A blue car with mass m c = 511 kg is moving east with a speed of v c = 22 m/s an

Question

A blue car with mass mc = 511 kg is moving east with a speed of vc = 22 m/s and collides with a purple truck with mass mt = 1379 kg that is moving south with an unknown speed. The two collide and lock together after the collision moving at an angle of = 51° South of East

2)

What is the magnitude of the initial momentum of the truck?
kg-m/s

3)

What is the speed of the truck before the collision?
m/s

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 magnitude of the TOTAL momentum of the system before and after the collision:

pi = pf

pi > pf

pi < pf

Explanation / Answer

2) Apply conservation of momentum in x-direction

mc*vc = (mc+mt)*v*cos(55)

v*cos(55) = mc*vc/(mc + mt) ---(1)

Apply conservation of momentum in y-direction.

-mt*vt = -(mc+mt)*v*sin(55)

v*sin(55) = mt*vt/(mc + mt) ---(2)

take equation (2)/(1)

tan(55) = mt*vt/(mc*vc)

vt = mc*vc*tan(55)/mt

= 511*22*tan(55)/1379

= 11.64 m/s

Pt = mt*vt

= 1379*11.64

= 16052 kg.m/s

3) vt = 11.64 m/s

4) Pf = sqrt(pc^2 + pt^2)

= sqrt( (mc*vc)^2 + pt^2)

= sqrt((511*22)^2 + 16052^2)

= 19597 kg.m/s

5) vf = pf/(mc + mt)

= 19597/(511 + 1379)

= 10.37 m/s

6) Pi = Pf

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