Example 9.5 Carry Collision Insurance! An 1810 kg car stopped at a traffic light
ID: 1349816 • Letter: E
Question
Example 9.5 Carry Collision Insurance! An 1810 kg car stopped at a traffic light is struck from the rear by a 905 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 26.9 m/s before the collision, what is the velocity of the entangled cars after the collision? SOLVE IT 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 the 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. Pi = m|v, = (905 kg) (26.9 m/s) = 2.43 × 104 kg·m/s Pt = (m, + m2) vf = (2715 kg) Evaluate the initial momentum of the system: Evaluate the final momentum of the system: Equate the initial and final momenta and solve for ve: 2.43 x 104 kq m/s 2715 kg m/s Pi m1 2 18.95Explanation / Answer
C) if the smaller car is moving backward with 2.1 m/s to the right after the collision
Let the velcoity of the larger car be v m/s to the left
by the conservation of momentum
m1u1 + m2 u2 = m1v1 + m2v2
905*26.9 + 0 = 905*(-2.1) + 1860 * v2
v2 = 14.11 m/s
D) loss of kinetic energy
KE(initial ) = 1/2 m1u1^2 + 1/2 m2u2^2
= 1/2 * 905*26.9^2 + 0 = 327.43kJ
KE(final) = 1/2 m1v1^2 + 1/2 m2v2^2
= 1/2 ( 905*(-2.1)^2 + 1860 * (14.11)^2)
= 186.9kJ
loss of KE = KE(initial) - KE(final)
= 327.43 - 186.9
= 140.5 kJ
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