A blue car with mass mc = 490 kg is moving east with a speed of vc = 22 m/s and

Question

A blue car with mass mc = 490 kg is moving east with a speed of vc = 22 m/s and collides with a purple truck with mass mt = 1243 kg that is moving south with a speed of vt = 12 m/s . The two collide and lock together after the collision.
1)What is the angle that the car-truck combination travel after the collision? (give your answer as an angle South of East)
2)What is the magnitude of the momentum of the car-truck combination immediately after the collision?
3)What is the speed of the car-truck combination immediately after the collision?

Explanation / Answer

SOLUTION: (a) Mass of the blue car, mc = 490 kg speed of the blue car, vc=22 m/s (to wards East) Mass of the purple car, mt=1243 kg Speed of the purple car, vt=12 m/s (to wards south) let, speed of the car-truck after the colision takes place, vct Momentum of the blue car along East, pc=mcvc Momentum of the purple car along south, pt=mtvt After the collision takes place, let, be the angle (south of East) that the car-truck combination travel. Momentum of the blue car, pc'=(mc+mt)vct cos   (East) Momentum of the purple car, pt'=(mc+mt)vct sin    (south) Apply conservation of momentum, (equating east,south components seperately) pc=pc'   (East components) mcvc=(mc+mt)vct cos           ......(1) pt=pt' mtvt=(mc+mt)vct sin              ......(2) divide eqn(2) by (1) mtvt /mcvc =(mc+mt)vct sin /(mc+mt)vct cos     tan=mtvt /mcvc       =(1243 kg)(12 m/s)/(490 kg)(22 m/s)       =54.14o _______________________________________________________________ _______________________________________________________________ (b) Magnitude of the momentum of the car-truck combination Magnitude of the momentum of the car-truck combination immediately after the collision,    P= ([(mc+mt)vct cos]2+[(mc+mt)vct sin]2)      =([(mc+mt)vct from equation (1), mcvc=(mc+mt)vct cos vct =(mcvc) /(mc+mt) cos        =(490 kg)(22 m/s)/(490 kg+1243 kg)cos54.14o        =10.618 m/s thus,     P=(mc+mt)vct       =(490 kg+1243 kg)(10.618 m/s)       =18401.97 kgm/s _____________________________________________________________ _____________________________________________________________ _____________________________________________________________ _____________________________________________________________ (c) As from part (b), Speed of the car-truck combination immediately after the collision is, vct= 10.618 m/s

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