1) When cars are equipped with flexible bumpers, they will bounce off each other
ID: 1350506 • Letter: 1
Question
1) When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1850 kg car traveling to the right at 1.40 m/s collides with a 1450 kg car going to the left at 1.10 m/s . Measurements show that the heavier car's speed just after the collision was 0.260 m/s in its original direction. You can ignore any road friction during the collision.
Part A
What was the speed of the lighter car just after the collision?
Part B
Calculate the change in the combined kinetic energy of the two-car system during this collision.
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2) James and Ramon are standing 20.0 m apart on the slippery surface of a frozen pond. Ramon has mass 54.0 kg and James has mass 96.0 kg . Midway between the two men a mug of their favorite beverage sits on the ice. They pull on the ends of a light rope that is stretched between them. Ramon pulls on the rope to give himself a speed of 0.70 m/s .
Part A
What is James's speed?
Express your answer using two significant figures.
Explanation / Answer
apply from the law of conservation of momentum
m1u1 -m2u2 = m1v1 + m2v2
m2v2 = (1850 * 1.4) -(1450 * 1.1) -(1850 * 0.260)
m2v2 = 514
v2 = 514/1450
v2 = 0.354 m/s
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Initial KE of car1 = 0.5 mv^2 = 0.5* 1850 * 1.4*1.4 = 1813 J
INitial KE f car 2 = 0.5* m2v2^2 = 0.5* 1450 * 1.1*1.1 = 877.25 J
total KEi = 1813+ 877.25 = 2690.25 Joules
total final KE = (0.5* 1850 * 0.26^2) + (0.5*1450 * 0.354^2) = 153.38 J
change in KE = 2690.25 - 153.38 = 2536.87 JOules
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KE = PE
0.5 mv^2 - 0.5 mu^2 = Gm1m2(1/r1)
0.5 * mu^2 = ( 6.67 e -11 * 54 * 96*2)/20 + (0.5 * 96* 0.7* 0.7)
u^2 = 23.51 * 2/54
u = 0.933 m/s
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