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One train (22,680 kg) moving east at 170 km/h collides with a car (1,200 kg) that was moving north at 5 km/h. They get attached to each other, moving in the direction of the train. What is the final velocity of the objects and the type of collision? If the collision happened in 1.2X10-3 seconds, what was the force? How much kinetic energy was lost?

Answer :

abidemiokin
  • The final velocity of the objects is 229.82km/hr
  • Since energy was lost after the collision, the type of collision that occurs is an elastic collision.
  • The required force is 1,090,918,000N
  • The amount of Kinetic energy lost is 12,401,976.656Joules

According to the law of collision'

m1u1 + m2u2 = (m1+m2)v

Given the following:

m1 = 22,680kg

u1 =170km/h

m2 = 1200kg

u2 = 5km/hr

Get the final velocity "v"

22680(170) + 1200(5) = (22680 + 1200)v

3855600 + 6000 = 12880v

3,861,600 = 12880v

v = 3,861,600/12,880

v = 229.82km/hr

Hence the final velocity of the objects is 229.82km/hr

Since energy was lost after the collision, the type of collision that occurs is an elastic collision.

According to Newton's second law, the formula for calculating the force is expressed as;

F = ma

F = m(v-u)/t

F = 22680+1200(229.82-175)/t

Ft = 23880(54.82)

F = 1,309,101.6/0.0012

F = 1,090,918,000N

Hence the required force is 1,090,918,000N

KE lost = Kinetic energy after collision - Kinetic energy before collision

Kinetic energy after collision = 1/2 * 12880 * 229.82²

Kinetic energy after collision = 340,142,976.656 Jolues

Kinetic energy before collision = 1/2 * 22680(170)²+ 1/2*1200(5)²

Kinetic energy before collision = 327,726,000 + 15000

Kinetic energy before collision = 327,741,000

Kinetic energy lost = 340,142,976.656 - 327,741,000 =  12,401,976.656Joules

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