A(n) 1300-kg car going at 6.32 m/s in the positive x direction collides with a 3

Question

A(n) 1300-kg car going at 6.32 m/s in the positivex direction collides with a 3700-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 x component of the acceleration of the truck during the collision?

C.)Are the ratio of accelerations and the ratio of inertias related the way you expect them to be? The inertias of the car and the truck are mc and mt respectively.

1.) Yes, the ratio of the accelerations acx/atx is equal to mc/mt, as expected. 2.) No, the ratio of the accelerations acx/atx is equal to mc/mt, but it was expected to be mt/mc. 3.) No, the ratio of the accelerations acx/atx is equal to mt/mc, but it was expected to be mc/mt. 4.) Yes, the ratio of the accelerations acx/atx is equal to mt/mc, as expected.

Explanation / Answer

Mass of car mc = 1300 Kg
Mass of truck mt = 3700 Kg
Initial Velocity of Car vc = 6.32 m/s
Initial Velocity of Truck vt = 0

In an inelastic collision, the two objects move together as one object after the collision.
Let us assume they move with velocity v.
Using Momentum Conservation-

mc vc + mt vt = mc v + mc v
1300 * 6.32 + 0 = 1300 * v + 3700 * v
v = 8216/5000 m/s
v = 1.64 m/s

A.)
Acceleration of the car during the collision
Initial Velocity = 6.32
Final Velocity = 1.64
time = 0.203 s

V = u + a*t
1.64 = 6.32 + a * 0.203
a = (1.64 - 6.32) / 0.203 m/s^2
a = -23.1 m/s^2

Car will decelerate = 23.1 m/s^2

B.)
Acceleration of the truck during the collision
Initial Velocity = 0
Final Velocity = 1.64
time = 0.203 s

V = u + a*t
1.64 = 0 + a * 0.203
a = 1.64 / 0.203 m/s^2
a = 8.1 m/s^2

Truck will accelerate, a = 8.1 m/s^2

C.)
As the force is acting on both car and truck which is same in magnitude but in opposite direction.

Therefore,
acx/atx = Yes, the ratio of the accelerations acx/atx is equal to mc/mt, as expected

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