Question

A 110.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 8900.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.
Required:
a. What is the velocity of the truck right after the collision?
b. What is the change in mechanical energy of the car?

Answer 1

Answer:

Explanation:

Using the law of conservation of momentum which states that the sum of momentum of the bodies before collision is equal to the sum of momentum of bodies after collision.

Momentum = Mass*velocity

BEFORE COLLISION

The momentum of a 110.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction = 110*25 = 2750kgm/s

The momentum of a 8900.0 kg truck with a speed of 20.000 m/s in an easterly direction = 8900*20 = 178000kgm/s

Sum of momentum before collision = 2750 + 178000 = 180,750 kgm/s

AFTER COLLISION

The momentum of the car will be 110*18 = 1980kgm/s

The momentum of the truck = 8900v where v is the velocity of the truck after collision.

Sum of momentum after collision = 1980 + 8900v

Applying the conservation law;

180750 = 1980 + 8900v

8900v = 180750-1980

8900v = 178770

v = 178770/8900

v = 20.09m/s

Velocity of the truck after collision is 20.09m/s

Note that the collision is inelastic i.e the body moves with different velocities after collision

b) The mechanical energy experienced by the bodies is kinetic energy.

Kinetic energy = 1/2mv²

Sum of the Kinetic energy before collision = 1/2(110)*25²+1/2(8900)*20²

= 34375 + 1780000

= 1,814,375Joules

Sum of kinetic energy after collision = 1/2*(110)*18²+1/2(8900)*20.09²

= 17820+1796056.045

= 1,813,876.045Joules

Change in mechanical energy =  1,813,876.045Joules - 1,814,375Joules

= -498.955Joules