Chapter 9
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Chapter 9. Momentum & Its Conservation. Determining Impulse. F = ma a = D v / D t. Thus. F = m D v / D t or F D t = m D v. Impulse. The product of a force times the amount of time the force is applied. F D t. Determining Momentum. D v = v f – v i thus m D v = m v f – m v i.

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Chapter 9

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Chapter 9

Chapter 9

Momentum & Its Conservation


Determining impulse

Determining Impulse

F = ma

a = Dv/Dt


Chapter 9

Thus

F = mDv/Dt or

FDt = mDv


Impulse

Impulse

  • The product of a force times the amount of time the force is applied.

  • FDt


Determining momentum

Determining Momentum

Dv = vf – vi

thus

mDv = mvf – mvi


Momentum p

Momentum (p)

  • The product of mass times velocity

  • p = mv


Change in momentum

Change in Momentum

Dp = mDv


F d t m d v

FDt = mDv

  • Impulse = momentum change


F d t m d v m v f m v i p f p i

FDt = mDv = mvf-mvi= pf-pi


The equation below is called the impulse momentum theorem f d t p f p i

The Equation below is called the Impulse-Momentum TheoremFDt = pf-pi


A 750 kg car is traveling east at 180 km hr calculate the magnitude direction of its momentum

A 750 kg car is traveling east at 180 km/hr. Calculate the magnitude & direction of its momentum.


Chapter 9

A 250 kg car is traveling east at 360 km/hr. Calculate the magnitude & direction of its momentum.


Chapter 9

A 250 kg car collides with a 10.0 Mg shed & remains in contact with the shed for 0.500 s. Calculate the force of the collision & the impulse imparted onto the shed.


A force of 25 n is applied to a 5 0 kg object for 5 0 seconds calculate impulse d p d v

A force of 25 N is applied to a 5.0 kg object for 5.0 seconds. Calculate: impulse, Dp & Dv:


Chapter 9

A force of 75 N is applied to a 5.0 kg object for 15.0 seconds. Calculate: impulse, Dp & Dv:


A 250 kg sled is accelerated from 6 0 m s to 18 m s over 120 s calculate a p i p f d p impulse

A 250 kg sled is accelerated from 6.0 m/s to 18 m/s over 120 s. Calculate: a, pi, pf, Dp, & impulse


Chapter 9

A 150 g ball pitched at 40.0 m/s is batted in the opposite direction at 40.0 m/s. Calculate: Dp, & impulse


Chapter 9

A 60.0 kg man drives his car into a tree at 25 m/s. The car comes to rest in 0.20 s. Calculate: Dp & F on the man.


Chapter 9

Calculate the momentum change when a 100.0 kg block accelerates for 10.0 s down a 37o incline with a frictional coefficient of 0.25


Conservation of momentum

Conservation of Momentum

  • In a closed system, momentum is conserved

  • pf = pi or p1 = p2


Chapter 9

Conservation of Momentum

  • In collisions, momentum is conserved

  • (p1 + p2)b = (p1 + p2)a


Chapter 9

Book Notation of Momentum

(p1 + p2)b = (p1 + p2)a

(pA + pB)1 = (pA + pB)2

pA1 + pB1 = pA2 + pB2


Chapter 9

Book Notation of Momentum

pA1 + pB1 = pA2 + pB2

mAvA1 + mBvB1 =

mAvA2 + mBvB2


Chapter 9

Collision Momentum

mAvA + mBvB =

mAvA’ + mBvB’


Chapter 9

A 200. Mg freight car moving at 2.5 m/s collides with the same sized car at rest where they remain connected. Calculate vf:


Chapter 9

A 125 g hockey puck moving at 40.0 m/s is caught in a glove by a 75 kg goalie. Calculate vf of the goalie.


Chapter 9

A 35 g bullet strikes a 2.5 kg stationary block at 750 m/s. The bullet exits the block at 350 m/s.Calculate vf of the block.


Chapter 9

A 250 g ball at 4.0 m/s collides head on with a 1.0 kg ball 2.0 m/s. the 250 g ball bounced backwards at 5.0 m/s. Calculate vf of the other.


Chapter 9

A 750 g ball at 4.0 m/s collides head on with a 1.0 kg ball 5.0 m/s. The 750 g ball bounced backwards at 8.0 m/s. Calculate vf of the other.


Chapter 9

A 25 g ball at 40.0 m/s collides head on with a 2.0 kg ball 2.0 m/s. the 25 g ball bounced backwards at 50.0 m/s. Calculate vf of the other.


Chapter 9

A 250 g ball at 4.0 m/s collides head on with a 2.0 kg ball 5.0 m/s. the 250 g ball bounced backwards at 40.0 m/s. Calculate vf of the other.


Chapter 9

A 1.0 kg bat swung at 50.0 m/s strikes a 250 g ball thrown at 40.0 m/s. The bat continues at 10.0 m/s. Calculate vf of the ball.


Explosion momentum

Explosion Momentum

  • The momentum before the explosion must = the momentum after the explosion.

  • The momentum before the explosion = 0


Chapter 9

Explosion Momentum

  • pA = pB

  • pB = 0 thus

  • pA = 0


Chapter 9

Explosion Momentum

  • The summation of all parts after the explosion = 0


Chapter 9

Explosion Momentum

mAvA + mBvB + etc = 0


Chapter 9

Explosion Momentum with only 2 parts

mAvA + mBvB

= 0


Chapter 9

Explosion Momentum with only 2 parts

mAvA = -mBvB


A 50 0 kg gun fired a 150 g bullet at 500 0 m s calculate the recoil velocity of the gun

A 50.0 kg gun fired a 150 g bullet at 500.0 m/s. Calculate the recoil velocity of the gun.


Chapter 9

A 500.0 Mg cannon fired a 150 kg projectile at 1500.0 m/s. Calculate the recoil velocity of the gun.


Chapter 9

A 250 g cart is connected to a 1.5 kg cart. When disconnected, a compressed spring pushes the smaller cart 4.0 m/s east. Calculate the velocity of the larger cart.


Chapter 9

A 2.0 kg block is tied to a 1.5 kg block. When untied, a compressed spring pushes the larger block 6.0 m/s east. mblock = 0.25 Calculate: vi, a, t, d for the smaller block


Chapter 9

A 5.0 kg block is tied to a 2.0 kg block. When untied, a compressed spring pushes the larger block 1.0 m/s east. mblock = 0.20 Calculate: vi, a, t, d for the smaller block


Two dimensional collisions

Two Dimensional Collisions


Chapter 9

A 5.0 kg ball moving at 40.0 m/s collides with a stationary 2.0 kg. The 2.0 kg ball bounced at a 30o angle from the path at 50.0 m/s. Calculate vf of the other.


Chapter 9

A 2.0 kg ball is dropped from a 14.7 m high ledge collides with a stationary 10.0 kg ball hanging at a height of 9.8 m. The 2.0 kg ball bounced straight up at 4.9 m/s. Calculate vi, vf, & tair of the 10 kg ball.


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