Momentum Part 3Momentum Part 3

By: Heather BrittonBy: Heather Britton

Elastic CollisionsElastic Collisions

•Elastic collisions are a special type of collisions that do not often occur in everyday experience

•Elastic collision - one in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision

•Not only is momentum conserved, but kinetic energy is also conserved

•Elastic collisions are a special type of collisions that do not often occur in everyday experience

•Elastic collision - one in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision

•Not only is momentum conserved, but kinetic energy is also conserved

Elastic CollisionsElastic Collisions

•p1o + p2o = p1 + p2

•(mv)1o + (mv)2o = (mv)1 + (mv)2

•KE1o + KE2o = KE1 + KE2

•(1/2)(mv2)1o + (1/2)(mv2)2o = (1/2)(mv2)1 +(1/2)(mv2)2

•p1o + p2o = p1 + p2

•(mv)1o + (mv)2o = (mv)1 + (mv)2

•KE1o + KE2o = KE1 + KE2

•(1/2)(mv2)1o + (1/2)(mv2)2o = (1/2)(mv2)1 +(1/2)(mv2)2

Elastic CollisionsElastic Collisions

•Solving problems concerning collisions will involve using simultaneous equations

•With elastic collisions we can derive an equation to represent the velocities before and after the event

•We will use po = p and KEo = KE

•Solving problems concerning collisions will involve using simultaneous equations

•With elastic collisions we can derive an equation to represent the velocities before and after the event

•We will use po = p and KEo = KE

Elastic CollisionsElastic Collisions

•(1/2)(mv2)1o + (1/2)(mv2)2o = (1/2)(mv2)1 +(1/2)(mv2)2

•Canceling out the (1/2) gives

•(mv2)10 + (mv2)2o = (mv2)1 + (mv2)2

•Rearranging gives

•(mv2)1o - (mv2)1 = (mv2)2 - (mv2)2o

•(1/2)(mv2)1o + (1/2)(mv2)2o = (1/2)(mv2)1 +(1/2)(mv2)2

•Canceling out the (1/2) gives

•(mv2)10 + (mv2)2o = (mv2)1 + (mv2)2

•Rearranging gives

•(mv2)1o - (mv2)1 = (mv2)2 - (mv2)2o

Elastic CollisionsElastic Collisions

•Factoring out the m gives

•m1(v1o2 - v1

2) = m2(v22 - v2o

2)

•Factoring the squares gives

•m1(v1o - v1)(v1o + v1) = m2(v2 - v2o)(v2 + v2o)

•Factoring out the m gives

•m1(v1o2 - v1

2) = m2(v22 - v2o

2)

•Factoring the squares gives

•m1(v1o - v1)(v1o + v1) = m2(v2 - v2o)(v2 + v2o)

Elastic CollisionsElastic Collisions

•Now lets play with the conservation of momentum equation

•(mv)1o + (mv)2o = (mv)1 + (mv)2

•Rearranging gives

•(mv)1o - (mv)1 = (mv)2 - (mv)2o

•Now lets play with the conservation of momentum equation

•(mv)1o + (mv)2o = (mv)1 + (mv)2

•Rearranging gives

•(mv)1o - (mv)1 = (mv)2 - (mv)2o

Elastic CollisionsElastic Collisions

•Factoring out the m gives

•m1(v1o - v1) = m2(v2 - v2o)

•Now divide the conservation of energy equation by the conservation of momentum equation to get

•[m1(v1o - v1)(v1o + v1) = m2(v2 - v2o)(v2 + v2o)] / [m1(v1o - v1) = m2(v2 - v2o)]

•Factoring out the m gives

•m1(v1o - v1) = m2(v2 - v2o)

•Now divide the conservation of energy equation by the conservation of momentum equation to get

•[m1(v1o - v1)(v1o + v1) = m2(v2 - v2o)(v2 + v2o)] / [m1(v1o - v1) = m2(v2 - v2o)]

Elastic CollisionsElastic Collisions

•Canceling out and simplifying gives

•v1o + v1 = v2 + v2o

•Rearrange again to arrive at our destination

•v1o - v2o = -(v1 - v2)

•This is only valid for an elastic collision

•Canceling out and simplifying gives

•v1o + v1 = v2 + v2o

•Rearrange again to arrive at our destination

•v1o - v2o = -(v1 - v2)

•This is only valid for an elastic collision

Example 6Example 6

•A 0.450 kg ice puck, moving east with a speed of 3.7 m/s, has a head-on collision with a 0.900 kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed and direction of each object after the collision?

•A 0.450 kg ice puck, moving east with a speed of 3.7 m/s, has a head-on collision with a 0.900 kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed and direction of each object after the collision?