Post on 15-Oct-2020
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
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′ v =m1
(m1 + m2 )v1
Inelastic Collisionequal masses
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′ 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
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′ v 1′ v 2
=m2
m1
Inverse Relationship
Reverse Explosionv2before v1
after m1 m2
p′total = 0
ptotal = 0
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′ 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.