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```Physics 221, February 23
Key Concepts:
•Definition of momentum and impulse
•Conservation of momentum
•The center of mass
•Rockets
Linear momentum
Momentum:
p = mv
(vector)
Rate of change: ∆p /∆t = m∆v/∆t = ma = F
Fx = ∆px/ ∆t, Fy = ∆py/ ∆t
Impulse:
I = ∆p = pf – pi = F∆t
Kinetic energy: Ekin = ½mv2 = p2/(2m)
p = (2m* Ekin)1/2
(vector)
Two objects have masses m1 and m2, respectively. If m2 = 4m1, and
both have the same kinetic energy, which has more momentum?
For example:
or
or ….
1. Object 1 with mass m1 .
2. Object 2 with mass m2 .
3. Their momenta are the
same.
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Extra credit: A piece of clay with mass m = 0.01 kg collides with
the floor at speed of 2 m/s and sticks. The collision takes 0.01 s.
The force the piece of clay experiences during the collision is
(assume a constant force during the collision):
1.
2.
3.
4.
5.
0N
1N
2N
4N
8N
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A ball (mass 0.40 kg) is initially moving to the left at 30 m/s. After hitting the
wall, the ball is moving to the right at 20 m/s. What is the impulse of the net
force on the ball during its collision with the wall?
1.
2.
3.
4.
5.
20 kg m/s to the right
20 kg m/s to the left
4 kg m/s to the right
4 kg m/s to the left
None of the above
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You have heard the tried and true phrase “it is like running in to a brick wall”.
Now it is time to dig a little deeper and modify the phrase. Assume you are in
a car driving and come in contact with this proverbial brick wall. Your car can
do three things after the strike: it can go through, come to stop or bounce
back. Select the option which will be the most dangerous to you from an
impulse point of view.
1.
2.
3.
4.
Going through the wall.
Coming to a stop.
Bouncing back.
All options are equally
dangerous.
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Conservation of momentum
For a system of objects, a component of the momentum along a chosen
direction is constant, if no net outside force with a component in this
chosen direction acts on the system.
In collisions between isolated objects momentum is always conserved.
m1v1i + m2v2i = m1v1f + m2v2f
Kinetic energy is only conserved in elastic collisions.
(1/2)m1v1i2 + (1/2)m2v2i2 = (1/2)m1v1f2 + (1/2)m2v2f2
I explosions or disintegrations momentum is conserved.
(∑mivi )before = (∑mivi )after
Kinetic energy is not conserved.
Stored potential energy is converted into ordered or disordered kinetic
energy.
Extra credit: Two objects with different masses collide and stick to each
other. Compared to before the collision, the system of two objects after the
collision has
1.
2.
3.
4.
5.
the same total momentum and
the same total kinetic energy.
the same total momentum but
less total kinetic energy.
less total momentum but the
same total kinetic energy.
less total momentum and less
total kinetic energy.
not enough information given to
decide.
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Block A has mass 1.00 kg and block B has mass 3.00 kg. The blocks collide and
stick together on a level, frictionless surface. After the collision, the kinetic
energy (KE) of block A is
1.
2.
3.
4.
5.
1/9 the KE of block B.
1/3 the KE of block B.
3 times the KE of block B.
9 times the KE of block B.
the same as the KE of block B.
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You have a mass of 60 kg. You are standing on an icy pond, when your
“friend” throws a 10 kg ball at you with speed 7 m/s.
If you catch the ball, how fast will you be moving?
1.
2.
3.
4.
5.
0 m/s
1 m/s
2 m/s
3.5 m/s
7 m/s
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A 1 kg block slides along a horizontal frictionless surface with speed v =1 m/s. It collides with
another 1 kg stationary block. After the collision the two blocks stick together and slide into a
Hooke's-law spring with spring constant k = 100 N/m. What is the maximum compression (in m)
of the spring?
1.
2.
3.
4.
0.5 m
25 cm
7 cm
5 mm
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Momentum conservation in the collision:
m*v = 2m*v’, 1kg*1m/s = 2kg*v’, v’ = 0.5 m/s.
The 2 block combination has kinetic energy:
½ mv2 = ½*2kg*(0.5 m/s)2 = 0.25 J
This kinetic energy is converted into elastic potential energy.
0.25 J = ½ k(∆x)2 = ½ 100 N/m (∆x)2
(∆x)2 = (0.25/50) m2, ∆x = 7 cm
Demonstrations
Collisions and conservation of momentum
Astroblaster
Center of mass
• The center of mass (CM) of a system moves as if the total
mass of the system were concentrated at this special point.
• It responds to external forces as if the total mass of the
system were concentrated at this point.
• The total momentum of the system only changes, if external
forces are acting on the system.
• The center of mass of the system only accelerates, if external
forces are acting on the system.
• Coordinates of the center of mass (CM):
Extra credit:
Two particles of masses 2 g and 6 g are separated by a distance of 6
cm. The distance of their center of mass from the heavier particle is
•
•
•
•
•
1.5 cm
2 cm
3 cm
4 cm
4.5 cm
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A radioactive nucleus of mass M moving along the positive x-direction
with speed v emits an α-particle of mass m. If the α-particle proceeds
along the positive y-direction, the centre of mass of the system (made
of the daughter nucleus and the α-particle) will
• remain at rest .
• move along the positive
x-direction with speed less than v.
• move along the positive x-direction
with speed greater than v.
• move in a direction inclined to the
positive x-direction .
• move along the positive x-direction
with speed equal to v.
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The rocket principle
System consisting of many parts:
no external force  no acceleration of the CM
But different parts of the system can accelerate with respect to
the CM, as long as the total momentum of the system is
constant.
Examples:
Extra Credit: A 55 kg physics student is at rest on a 5 kg sled that
also holds a chunk of ice with a mass of 1.5 kg. The student
throws the ice horizontally with a speed of 12 m/s relative to the
ground. If the sled slides over a frozen pond without friction,
how fast (in m/s) are the sled and student traveling with respect
to the ground after throwing the chunk of ice?
1.
2.
3.
4.
5.
0.3 m/s
0.327 m/s
5 m/s
0.2 m/s
0.218 m/s
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