An 1640 kg car stopped at a traffic light is struck from the rear by a 820 kg ca

Question

An 1640 kg car stopped at a traffic light is struck from the rear by a 820 kg car. The two cars become entangled, moving along the same path as that of the originally moving car. If the smaller car were moving at 25.6 m/s before the collision, what is the velocity of the entangled cars after the collision?

(a) What is the loss of kinetic energy (Ki - Kf) in the situation described in the example?
kJ

(b) What if the 820 kg car actually moves backwards with a speed of 1.9 m/s right after the collision instead of having a perfectly inelastic collision. What is the velocity of the heavier car immediately after the collision? Use the same convention for positive direction as defined in the example.
m/s

(c) What is the loss of kinetic energy in this case?
kJ

Explanation / Answer

momentum conservation ,

1640*(0)+820*(25.6)=(1640+820)* Vf

Vf= 8.533 m/s,

loss in K.E= 1/2*820 *(25.6)2 -1/2*(820+1640)*(8.5333)2 = 177.04 KJ

again,

momentum conservation will give you,

1640*(0)+820*(25.6)=820*(-1.9)+1640*(Vf)

Vf of the heavier car = 13.75 m/s

loss in K.E= 1/2*820 *(25.6)2 - [1/2*820*(1.9)2 +1/2*1640*(13.75)2 ] =112.186 KJ

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