A blue car with mass m c = 403 kg is moving east with a speed of v c = 21 m/s an

Question

A blue car with mass mc = 403 kg is moving east with a speed of vc = 21 m/s and collides with a purple truck with mass mt = 1233 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?

2) What is the magnitude of the initial momentum of the truck?

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?

5) What is the speed of the car-truck combination immediately after the collision?

6) Compare the initial and final kinetic energy of the total system before and after the collision:

KEi = KEf

KEi > KEf

KEi < KEf

Explanation / Answer

1) momentum = mass x velocity

momentum of car = 403 x 21 = 8463 kg.m/s .....along east

2) momentum of truck = 1233 x 13 =16029 kg.m/s   ....along south

3) using momentum conservation,

initial momentum = final momentum

8463i + 16029(-j) = (403 + 1233)v

v = 5.17i - 9.80 j

@ = tan^-1 ( 9.80 / 5.17) = 62.19 deg

4)
magnitude of momentum = sqrt(8463^2 + 16029^2) = 18125.98 kg.m/s

5) speed = sqrt ( 5.17^2 + 9.80^2) =11.08 m/s

6) KEi = 403 x21^2 /2   + 1233x13^2 /2 = 193050 J

KEf = (403 + 1233) x 11.08^2 = 100422.92 J

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