A 42,000kg train is moving at 2.8 m/s collides with a second train with mass 60,

Question

A 42,000kg train is moving at 2.8 m/s collides with a second train with mass 60,000 kg moving at 0.90 m/s in the opposite direction on same track. They both come together. A. What is their speed after collision B. If collision time was 8.2x10^-3 s, find the average force exerted on the 60,000 train A 42,000kg train is moving at 2.8 m/s collides with a second train with mass 60,000 kg moving at 0.90 m/s in the opposite direction on same track. They both come together. A. What is their speed after collision B. If collision time was 8.2x10^-3 s, find the average force exerted on the 60,000 train A. What is their speed after collision B. If collision time was 8.2x10^-3 s, find the average force exerted on the 60,000 train

Explanation / Answer

As it is a elastic collision,speed of seperation = speed of approach.v2-v1/u1-u2=1

v2-v1/2.8-(-0.9)=1

that is v2-v1=3.7--------(Eqn 1)

also m(2.8)+m(-0.9)=m(v1+v2). (conservation of momentum)
solving we get,v1+v2=1.9---------(eqn 2)

from equations 1 and 2 ,
v1=0.9 m/s
v2=2.8 m/s

F =ma =m v/t = 60000* 2.8 *8.2 x10^-3 =1377.6 N

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