A blue car with mass m c = 473 kg is moving east with a speed of v c = 22 m/s an
ID: 1286084 • Letter: A
Question
A blue car with mass mc = 473 kg is moving east with a speed of vc = 22 m/s and collides with a purple truck with mass mt = 1274 kg that is moving south with a speed of vt = 13 m/s . The two collide and lock together after the collision.
1)
What is the magnitude of the initial momentum of the car?
kg-m/s
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2)
What is the magnitude of the initial momentum of the truck?
kg-m/s
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3)
What is the angle that the car-truck combination travel after the collision? (give your answer as an angle South of East)
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4)
What is the magnitude of the momentum of the car-truck combination immediately after the collision?
kg-m/s
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5)
What is the speed of the car-truck combination immediately after the collision?
m/s
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6)
Compare the initial and final kinetic energy of the total system before and after the collision:
KEi = KEf
KEi > KEf
KEi < KEf
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Explanation / Answer
(1) magnitude of the initial momentum of the car = mass of car * velocity of car = (473*22) = 10406 kg m/s
(2)magnitude of the initial momentum of the truck = mass of truck * velocity of truck =(1274*13) = 16562 kg m/s
(3) After collision (car+truck) will move in 32.11 degree south of east.
(4) after collision magnitude of combined will remain same (from momentum conservation) but direction will change.
(5) from momentum conservation,
(wt of car + wt of truck) * v = (473*22) south + (1274*13) east
v= 5.95 south + 9.48 east so speed will be sqrt(5.95^2 + 9.48^2)=11.19 m/s
(6) K.E= (1/2) *m * v^2
after calculation we find KEi > KEf
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