Question

Four people, each has a mass of 65kg, get in a car with a mass of 1700kg.

1) What impulse is required to accelerate the car from rest to a steady speed of 75 km/h on a flat road? Drag and friction can be ignored.

2) Assume the car hits a stationary 600kg moose. The car was still traveling at 75 km/h before the collision, and the collision is perfectly inelastic, what is the speed of the car+moose (in km/h) immediately after the collision?

3) If the collision takes place over 120 milliseconds, what is the average force on the car?

Please help with this. Thanks!!

Answer 1

Part A.

Impulse = Change in momentum

J = dP = Pf - Pi

J = m*Vf - m*Vi = m*(Vf - Vi)

Vi = initial speed of car = 0 m/sec

Vf = final speed of car = 75 km/hr = 75*5/18 = 20.83 m/sec

m = mass of car + people = 1700 + 4*65 = 1960 kg

So,

J = 1960*(20.83 - 0) = 40826.8 kg.m/sec

Part B.

Using momentum conservation:

Pi = Pf

m1*u1 + m2*u2 = (m1 + m2)*V

V = (m1*u1 + m2*u2)/(m1 + m2)

m1 = 1960 kg, m2 = 600 kg

u1 = 75 km/hr = 20.83 m/sec & u2 = 0 m/sec

So,

V = (1960*20.83 + 600*0)/(1960 + 600)

V = 15.95 m/sec = 15.95*18/5 = 57.42 km/hr = final speed of car + moose

Part C.

We know that:

Change in momentum of car = F_avg*dt

F_avg = dP/dt = (Pf - Pi)/dt = m*(Vf - Vi)/dt

Vi = initial speed of car before collision = 20.83 m/sec

Vf = final speed of car after collision = 15.95 m/sec

So,

F_avg = 1960*(15.95 - 20.83)/(120*10^-3) = -79706.67 N

F_avg = -7.97*10^4 N