Momentum and Impulse

Conservation Laws

Impulse

Fra

cket

on

ball

ImpulseWhen the racket hits the ball

F(t)racket on ball = – F (t) ball on racket

Time interval from ti to tf is the same

Fr on b = mb (dvb/dt)

Fr on b dt = mb dvb

Define “Impulse”

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J on ball = Fr on bdtti

t f

∫

Momentum

Fr on b dt = mb dvb = d (mb vb)

Mass x Velocity = Momentum

pb = mb vb

Fr on b dt = d pb

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J on ball = Fr on bdt = dpb

ti

tf

∫ti

t f

∫ = pf − pi

Impulse Momentum

The impulse on an object in a time interval equals the object’s change of momentum.

Average ForceMaximum Force

Fmax

The area under the triangle is the same as the area in part (a)

Conservation of Momentum

The momentum change of ball 1is equal and opposite to

the momentum change of ball 2

The total momentum isalways the same

Inelastic Collisionv1

m1 m2before collision

after collision

v’

stuck together

Use conservation of momentum to find v’

Inelastic Collision1-D

before

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ptotal = m1v1

after

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′ p total = (m1 + m2 ) ′ v =

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m1v1 = (m1 + m2 ) ′ v

€

′ v =m1

(m1 + m2 )v1

Inelastic Collisionequal masses

€

′ v =m1

(m1 + m2 )v1

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m1 = m2

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′ v =m1

(2m1)v1 =

v1

2

Inelastic Collisionm2 =2 m1

v’=?

Explosions

v′2

m1 m2before “explosion”

v′1

after “explosion”ptotal = 0

p′total = 0

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m1 ′ v 1 + m2 ′ v 2 = 0

€

′ v 1′ v 2

=m2

m1

Inverse Relationship

Reverse Explosionv2before v1

after m1 m2

p′total = 0

ptotal = 0

€

′ v 1′ v 2

=m2

m1

Velcro

Note that if you sit on one of the blocks, the “explosion” looks like a collision:

Change of reference frame.

Explosions in 2-D

Find v3 and θ

unknowns

Explosions in 2-D

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pix = p1x + p2x + p3x

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piy = p1y + p2 y + p3 y = 0

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p3 y = −( p1y + p2 y )

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p3x = pix − ( p1x + p2x )

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pix = MVix = (10g)(2.0m/s) = 20 g•m/s

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p2x = (3g)(−10.0m/s)

= −30 g•m/s

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p1x = (3g)(12.0m/s) cos 40°

= 27.6 g•ms

0

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p1y = (3g)(12.0m/s) sin 40¡

= 23.1 g•m/s

= 22.4 g•m/s

= –23.1 g•m/s

Accountancy 101

Find θ and v3.