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A 5.00 kg ball (ball 1), moving to the right at a velocity of +3.00 m/s on a frictionless table, collides head-on with a stationary 8.00 kg ball (ball 2). Find the final velocities of the balls if the collision is as specified below.
(a) a completely elastic collision
ball 1
______ m/s
ball 2
______m/s
(b) a completely inelastic collision
ball 1
_______m/s
ball 2
______m/s

what is the difference between completly inelastic and elastic?

2006-12-06 06:41:43 · 3 answers · asked by lifewithgooli 1 in Science & Mathematics Physics

3 answers

Elastic collisions mean the objects bounce off of one another, and there is no energy lost.

Completely inelastic collisions mean the objects stick together after the collision.

This is a problem of momentum conservation: the initial momentum equals the final momentum.

Elastic collision:

Since only one object is moving before the collision the initial momentum is given by:

mass of ball times its velocity = 5.00*3.00 = 15.0 kgm/s.

Therefore, the momentum after the collision will also equal 15.0 kgm/s. You do have a slight problem in that, you don't know either of the ball's speed after, but you do know this:

5.00V1' + 8.00V2' = 15.0

Where V1' is the final velocity of ball 1, and V2' is the final velocity of ball2.

Luckily, you can bring in energy conservation and actually solve for two equations that will give you the final velocities of each ball. They are given below:

V1' = (m1-m2)/(m1+m2)*V1

V2' = (2m1)/(m1+ m2)*V1

These two equations only work if one of the masses is at rest before the collision. Anyway, if you plug in the numbers, you'll find the final velocities.

Completely Inelastic collision:

This one is easier, because after the collision, the two masses are stuck together, so they are in essence one mass. This means the momentum conservation equation becomes:

m1V1 = ( m1 + m2) V'

V' is the final velocity of the two masses stuck together.

Hope that helps.

2006-12-06 07:08:19 · answer #1 · answered by phyziczteacher 3 · 0 0

In elastic collisions, the 2 bounce apart with no loss of system KE.

In completely inelastic collisions, the 2 objects stick together like a pie thrown in the face of a politician. Or 2 cars collide, stick together, and continue to move with some new identical velocity. KE is lost in this case to work done changing the shape of the cars. KE is not conserved but momentum is conserved.

The equations:
elastic collisions:
v1 = (m1-m2)*u1/(m1+m2) + 2*m2*u2/(m1+m2)
v2 = 2*m1*u1/(m1+m2) + (m2-m1)*u2/(m1+m2)

completely inelastic collisions:
v = (m1*u1 + m2*u2)/(m1+m2)

The u's are velocities before the collision, the v's are after.

2006-12-06 15:11:32 · answer #2 · answered by sojsail 7 · 0 0

69

2006-12-06 14:48:32 · answer #3 · answered by Greg W 2 · 0 1

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