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A 2,000 kg car traveling north at a velocity of 45m/s strikes 2,200 Kg car traveling...

A 2,000 kg car traveling north at a velocity of 45m/s strikes 2,200 Kg car traveling west at 50 m/s. The two vehicles strike each other and stick together. As a single combined mass they slide across the pavement which has a coefficient of kinetic friction 0.07. What is the displacement from the point of impact to the final resting place of the cars? (How long does it take for them to come to rest?)







A 200 Kg ball is moving west at 20m/s. It elastically strikes a stationary 300Kg ball. Ignore any effect of friction, what is the final velocity of the balls.

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Answer #1

1) In X direction

=> - 2200 * 50   = (2000 + 2200) * vx

=> vx = - 26.19 m/sec

In Y direction

=> 2000 * 45 = (2000 + 2200) * vy

=> vy = 21.428 m/sec

=> Resultant velocity = 33.83 m/sec

=>   displacement from the point of impact to the final = (33.83)2/(2 * 0.07 * 9.8)

                                                                         = 834.16 m

=>    long does it take for them to come to rest   =   (33.83)/(0.07 * 9.8)

                                                                          = 49.31 sec

2)   Here,   - 200 * 20 = 200v1 + 300v2

Also,    -20 = v2 - v1

=> final velocity of 200 kg ball = 4 m/sec     (towards east)

=>   final velocity of 300 kg ball = - 16 m/sec    (towards west)

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