An 1820 kg car stopped at a traffic light is struck from the rear by a 910 kg ca

Question

An 1820 kg car stopped at a traffic light is struck from the rear by a 910 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 27.2 m/s before the collision

Conceptualize This kind of collision is easily visualized, and one can predict that after the collision both cars will be moving in the same direction as that of the initially moving car. Because the initially moving car has only half the mass of the stationary car, we expect the final velocity of the cars to be relatively Small Categorize We identify the system of two cars as isolated and apply the impulse approximation during th short time interval of the collision. The phrase "become entangled" tells us to categorize the collision as perfectly inelastic. Analyze The magnitude of the total momentum of the system before the collision is equal to that of the smaller car because the larger car is initially at rest. Evaluate the initial momentum of the pi mivi (910 kg) (27.2 m/s) system: 2.48 x 10 kg m/s pf (m1 m2) vf (2730 kg)vf Evaluate the final momentum of the system: Equate the initial and final momenta and Di 2.48 x 10 kg m/s vf solve for vf m1 m 2730 kg m/s 8.79 Because the final velocity is positive, the direction of the final velocity of the combination is the same as the velocity of the initially moving car as predicted. The speed of the combination is also much lower tha the initial speed of the moving car. MASTER I HINTS GETTING STARTED M STUCK (a) What is the loss of kinetic energy (Ki K0 in the situation described in the example? (b) What if the 910 kg car actually moves backwards with a speed of 1.2 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?

Explanation / Answer

a) Ki = 1/2 (910) (27.2)2 = 336627.2 J
Kf = 1/2 ( 910 + 1820) (8.79)2 = 105465.5 J
Loss of KE = Ki - Kf = 231161.7 J

b) Pi = 2.48x104 kg m/sec
Final mometum of lighter car = - 910 x 1.2 = -1092 kg m/sec
Let velocity of heavier car be V
2.48x104 = 1820 V - 1092
V = 14.23 m/s

c) Final KE of lighter car = 1/2 ( 910) (1.2)2 = 655.2 J
  Final KE of heavier car = 1/2 ( 1820) (14.23)2 = 184268.5 J
Loss of KE = Ki - ( sum of KE of both cars)
   =  336627.2 - 655.2 - 184268.5 = 151703.5 J

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