Download Chris Khan 2008 Physics Chapter 9 Linear momentum is defined as

Chris Khan 2008
Physics Chapter 9
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Linear momentum is defined as the product of mass and velocity of an object.
o P = mv
When dropping an item, the momentum before hitting the floor is –mv since the velocity is in the negative y-direction.
If the item lands on the floor, final momentum is 0. If the item bounces, the final momentum is mv.
o When dropping a beanbag, ∆P = Pf – Pi = 0 – m(-v) = mv.
o When dropping a rubber ball, ∆P = P f – Pi = mv – m(-v) = 2mv.
Since momentum is a vector quantity, the total momentum of a system is the vector sum of the momenta of all the
objects.
o If two 4-kg ducks and a 9-kg goose swim towards bread, swimming at 1.1 m/s and 1.3 m/s, what is the
magnitude and direction of the total momentum? Since the ducks are coming from the top and the left and the
goose is coming from the bottom, we have to find x and y vectors for momentum. The x-direction is equal to
mv = (1.10 m/s)(4 kg) = 4.40 kg m/s (towards the right). The y-direction is equal to mvg – mvd = (1.3 m/s)(9 kg)
–
(1.1
m/s)(4
kg)
=
11.7
–
4.40
=
7.30
kg
m/s.
Therefore,
Newton’s Second Law can be written as F = P/t. Therefore, ∑F = m∆v/∆t. In addition, a = ∆v/∆t.
Impulse is found by doing I = Fav∆t. Since Fav = ∆P/∆t, Fav∆t = ∆P. Therefore, impulse is given as the change in
momentum  I = ∆P.
o If you bunt a 0.144-kg baseball, initially moving towards home plate with a speed of 43 m/s, with an average
force of 6.50 x 103 N for 1.30 ms, what is the final speed of the ball? First, relate change in momentum to
impulse: ∆P = Pf – Pi = I = Fav∆t. Therefore,
Next, calculate the
impulse:
With these results, we can find Pf:
Use these with P = mv to find vf = 2.26/0.144 =
15.7 m/s.
Jumping for joy, a 72-kg balla jumps upward with a speed of 2.1 m/s. What is the impulse experienced by him?
Before he jumps, the floor exerts a force of mg on the contestant. What additional average upward force does
the floor exert if the contestant pushes down on it for 0.36s during the jump? First, I = ∆P = P f – Pi = mvf (since
v1 = 0). I = mvf = (72)(2.1)y = (150 kg m/s)y. To find average force in terms of the impulse I and the time
interval ∆t, Fav = I/∆t = (150 kg m/s)y / 0.36 s = (420 N)y.
The Law of Conservation of Momentum says that if the net force acting on an object is zero, its momentum is
conserved  Pf = Pi.
o Two groups of canoeists meet in the middle of a lake when a person in canoe 1 pushes on canoe 2 with 46 N to
separate the canoes. If the mass of canoe 1 is 130 kg and the mass of canoe 2 is 250 kg, what is the momentum
of each canoe after 1.2 s of pushing? First, find a using a2x = F/m = 46/250 = 0.18 m/s2 and a1x = F/m = -46/130
= -0.35 m/s2. Now, find v after 1.2 s using v = at. This tells us that v1x = -0.42 m/s and v2x = 0.22 m/s. Using
this, P1x = m1v1x = (130)(-0.42) = -55 kg m/s. Also, P2x = m2v2x = (250)(0.22) = 55 kg m/s.
Recoil is the effect of moving in the opposite direction of the force you exerted.
Collisions are the acts of two objects hitting each other when the external forces are 0 or negligible. With inelastic
collisions, the momentum is conserved (P f = Pi). In completely inelastic collisions, the objects stick together after the
collision.
Inelastic Collision Equations
o After a collision when objects move together with a common velocity, we can find this velocity:
o
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At the Superbowl, Tom Brady (95-kg) runs at 3.75 m/s and gets sacked by Tedy Bruschi (111-kg) running at
4.10 m/s. They stick together, so what is their velocity after the collision? What are their initial and final
kinetic energies?
First, set Pf = Pi, so m1v1 + m2v2 = (m1 + m2)vf.
Now, find vf:
Now to find their kinetic energies, Ki = ½ m1v12 + ½ m2v22 and Kf = ½(m1 + m2)vf2.
o
A Mercedes of 950-kg going 16 m/s collides with a beat-down Chevrolet of 1300-kg going at 21 m/s in the
same intersection. They collide and stick together. Find the speed and direction of the wrecked vehicles after
the collision. First, if you think of it as a triangle, the x component is m1v1, the y component is m2v2 and the
hypotenuse is (m1 + m2)vf. Therefore, to find the x component of the momentum, m1v1 = (m1 + m2)vf cos θ and
to solve for the y component, m2v2 = (m1 + m2)vf sin θ. We can solve for this algebraically, but now, divide the
y
equation
by
the
x
equation
and
get
Now,
o
can
find
the
speed
using
the
x-direction
equation
for
momentum:
In an elastic collision where m1 is moving with vo initially and m2 is at rest initially, their velocities are:
o
o
we
An apple of 0.130-kg and an orange of 0.160-kg collide with speeds of 1.11 m/s and 1.16 m/s, respectively.
The orange makes an angle of 42o after collision with respect to its original direction of motion. What is the
final speed and direction of the apple assuming an elastic collision? First, find KE i. then KEf in terms of v1f,
then
set
the
KE’s
equal:
The center of mass is one point of any system that has special significance.
o
o
o
The same goes for the YCM coordinate.
Find the center of mass when an arm is held perpendicular to the table when the upper arm has a mass of 2.5-kg
and the CM is 0.18m above the elbow and the lower arm has a mass of 1.6 kg and has a CM 0.40m to the right
of the elbow and the hand has a mass of 0.64-kg and a center of mass 0.40m to the right of the elbow.
To find velocity of the center of mass, replace the x’s with v’s. To find acceleration of the center of mass, replace the x’s
with a’s.
o An air cart of mass m and speed vo moves towards a second, identical air cart at rest. When the carts collide
they stick together and move as one. Find the velocity of the center of mass of this system before and after the
collision. First, use the equation to find the velocity of the mass before collision and then use momentum
conservation in the x-direction to find the speed after the collision. Then calculate the velocity of the center of
mass
of
the
two
carts
after
the
collision.