A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 3

Question

A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 3500-kg truck at rest. The collision is totally inelastic and takes place over an interval of 0.203 s. Assume that no brakes are applied during the collision and the car strikes the rear of the truck. Neglect the friction between the vehicles and the ground. A) What is the average x component of the acceleration of the car during the collision? B)What is the average component of the acceleration of the truck during the collision? C)Are the ratio of the accelerations and the ratio of the inertias related to the way you expect them to be? the inertias of the car and the truck are mc and mt respectively?

Explanation / Answer

A)

Initial velocity of the car u_c = 6.32 m/s

initial velocity of the truck u_t = 0

final velocity of the car v_c = final velocity of the truck v_t because the collision is inelastic.

From momentum conservation, 1400*6.32+0 = (1400+3500)*v_c

v_c = v_t = 1.81 m/s

So change in momentum of car = 1400*(1.81-6.32) = -6314kg m/s

But change in momentum is equal to the impulse or force times the time interval

hence average force = 6314/0.203 = 31103.45 N

Hence acceleration a_c = 22.22 m/s^2

B)

Average force is same for both hence the average acceleration of the truck a_t = 31103.45/3500 = 8.89 m/s^2

C)

Yes the mass and the acceleration are invertially related which is logically true.

The ratio of Inertia = mc/mt = 0.4

the ratio of acceleration = 2.5

so its true.

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