Question

A small car with mass 940. kg traveling with a velocity of 34.0 m/s collides head-on with a large car with mass 2,980. kg traveling in the opposite direction at a velocity of-29.3 m/s. The two cars stick together. The duration of the collision is 94.7 ms. What acceleration (in units of g) do the occupants of the small car experience? What acceleration (in units of g) do the occupants of the large car experience? (Use the direction convention that positive values represent velocities and accelerations to the right and negative values represent velocities and accelerations to the left.) occupants of small car occupants of large car-

Answer 1

As given in the problem -

m1 = 940 kg

m2 = 2980 kg

Initial velocity of small car, v1 = 34.0 m/s

Initial velocity of large car, v2 = -29.3 m/s

Suppose the combined velocity of the two cars after collision = V m/s

Apply conservation of momentum -

Initial momentum (Pi) = Final momentum (Pf)

=> m1*v1 + m2*v2 = (m1+m2)*V [Two cars stick togather after the collision]

=> 940*34.0 + 2980*(-29.3) = (940+2980)*V

=> V = (31960 - 87314) / 3920 = -14.12 m/s

(a) Rate of change of momentum of the small car = [m1*v1 - m1*V] / t = Fs

This will be equal to the force experienced by the smaller car.

So, the acceleration of the occupants of the smaller car = Fs / m1

= (v1 - V) / t = (34 - (-14.12)) / 0.0947

= 508.13 m/s^2

In the units of 'g', the answer = 508.13 / 9.81 = 51.80g (Answer)

(b) Like above -

acceleration of the occupants of the larger car = Fl / m2

= (v2 - V) / t = (-29.3 - (-14.12)) / 0.0947

= 160.3 m/s^2

In the units of 'g', the answer = 160.3 / 9.81 = 16.3 g (Answer)