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.
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.
v: laboratory reference frame velocity
u: center of mass reference frame velocity
1: block 1
2: block 2
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
v1f = 2*vcm - v1i
v2f = 2*vcm - v2i
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.