Two blocks move along a linear path on a nearly frictionless air track. One bloc

Question

Two blocks move along a linear path on a nearly frictionless air track. One block, of mass 0.103 kg, initially moves to the right at a speed of 5.40 m/s, while the second block, of mass 0.206 kg, is initially to the left of the first block and moving to the right at 6.80 m/s. Find the final velocities of the blocks, assuming the collision is elastic.

velocity of the 0.103 kg block =

velocity of the 0.206 kg block =

Explanation / Answer

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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