Lecture series for Conceptual Physics 8 th Ed.. Momentum p 82 momentum =mass x velocity p =m x v...

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Case 1 : Increasing Momentum p85 Increasing the force…hitting it harder. and / or Increasing the time of contact. The force on the golf ball builds up to a huge amount then decreases. We’ll use the average force. F t F tF t

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Lecture series for Conceptual Physics 8th Ed.

Momentum p 82

momentum = mass x velocity

p = m x v

Impulse p83

impulse = force x timeI = F x t

Together:

F x t = m x vImpulse = change in momentum

Case 1 : Increasing Momentum p85

Increasing the force…hitting it harder.

and / orIncreasing the time of contact.

The force on the golf ball builds up to a huge amount then decreases.

We’ll use the average force.

F t

F t

Case 2: Decreasing Momentum Over a Long Time p85

Long means gentle, gradual, softly.

A large momentum can be reduced by a small force

in a long time.

A large momentum can by reduced by a large force in a short time.

Not so gently.

m v = F t

Case 3: Decreasing Momentum Over a Short Time

A flower pot falling on your head is bad enough. But,if the pot bounces, it’ll really hurt!

First, there’s the impulse required to stop the pot.

Then, there’s the impulse in the other direction to throw it back.

When something bounces, it hits twice as hard.

Figure 5-8 the Pelton water wheel.

Conservation of Momentum p 88

The momentum before a collision

equals the momentum after a collision.

BEFORE:p = m v = m x 0 = 0

AFTER:m (-v) + m v = 0

Momentum is conserved…it is the same before and after.

Collisions p 90 pbefore = pafter

These are elastic collisions

More on Collisions

Notice that these box cars stick together.

These are inelastic collisions.

pbefore = pafter

(m x 10 m/s)before = (2m x v)after

vafter = 5 m/s

Before

During

After

More inelastic collisions p91

Before: m v + m (-v) = 0

After: m 0 + m 0 = 0

Before: m vA + m vB = k

After: v (m + m) = k

Fig 5-12 p91

m v + m (-v) = m 0 + m 0 m vA + m vB = v (m + m)

1 m/s

m = 5kg

4 m/s

m = 1 kg

Mv + m(-v) = (M+m)v

(5kg)(1m/s) + (1kg)(-4m/s) = (5kg + 1kg)v

v = 1/6 = 0.17 m/s

What’s the velocity of the fish after lunch?

We’ll do the more complicated collisions on the board.

The End

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