You have an air-track that has been leveled. You also have two photogates and tw

Question

You have an air-track that has been leveled. You also have two photogates and two air-cars. Each air-car is 13.0cm long. the heavier of the air-cars is placed at rest in the middle of the track and is called the target car. The lighter of the air-cars is given an initial velocity and it travels through one photogate , collideds with the target car, and then returns through the same photogate. this incoming car blocks the photogate for .250 seconds on its way to the collision, then, it blocks the photogate for 1.00 seconds after the collision. the target car travels through the second photogate and blocks it for .500 seconds. A.) what is the ratio of the mass of the target car to the mass of the incoming car? B.) what is the fractional energy loss in the collision?

Explanation / Answer

A)Initial velocity of lighter car = 0.13/0.25 = 0.52 m/s

Final velocity of lighter car = -0.13/1 = -0.13 m/s

Final velocity of the target car = 0.13/0.5 =0.26 m/s

By momentum conservation,

initial momentum = final momentum

m1 ×0.52 + m2×0 = m1×(-0.13) + m2×(0.26)

0.65 m1 = 0.26m2

m2/m1 =0.65/0.26 = 2.5

B) Let mass of lighter be 1 kg and that of target be 2.5kg

initial energy = 0.5×1×0.52×0.52 =0.1352

Final energy = 0.5×1×0.13×0.13 + 0.5×2.5×0.26×0.26

= 0.00845 + 0.0845 =0.09295

Energy lost =initial - final=0.1352 -0.09295=0.04225

Factoral energy lost =0.04225/0.1352 = 0.3125 or 31.25%

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