A train car with mass m1 = 663.0 kg is moving to the right with a speed of v1 =

Question

A train car with mass m1 = 663.0 kg is moving to the right with a speed of v1 = 8.8 m/s and collides with a second train car. The two cars latch together during the collision and then move off to the right at vf = 5.6 m/s. 1) What is the initial momentum of the first train car? 2) What is the mass of the second train car? 3) What is the change in kinetic energy of the two train system during the collision? 4)Now the same two cars are involved in a second collision. The first car is again moving to the right with a speed of v1 = 8.8 m/s and collides with the second train car that is now moving to the left with a velocity v2 = -6.2 m/s before the collision. The two cars latch together at impact. What is the final velocity of the two-car system? (A positive velocity means the two train cars move to the right

Explanation / Answer

1) intital momentun of 1st train is P1=m1*v1

P=563*8.8= 4954.4 N.s

2) conservation of linear momentum

m1v1+m2v2= (m1+m2)v

563*8.8+0=(563+m2)*5.5

563+m2=900.8

m2= 900.8-563= 337.8 kg

the mass of 2nd train car is 337.8 kg

3) initial KE= 0.5*563*8.8^2=21799.36 J

final KE= 0.5*900.8*5.5^2=13624.6 J

chance in KE= initialKE - final KE

KE= 21799.36-13624.6

KE=8174.76 J

4) m1v1-m2v2=(m1+m2)*v

563*8.8- 337.8*6.2=900.8*v

2860.04/900.8=v

v=3.175 m/sec

5)after collision the momentum decreases so Pinitial> Pfinal

Pinitial=563*8.8= 4954.4 N.s

Pfinal= 563*5.5= 3096.5 N.s [case1]

Pfinal= 563*3.175= 1787.525 N.s [case 2]

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