Question

A race car collided with race track wall.Just before the collision,he is travelling at speed v1= 70m/s along a straight line at 30 degrees from the wall. Just after collision he is travelling at speed 50m/s along a straight line at 10 degree from the wall.His mass is 80kg.(a) the impulse J of the driver due to collision
(b)the collision last for 14ms.What is the magnitude of the average force on the driver during collision

Answer 1

The change of impulse can be calculated as the sum of changes in parallel and perpendicular to the wall directions.

$J_{\text{par}} = m(v_2\cos\alpha_2 - v_1\cos\alpha_1) = 80\,\mathrm{kg}\cdot(50\,\mathrm{m/s}\cdot\cos10^\circ - 70\,\mathrm{m/s}\cdot\cos30^\circ ) = -911\,\mathrm{kg\cdot m/s} \\ J_{\text{perp}} = m(v_2\sin\alpha_2+ v_1\sin\alpha_1) = 80\,\mathrm{kg}\cdot(50\,\mathrm{m/s}\cdot\sin10^\circ+ 70\,\mathrm{m/s}\cdot\sin30^\circ ) =3495\,\mathrm{kg\cdot m/s}.$

(a) The total change of impulse of the driver can be calculated by means of the Pythagorean theorem:

$J = \sqrt{J_{\mathrm{par}}^2 + J_{\mathrm{perp}}^2} = 3611\,\mathrm{kg\cdot m/s}.$

(b) The change of impulse obtained above is due to the force from interaction with the wall. So the average force can be calculated as

$F_{\text{av}} = \dfrac{J}{\Delta t} = \dfrac{3611\,\mathrm{kg\cdot m/s}}{0.014\,\mathrm{s}} \approx 2.6\cdot10^5\,\mathrm{N}.$