At a race track, a car of mass 2000 kg crashes into a concrete wall at a speed of 77 m/s.

Question

a. The car crashes into the wall, stopping in 0.08 seconds. What force is applied to the car?

b. To make things safer, the race track installs barrels of sand along the wall. They want to reduce the force on a 2000 kg car crashing at 77 m/s to 500,000 N. By what factor must the barrels increase the time of the crash?

Answer 1

The key to this problem is the impulse-momentum theorem that says the change in momentum is equal to the impulse applied.

Momentum is usually identified as P and equals mass times velocity So:

P = mv and the change in momentum (ΔP) is the final momentum minus the initial momentum or:

ΔP = m₂v₂ - m₁v₁

Impulse is usually identified as J is applied force times the time it is applied or J = FΔt

Since the change in momentum is equal to the impulse applied, we can say:

ΔP = J

m₂v₂ - m₁v₁ = FΔt

In this case we want to know F so we can solve this equation for F:

F = (m₂v₂ - m₁v₁)/Δt

Plugging in the numbers we get:

F = (2000*0 - 2000*77)/0.08

F = -1,925,000 N (the negative just means the force is applied in the opposite direction as the velocity)

For part b, lets go back to our equation m₂v₂ - m₁v₁ = FΔt. We want to add in a multiplication factor (lets call it K) to multiply the time by so that the force goes down. So ket's make the equation be:

m₂v₂ - m₁v₁ = FΔtK

Solving for K, we get:

K = (m₂v₂ - m₁v₁)/(FΔt)

Plugging in the numbers (using F = 500,000) we get"

K = (2000*0 - 2000*77)/(-500000*0.08)

K = 3.85 so the barrels must increase the time of the crash by a factor of 3.85 in order to get the applied force down to 500,000N