A train car with mass m1 = 563.0 kg is moving to the right with a speed of v1 =

Question

A train car with mass m1 = 563.0 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 = 5.5 m/s.
1) What is the initial momentum of the first train car?
kg-m/s
2) What is the mass of the second train car?
kg
3) What is the change in kinetic energy of the two train system during the collision?
J

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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