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- #1

#### Bri

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The mass of the blue puck in the figure is 20.0% greater than the mass of the green one. Before colliding the, the pucks approach each other with momenta of equal magnitudes and opposite directions, and the green puck has an initial speed of 10.0 m/s. Find the speeds of the pucks after the collision if half the kinetic energy is lost.

Code:

`[COLOR=Green] / / / AO-->--- [/COLOR][COLOR=Blue]---<--O B / / /[/COLOR]`

Angles A and B are 30 degrees.

I calculated the initial velocity of the blue puck to be 25/3 m/s. I've tried setting the initial Kinetic energy in the x and y directions equal to two times the final Kinetic energy in the x and y directions and solving the system of equations, but it doesn't work out.

Could someone please tell me how I should set this up so I can solve it?

Thanks.

## Answers and Replies

- #2

#### Doc Al

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You need to apply:

(1) conservation of momentum

(2) the fact that the *total * KE after the collision is half what it was before the collision.

- #3

#### Bri

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Do I need conservation of momentum to solve for the final speeds?

Should I be separating it into two equations for KE and Momentum, one for in the x direction and one for in the y direction? And wouldn't I end up with 4 equations then? What would I do with them?

I used momentum to find the initial velocity of the blue puck. So far I've been using these equations to try to solve this...

KE(Blue, initial, x direction) + KE(Green, initial, x direction) = 2*KE(Blue, final, x direction) + 2*KE(Green, final, x direction)

KE(Blue, initial, y direction) + KE(Green, initial, y direction) = 2*KE(Blue, final, y direction) + 2*KE(Green, final, y direction)

I simplified those and solved the system.

Last edited:

- #4

#### Doc Al

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Bri said:

See Also9.3 Conservation of Linear Momentum – General Physics Using Calculus I9.3 Conservation of Linear Momentum - University Physics Volume 1 | OpenStaxChapter 4 Test - QuizDo I need conservation of momentum to solve for the final speeds?

Should I be separating it into two equations for KE and Momentum, one for in the x direction and one for in the y direction? And wouldn't I end up with 4 equations then? What would I do with them?

It's much simpler than all that. You have two unknowns (the final speeds of the pucks), but you also have two equations:

(1) conservation of momentum: What's the total momentum of the system?

(2) the fact that the *total * KE after the collision is half what it was before the collision

That's all you need. (Here's a hint: Let the mass of the green puck = m; then the mass of the blue puck = 1.2m. You know the initial speeds of the pucks so you should be able to figure out the initial KE, at least in terms of m.)

- #5

#### Bri

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Ok, so I used:

m(b) = 1.2m(g)

v(bi) = 25/3 m/s

v(gi) = 10 m/s

m(b)v(bi) + m(g)v(gi) = m(b)v(bf) + m(g)v(gf)

1.2m(g)v(bi) + m(g)v(gi) = 1.2m(g)v(bf) + m(g)v(gf)

1.2v(bi) + v(gi) = 1.2v(bf) + v(gf)

20 = 1.2v(bf) + v(gf)

.5m(b)v(bi)^2 + .5m(g)v(gi)^2 = m(b)v(bf)^2 + m(g)v(gf)^2

275/3 = 1.2v(bf)^2 + v(gf)^2

I solved for v(gf) in the momentum equation and put it in the kinetic energy equation and set the equation equal to zero to get

0 = 2.64v(bf)^2 - 48v(bf) + 925/3

There's no solution.

- #6

#### donjennix

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If I didn't make any math errors, the KE before the collision is 65*m(g)

So the KE after the collision is 32.5*m(g) which will be equal to

.5*m(g)*v(g2)^2 + .5*(1.2*m(g))*v(b2)^2

Note that the m(g) terms cancel out leaving

32.5 = .5*v(g2)^2 + .5*1.2*v(b2)^2

constrained by m(g)*v(g2) + 1.2*m(g)*v(b2) = 0

(post momentum = pre momentum = 0)

this is the same as saying v(g2) = -1.2*v(b2) ---- so it you stick the value of v(g2) in terms of v(b2) into the Post-KE expression, don't you get v(b2) and thus v(g2) or have I missed something?

Last edited:

- #7

#### Doc Al

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Bri said:

Ok, so I used:

m(b) = 1.2m(g)

v(bi) = 25/3 m/s

v(gi) = 10 m/s

Realize that the pucks move in opposite directions.

m(b)v(bi) + m(g)v(gi) = m(b)v(bf) + m(g)v(gf)

1.2m(g)v(bi) + m(g)v(gi) = 1.2m(g)v(bf) + m(g)v(gf)

1.2v(bi) + v(gi) = 1.2v(bf) + v(gf)

20 = 1.2v(bf) + v(gf)

Since "the pucks approach each other with momenta of equal magnitudes and opposite directions", what must be the total momentum?

- #8

#### Bri

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Woohoo! I got it now...

Had to set the momentum to 0, not 20. Setting one of their velocities to negative because of their opposite direction slipped past me.

Thanks so much for all the help!

- #9

#### dharmatech

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PSE 5E PRO 9.30

PSE 6E PRO 9.31

The mass of the blue puck in Figure P9.30 is 20.0% greater than the

mass of the green one. Before colliding, the pucks approach each other

with equal and opposite momenta, and the green puck has an initial

speed of 10.0 m/s. Find the speeds of the pucks after the collision if

half the kinetic energy is lost during the collision.

----------------------------------------------------------------------

The mass of the blue puck in Figure P9.30 is 20.0% greater

than the mass of the green one.

(1) m2 = m1 + 0.2 m1

m2 = m1 ( 1 + 0.2 )

m2 = m1 ( 1.2 )

m2 = 1.2 m1

Before colliding, the pucks approach each other with equal and

opposite momenta,

(2) pi1 = - pi2

pi1 = m1 vi1

pi2 = m2 vi2

the green puck has an initial speed of 10.0 m/s

(3) vi1 = 10

(4) pi = pf

pi = pi1 + pi2

pf = pf1 + pf2

pf1 = m1 vf1

pf2 = m2 vf2

(5) Ef = 1/2 Ei

Ei = Ki1 + Ki2

Ef = Kf1 + Kf2

Ki1 = 1/2 m1 vi1^2

Ki2 = 1/2 m2 vi2^2

Kf1 = 1/2 m1 vf1^2

Kf2 = 1/2 m2 vf2^2

----------------------------------------------------------------------

(2): pi1 = - pi2

m1 vi1 = - m2 vi2

m1 vi1 = - (1.2 m1) vi2

vi1 = - 1.2 vi2

vi2 = - vi1 / 1.2

vi2 = - 8.33

(4): pi = pf

pi1 + pi2 = pf1 + pf2

(- pi2) + pi2 = pf1 + pf2

0 = pf1 + pf2

0 = m1 vf1 + m2 vf2

0 = m1 vf1 + (1.2 m1) vf2

0 = vf1 + (1.2) vf2

vf2 = - vf1 / 1.2

----------------------------------------------------------------------

(5): Ef = 1/2 Ei

Kf1 + Kf2 = 1/2 (Ki1 + Ki2)

1/2 m1 vf1^2 + 1/2 m2 vf2^2 = 1/2 (1/2 m1 vi1^2 + 1/2 m2 vi2^2)

m1 vf1^2 + m2 vf2^2 = 1/2 m1 vi1^2 + 1/2 m2 vi2^2

m1 vf1^2 + (1.2 m1) vf2^2 = 1/2 m1 vi1^2 + 1/2 (1.2 m1) vi2^2

vf1^2 + (1.2) vf2^2 = 1/2 vi1^2 + 1/2 (1.2) vi2^2

vf1^2 + (1.2) (- vf1 / 1.2)^2 = 1/2 vi1^2 + 1/2 (1.2) vi2^2

vf1^2 + vf1^2 / 1.2 = 1/2 vi1^2 + 1/2 (1.2) vi2^2

vf1^2 ( 1 + 1 / 1.2 ) = 1/2 vi1^2 + 1/2 (1.2) vi2^2

vf1^2 = ( 1/2 vi1^2 + 1/2 (1.2) vi2^2 ) / ( 1 + 1 / 1.2 )

vf1 = sqrt[ ( 1/2 vi1^2 + 1/2 (1.2) vi2^2 ) / ( 1 + 1 / 1.2 ) ]

vf1 = 7.07

----------------------------------------------------------------------

vf2 = - 5.89

----------------------------------------------------------------------

Also available at: https://gist.github.com/808567

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