Use the exact values you enter in previous answer(s) to make later calculation(s

Question

Use the exact values you enter in previous answer(s) to make later calculation(s).

A 1130 kg car traveling initially with a speed of 27.60 m/s in an easterly direction crashes into the back of a 9 550 kg truck moving in the same direction at 20.0 m/s. The velocity of the car right after the collision is 18.0 m/s to the east.

(a) What is the velocity of the truck right after the collision? (Give your answer to at least four significant figures.)

(b) What is the change in mechanical energy of the car-truck system in the collision?
J

Explanation / Answer

consider the conservation of the linear momentum

P = P0

mc v'c + mT v'T = mc vc + mT vT --->   v'T = ( mc vc + mT vT - mc v'c ) / mT   

v'T = ( 1130 kg * ( 27.60 m/s ) + 9550 Kg * ( 20.00 m/s ) - 1130 kg * ( 18.00 m/s ) ) / ( 9550 Kg )

v'T = ( 1130 kg * ( 27.60 m/s ) + 9550 Kg * ( 20.00 m/s ) - 1130 kg * ( 18.00 m/s ) ) / ( 9550 Kg )

v'T = 21.14 m/s

the direction is to EAST

(b)

the initial kinetic energy is,,, K0 = ( mc vc^{2} ) / 2 + ( mT vT^{2} ) / 2

then,,,, K0 = ( 1130 Kg * ( 27.60 m/s )^{2} ) / 2 + ( 9550 Kg * ( 20.00 m/s )^{2} ) / 2 = 2.340x10^{6} J

the final kinetic energy is,,, Kf = ( mc vc'^{2} ) / 2 + ( mT vT'^{2} ) / 2

then,,,, K0 = ( 1130 Kg * ( 18.00 m/s )^{2} ) / 2 + ( 9550 Kg * ( 21.14 m/s )^{2} ) / 2 = 2.317x10^{6} J

Hence ,,, K = Kf - K0 = 2.317x10^{6} J - 2.340x10^{6} J = -2.300x10^{4} J

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