Physics 220 Class Exercise October 11 1999
Two vehicles are travelling along a straight stretch of road. The first vehicle, mass m_{1 }= 1200 kg, abruptly stops. The vehicle behind, mass m_{2} = 1000 kg, is unable to stop in time and hits the first vehicle at a speed of 15 m/s. After the collision the first vehicle is moving forward at 10 m/s. Find the velocity of the second vehicle after the collision.
(a) Draw a diagram showing the two vehicles immediately before the collision.
(b) Draw a diagram showing the two vehicles immediately after the collision.
(c) Choose a coordinate system.
(d) Calculate the momentum of the system of two vehicles in the before and after states. How many components do you need to consider here? Use an algebraic symbol for any unknown quantities.
Momentum component 
Before 
After 

P_{x}

Vehicle 1 0 
Vehicle 2 m_{2}v_{2i} 
Total m_{2}v_{2i} 
Vehicle 1 m_{1}v_{1f} 
Vehicle 2 m_{2}v_{2f} 
Total m_{1}v_{1f} + m_{2}v_{2f} 
Set totals equal

m_{2}v_{2i} =

m_{1}v_{1f} + m_{2}v_{2f} 
Solve for the velocity of vehicle 2. Comment on your result.
v_{2f }= v_{2i}  (m_{1}/ m_{2}) v_{1f }= 15 m/s  1.2(10 m/s) = 15 m/s  12 m/s = 3 m/s.
After the collision, the second vehicle is moving in the positive xdirection at 3 m/s.
If you got everything done, then estimate the force exerted on the first car by the second.
We can use the impulsemomentum theorem applied to the first car:
I_{x }= p_{1,f}  p_{1,i} = m_{1}v_{1f}
Then we estimate the duration of the collision. If the car crumples about 0.1 m, then the duration is about (0.1 m)/(10 m/s) or about 0.01 s. Then the average force magnitude is about:
F » I_{x }/D t = (1200 kg)(10 m/s)/(0.01 s) = 1.2´ 10^{6} N.