1 2 3 4 5 6 7 A blue car with mass m c = 479 kg is moving east with a speed of v
ID: 1374056 • Letter: 1
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A blue car with mass mc = 479 kg is moving east with a speed of vc = 23 m/s and collides with a purple truck with mass mt = 1286 kg that is moving south with a speed of vt = 10 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
2)
What is the magnitude of the initial momentum of the truck?
kg-m/s
3)
What is the angle that the car-truck combination travel after the collision? (give your answer as an angle South of East)
A blue car with mass mc = 479 kg is moving east with a speed of vc = 23 m/s and collides with a purple truck with mass mt = 1286 kg that is moving south with a speed of vt = 10 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 2) What is the magnitude of the initial momentum of the truck? kg-m/s 3) What is the angle that the car-truck combination travel after the collision? (give your answer as an angle South of East) 4) What is the magnitude of the momentum of the car-truck combination immediately after the collision? kg-m/s 5) What is the speed of the car-truck combination immediately after the collision? m/s 6) Compare the initial and final kinetic energy of the total system before and after the collision:Explanation / Answer
Given that,
mass of the car = Mc= 479 Kg ; Velocity =Vc= 23 m/s
mass of the truck = Mt = 1286 Kg ; Velocity = Vt = 10 m/s
(1)The magnitude of initial momentum of the car P(ic) will be :
P(ic) = Mc x Vc = 479 x 23 = 11017 Kg - m/s
(2) Magnitude of initial momentum P(it) of the truck is ;
P(it) = Mt x Vt = 1286 x 10 = 12860 Kg.m/s
(3)Angle theta, after collision will be given by
Tan (theta) = Momentum of truck / mometum of car
(theta) = Tan -1 (12860 / 11017 ) = 49.41 degrees
direction will be, North of east
(4)The magnitude of car truck combination after collision P(final) will be given by :
P(final) = sqrt [ (Pic)2 + (Pit)2 ] = sqrt [ (11017)2 + (12860)2 ] = 16892.83 Kg.m/s
(5)Final velocity of the car truck combination will be V(f), and given by
V(f) = P(final) / (Mc + Mt) = 16892.83 / 1765 = 9.57 m/s
(6)Ke initial of the system was
KE(i) = 1/2 (Mc Vc2 + Mt Vt2) = 1/2 (479 x 23 x 23 + 1286 x 10 x 10) = 190995.5 Joules
KE(f) = 1/2 (Mc +Mt) (Vf)2 = 1/2 (479 + 1286) (9.57)2 = 80819.35 Joules
Hence KE(i) > KE(f)
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