Question
9 only
Two railroad cars have masses of m_1 = 1,000 kg and m_2 = 3,000 lb_m. The first car (m_1) is moving at a velocity of 10 m/sec and the second car (m_2) is moving at a velocity of 10 ft/sec. The first car overtakes the second car and couples with it. Calculate the final velocity of the two cars in ft/s. Two cars have the masses of m_1 = 4,000 lb_m and m_2 = 3,000 lb_m and are moving on a sheet of frictionless ice. The first car (m_1) is moving at a velocity of 1.5 ft/sec to the east and the second car (m_2) is moving at a velocity of 1.5 ft/sec to the west. The two cars collide and lock bumpers, staying together. Calculate the final velocity of the two cars.
Explanation / Answer
9. taking east as positive direction then,
v1o = 1.5 ft/s
v2o = - 1.5 ft/s
cars locked together then after colliison, they will move with same velocity v.
Applying momentum conservation for the collison,
m1 v1o + m2 v2o = (m1 + m2) v
(4000 x 1.5) + (3000 x -1.5) = (4000 + 3000) v
v = 0.214 ft/s
so after collision, they are moving with 0.214 ft/s to the east.