A railroad flatcar of weight W can roll without friction along astraight horizontal track. Initially, a man of weight w is standing on the car, which is moving to the right with speed v_0. What is the change in velocity of the car if the man runs to the left so that his speed relative to the car is v_rel?
I know that the man and the flatcar are moving inopposite directions. Would the answer be: v_f-v_m= mv/(m_flatcar - m_man)?
The mans change in momentum (-mv) relative to the original moving axis (speed v0) added to the change in car momentum is zero.
W(deltaV)+ w(-v)= 0
deltaV= w/W (v)
Physics - Sharon, Sunday, December 14, 2014 at 7:27pm
Let final velocity be v_1
Conservation of momentum implies that:
==> (m_car + m_man)v_0 = (m_man * v_rel) + (m_car + m_man)v_1
Solve this equation for v_1 - v_0 to get:
v_1 - v_0 = (m_man * v_rel)/(m_man + m_car)
so the change in velocity is:
(w * v_rel)/(w + W)