9 only Two railroad cars have masses of m_1 = 1,000 kg and m_2 = 3,000 lb_m. The

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.

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