momentum and impulse - simon fraser university · momentum, using the fact that newton's...
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Name ______________________ Date(YY/MM/DD) ______/_________/_______ St.No. __ __ __ __ __-__ __ __ __ Section__________
UNIT 8: ONE-DIMENSIONAL COLLISIONSApproximate Classroom Time: Three 100 minute sessions
In any system of bodies which act on each other, action and reaction, estimated by momentum gained and lost, balance each other according to the laws of equilibrium.
Jean de la Rond D'Alembert 18th Century
OBJECTIVES 1. To understand the definition of momentum and its vector na-ture as it applies to one-dimensional collisions.
2. To reformulate Newton's second law in terms of change in momentum, using the fact that Newton's "motion" is what we refer to as momentum .
3. To develop the concept of impulse to explain how forces act over time when an object undergoes a collision.
4. To use Newton's second law to develop a mathematical equa-tion relating impulse and momentum change for any object expe-riencing a force.
5. To verify the relationship between impulse and momentum experimentally.
6. To study the forces between objects that undergo collisions and other types of interactions in a short time period.
7. To formulate the Law of Conservation of Momentum as a theoretical consequence of Newton's laws and to verify it experi-mentally.
© 1992-94 Dept. of Physics & Astronomy, Dickinson College Supported by FIPSE (U.S. Dept. of Ed.) and NSF. Modified for SFU by N. Alberding, 2005.
Momentum and Impulse
Conservation Laws
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• A person to maintain the constant orbital speed in a tangential
direction
• A stop watch
• A bathroom scale (for determining cart & rider mass)
You should attach the spring scale between the centre post and the rope. The end of
the rope should be attached to the cart rider's belt or held by the rider. The second
rope should be attached to the cart in a direction which is perpendicular to the origi-
nal rope. See the diagram below for details.
✍ Activity 7-5a: Verifying the Fc Equation
Qualitatively
(a) If a puller applies a force on the cart which is always tangent
to the circle and which is just sufficient to overcome friction in
the cart and maintain the cart's motion at a constant speed,
what is the net force in a direction perpendicular to the circle?
Remember Newton's first law!!??
(b) You and members of your group should practice pulling each
of your members around at a constant speed. The rider doesn't
need the belt for this exercise. Instead, the rider should hold on
to the rope and close his or her eyes and concentrate on feeling
the centripetal force.
(c) When you hold on to the centre rope while riding on the cart,
what is the direction of the net force you feel on you (include con-
sideration of the forces in the radial direction and in the tangen-
tial direction.) Does the force you feel seem to increase as you
rotate faster? As you rotate in a smaller circle? Explain.
Workshop Physics II: Unit 7 – Applications of Newton's Laws Page 7-11
Author: Priscilla Laws with David Sokoloff and Ronald Thornton SFU 1097
© 1992-93 Dept. of Physics and Astronomy, Dickinson College Supported by FIPSE
(U.S. Dept. of Ed.) and NSF. Modified for SFU by N. Alberding, 2005—2007
Figure 7-4: a) Top view of the
centripetal force experiment.
Review Question
A puller applies a force on the cart which is always tangent to the circle and which is just sufficient to overcome friction in the cart and maintain the cart's motion at a constant speed. The force in a direction perpendicular to the circle is
A. outward away from the centre
B. zero
C. inward, towards the centre
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Momentum
A. It’s a scalar
B. It’s a vector
C. No, It’s superman!
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Impulse
A. It’s something else
B. It’s a scalar
C. It’s a vector
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Impulse
Fra
cket
on
ball
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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”
€
J on ball = Fr on bdtti
t f
∫
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Momentum
Fr on b dt = mb dvb = d (mb vb)
Mass x Velocity = Momentum
pb = mb vb
Fr on b dt = d pb
€
J on ball = Fr on bdt = dpb
ti
tf
∫ti
t f
∫ = pf − pi
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Impulse Momentum
The impulse on an object in a time interval equals the object’s change of momentum.
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Average ForceMaximum Force
Fmax
The area under the triangle is the same as the area in part (a)
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A ball hits the wall.What direction is the momentum
change?
A. To the right
B. To the left
C. Up
D. Down
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A. To the right
B. To the left
C. Up
D. Down
A ball hits the wall.What direction is the impulse on the
ball?
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Conservation of Momentum
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The momentum change of ball 1is equal and opposite to
the momentum change of ball 2
The total momentum isalways the same
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Inelastic Collisionv1
m1 m2before collision
after collision
v’
stuck together
Use conservation of momentum to find v’
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Inelastic Collision1-D
before
€
ptotal = m1v1
after
€
ʹ′ p total = (m1 + m2 ) ʹ′ v =
€
m1v1 = (m1 + m2 ) ʹ′ v
€
ʹ′ v =m1
(m1 + m2 )v1
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Inelastic Collisionequal masses
€
ʹ′ v =m1
(m1 + m2 )v1
€
m1 = m2
€
ʹ′ v =m1
(2m1)v1 =
v1
2
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Inelastic Collisionm2 =2 m1
v’=?
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Explosions
vʹ′2
m1 m2before “explosion”
vʹ′1
after “explosion”ptotal = 0
pʹ′total = 0
€
m1 ʹ′ v 1 + m2 ʹ′ v 2 = 0
€
ʹ′ v 1ʹ′ v 2
=m2
m1
Inverse Relationship
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Reverse Explosionv2before v1
after m1 m2
pʹ′total = 0
ptotal = 0
€
ʹ′ v 1ʹ′ 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.
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Explosions in 2-D
Find v3 and !
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unknowns
Explosions in 2-D
€
pix = p1x + p2x + p3x
€
piy = p1y + p2 y + p3 y = 0
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€
p3 y = −( p1y + p2 y )
€
p3x = pix − ( p1x + p2x )
€
pix = MVix = (10g)(2.0m/s) = 20 g•m/s
€
p2x = (3g)(−10.0m/s)
= −30 g•m/s
€
p1x = (3g)(12.0m/s) cos 40°
= 27.6 g•ms
0
€
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.