Answer to Question #144540 in Physics for Saad

Question

In a crash test, a car of mass 1200 kg collides with a wall and rebounds as in figure. The initial and final velocities of the car are v_i=-13 m/s and v_f = 2.3 m/s, respectively. If the collision lasts for 0.12 s, find the size and direction of the average force exerted on the car.

Answer 1

We can find the average force exerted on the car from the the impulse-momentum change equation:

$$F_{avg}\Delta t=\Delta p=m(v_f-v_i),$$

here, $F_{avg}$ is the average force exerted on the car, $\Delta t = 0.12\ s$ is the time at which the car is in contact with the wall, $\Delta p$ is the change in momentum, $m=1200\ kg$ is the mass of the car, $v_i=-13\ \dfrac{m}{s}$ is the initial velocity of the car and $v_f=2.3\ \dfrac{m}{s}$ is the final velocity of the car.

Then, from this equation we can find the average force exerted on the car:

$$F_{avg}=\dfrac{m(v_f-v_i)}{\Delta t},$$ $$F_{avg}=\dfrac{1200\ kg\cdot(2.3\ \dfrac{m}{s}-(-13\ \dfrac{m}{s}))}{0.12\ s}=1.53\cdot 10^5\ N.$$

The sign plus means that the average force exerted on the car directed in opposite direction to the initial motion of the car (away from the wall).

Answer:

$F_{avg}=1.53\cdot 10^5\ N,$ away from the wall.