Physics 1ALecture 8C

"Sometimes being pushed to the wall gives you the momentum necessary to get over it!”

--Peter de Jager

ExplosionsJust like collisions, explosions will also conserve momentum (as long as we define our system properly).

Let’s say that we have a bomb sitting on a table and then it explodes into two pieces. What is the final momentum of the system?

The final momentum of the system should be zero since there are no external net forces affecting the bomb.

The internal potential energy that was released by the bomb will not affect the momentum of the pieces.

ExplosionsExampleA 0.5kg cart and a 2.0kg cart are attached and are rolling forward with a speed of 2.0m/s. Suddenly a spring-loaded plunger pops out and blows the two carts apart from each other. The smaller cart shoots backward at 2.0m/s. What are the speed and direction of the larger cart?

AnswerFirst, you must define a coordinate system.Let’s say that original motion of the two trains is in the positive x-direction.

0.5kg 2.0kg

2.0m/s

MomentumLet’s also define the two trains (A and B) and the spring as a single system. Every force between them is now internal.

Initially, the momentum of the system is:pi = mAivAi + mBivBi

pi = (mA + mB)vi

pi = (0.5kg + 2.0kg)2.0m/s = 5.0kg(m/s)

Finally, the momentum of the system is:pf = mAfvAf + mBfvBf

pf = (0.5kg)(–2.0m/s) + (2.0kg)vBf

pf = -1.0kg(m/s) + (2.0kg)vBf

Answer

MomentumAre there any net external forces on the system?

No, the spring is part of our system (note: FN and Fg cancel out).

This means that momentum will be conserved such that:

pi = pf

5.0kg(m/s) = -1.0kg(m/s) + (2.0kg)vBf

(2.0kg)vBf = 6.0kg(m/s)vBf = [6.0kg(m/s)]/(2.0kg) = +3.0m/s

The speed of the larger cart is 3.0m/s and the positive sign indicates that it will be moving in the positive x-direction.

Answer

Elastic CollisionExampleAt a stop sign, a car of mass 1.50x103kg waits for a break in traffic. A truck of mass 2.00x103kg comes up from behind and hits the stopped car. Assuming the collision is elastic, how fast was the truck moving just before the collision if the car is pushed ahead into the intersection at 20m/s just after the collision?

AnswerFirst, you must define a coordinate system.Let’s choose the direction of motion of the truck as the positive x-direction.

Truck Carv

Elastic CollisionLet’s also define the car and the truck as a single system. Every force between them is now internal.

We will examine momentum of the system just before the collision and just after the collision.

Initially, the momentum of the system is:

Answer

Finally, the momentum of the system is:

Applying momentum conservation gives us:

Elastic CollisionSince this is an elastic collision, we can apply the relative velocity equation (mixing momentum and energy conservation - eq. 8.21)

Answer

Inserting the variables for this particular case, we should solve for the final velocity of the truck and reinsert that into the momentum equation.

Elastic CollisionGoing back to the momentum equation:

Answer

We need to solve for the initial velocity of the truck:

2D CollisionsThe harder cases are collisions (or explosions) in two dimensions (or three).

Just remember that perpendicular directions (x and y) are independent.

If for external forces, ΣFx = 0 and ΣFy = 0, then px and py are conserved.

Apply conservation of momentum separately to each direction.

CollisionsSo, in the case of the picture below, the initial momentum in the y-direction is zero.

Thus, the final momentum in the y-direction is also zero since there are no external forces in the y-direction.

This means that: m1v1f sinθ = m2v2f sinϕ.

Also, in the x-direction:

pi = m1v1i

pf = m1v1f cosθ + m2v2f cosϕ

2D CollisionsExampleA 10.0g golf ball initially traveling at 5.00m/s hits a 0.500kg stationary basketball off center. The two collide such that they finally travel at 50.0o with respect to one another. Find the final velocities of both balls.

AnswerFirst, you must define a coordinate system.Let’s say that original motion the golf ball is in the positive x-direction.

vi

20o

30o

MomentumLet’s also define the two balls as a single system. Every force between them is now internal.

Initially, the momentum of the system in the x-direction is:

Answer

For the final momentum of the system we need to break the final velocities into components.

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pix = pgx + pBx

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pix = 5ms( ) 0.01kg( ) + 0 = 0.05kgm sInitially, the momentum of the system in the y-direction is:

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piy = pgy + pBy = 0

vg

vgcosθ

vgsinθθ

MomentumAnswerFinally, the momentum of the system in the x-direction is:

Finally, the momentum of the system in the y-direction is:

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pfx = pgx, f + pBx, f

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pfx = mgvgx + mBvBx

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pfx = mgvg cos30° + mBvB cos20°

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pix = pfx

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0.05kgm s = mgvg cos30° + mBvB cos20°

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pfy = pgy, f + pBy, f

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pfy = mgvgy + mBvBy

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pfy = mgvg sin30°−mBvB sin20°

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piy = pfy

Apply momentum conservation:

Apply momentum conservation:

MomentumAnswer

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0 = mgvg sin30°−mBvB sin20°

Let’s solve for vg:

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mgvg sin30° = mBvB sin20°

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vg = vBmB sin20°mg sin30°

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vg = vB0.5kg( )sin20°

0.01kg( )sin30°= 34.2vB

MomentumAnswer

Putting this back into the y-direction results:

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vB =0.05kgm s0.766( )

= 0.0653ms

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vg = 34.2vB

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vg = 34.2 0.0653ms( ) = 2.23ms

Substitute this back into the x-direction:

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0.05kgm s = 0.01kg( ) 34.2vB( )cos30° + 0.5kg( )vB cos20°

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0.05kgm s = 0.296( )vB + 0.470( )vB

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0.05kgm s = 0.766( )vB

Varying MassIf the mass in our system is varying we get the following:

In a rocket, mass is being expelled out the tail; but if we consider the rocket and gas to be one system; then the net force is zero between them.

Here v is the velocity of the expelled gas, it is usually written, vex, and is constant relative to dv which is the change in velocity of the rocket.

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F ∑ =

d p dt

=d m v ( )

dt= v dm

dt+ m d v

dt

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0 = v dm

dt+ m d v

dt

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− v dm

dt= m d v

dt

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− v exdm = md v

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− v ex1m

dm = d v

Varying MassIntegrating both sides gives us:

Distributing the negative sign and applying log rules:

Also, the ratio dm/dt is called the rate of loss of rocket mass, usually designated by –R.

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−vex1mdm

M i

M f

∫ = dvvi

v f

∫

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−vex ln Mf( ) − ln Mi( )[ ] = v f − vi

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vex lnMi

M f

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = v f − vi

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v f = vi + vex lnMi

M f

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

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R v ex = thrust of rocket

For Next Time (FNT)

Finish HW for Chapter 8.

Start reading Chapter 10.