- Submitted By: mayamayamaya
- Date Submitted: 05/22/2014 11:31 AM
- Category: Miscellaneous
- Words: 257
- Page: 2

elastic collisions are those in which kinetic energy and momentum are

conserved.

In a perfectly elastic head on collision, two objects approach one another at 180º angles,

collide, then bounce back at 180º angles in opposite directions. Since the momentum

before the collision is equal to the momentum after the collision, and momentum is equal

to mass times velocity, the following equation holds true:

m1v1(initial) + m2v2(initial) = m1v1(final) + m2v2(final)

m1 = mass of object 1, m2 = mass of object 2

v1 = velocity of object 1, v2 = velocity of object 2

You can use the equation to solve problems such as the one below.

Two train cars approach one another on a straight track. Car 1 (mass = 20,000 kg) travels

east at a speed of 20 m/s. Car 2 (mass = 25,000 kg) travels west at a speed of –10 m/s. If

car 1 accelerates to a velocity of –25 m/s after the collision, what is the velocity of car 2

after the collision? Assume the collision is perfectly elastic. (Note: In this example, a

train’s velocity is positive if it travels east, and negative if it travels west.)

m1v1(initial) + m2v2(initial) = m1v1(final) + m2v2(final)

20,000 kg ∙ 20 m/s + 25,000 kg ∙ –10 m/s = 20,000 kg ∙ –25 m/s + 25,000 kg ∙ v2(final)

400,000

kg m

s

+ –250,000

kg m

s

= –500,000

kg m

s

+ 25,000 kg ∙ v2(final)

150,000

kg m

s

= –500,000

kg m

s

+ 25,000 kg ∙ v2(final)

+500,000

kg m

s

+500,000

kg m

s

kg m

650,000

s

25,000 kg

= 2(final) 25,000 kg

25,000 kg

v

26 m/s = v2(final)

If you substitute the value of v2(final) into the