A(n) 1200-kg car going at 6.32 m/s in the positive x direction collides with a 3

Question

A(n) 1200-kg car going at 6.32 m/s in the positivex direction collides with a 3500-kg truck at rest. The collision is totally inelastic and takes place over an interval of 0.203 s. Assume that no brakes are applied during the collision and the car strikes the rear of the truck. Neglect the friction between the vehicles and the ground.

a)What is the average x component of the acceleration of the car during the collision?

b)What is the average x component of the acceleration of the truck during the collision?

Explanation / Answer

Momentum is conserved .

Therefore, mass of car = mc = 1200 kg

Corresponding ,initial velocity is vc1 = 6.32 m/s

Mass of truck = mt= 3500 kg

Initial truck velocity = vt1 = 0 (rest)

So by conserving momentum,

mc*vc1 + mt*vt1 = mc*V + mt*V

1200*6.32 = (1200+3500)*V

V = 1.6 m/s of both the vehicles after collision (inelastic collision)

a) average acceleration of car = V/t = 1.6/.203= 7.8 m/s2

b ) answer is same for truck = 7.8 m/s2

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