Question

Suppose a child drives a bumper car head on into the side rail, which exerts a force of 3900 N on the car for 0.55 s. Use the initial direction of the cars motion as the positive direction.

What impulse, in kilogram meters per second, is imparted to the car by this force?

Find the horizontal components of the final velocity of the bumper car, in meters per second, if its initial velocity was 2.95 m/s and the car plus driver have a mass of 190 kg. You may neglect friction between the car and floor.

Find the horizontal components of the final velocity of the bumper car, in meters per second, if its initial velocity was 2.95 m/s and the car plus driver have a mass of 190 kg. You may neglect friction between the car and floor.

Answer 1

Part A.

We know that Impulse is given by:

Impulse = change in momentum = F_avg*dt

Given that average force exerted on car by side rail = -3900 N (Since bumper rail will exert force on opposite direction of motion of car)

dt = time interval = 0.55 sec

So,

Impulse = -3900*0.55

Impulse = -2145 kg.m/sec

Part B.

Now Since

Impulse = Change in momentum

J = dP = m*dV

J = m*(Vf - Vi)

Vi = Initial velocity of car = +2.95 m/s (Since initial direction of car's motion is positive)

m = mass of car = 190 kg

So,

Vf = Vi + J/m

Vf = +2.95 - 2145/190

Vf = -8.34 m/s = final velocity of car (-ve means in opposite direction)

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