2f = 2*m1*v1i/(m1 + m2)

BECAUSE you asked about speed, and you specified m1>m2, we can answer that as well. Just take magnitudes. Not really anything to do here, because both values will end up positive.

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