1. Momentum and Impulse

2. Pretest
A 200kg spacecraft collides with a 400kg spacecraft at 1.85m/s. The two crafts stick together.
How fast is the combined spacecraft travelling now?
What type of collision was this?
Was energy conserved?
A skateboarder(60kg) travelling 4m/s leaps forward off her skateboard(1.5kg) giving herself a speed of 4.7m/s.
What was the impulse on her?
What was the impulse on the skateboard?
How fast is the skateboard going after she jumps off?
What type of “collision” is this?

3. Introduction to Momentum
Momentum is related to how hard it is to stop something. The heavier something is, and the faster it is going, the more momentum it has.
Momentum (p) is the product of mass and velocity
This is normally written p = mv
(Bolded letters denote vectors)
p = mv = kg•m/s or kilogram meters per second
In energy kg•m2/s2 is a Joule (J). Momentum doesn’t have this type of combined unit name.
What is the momentum of a bee that weighs 10 grams and flies at 2 m/s up?
10g=.01kg
p=mv=.01x2= (0.02kgm/s, 90 degrees)
How does that compare to a tortoise that weighs 1kg and moves at .05m/s left?
p=mv=1x.05= (0.05kgm/s, 180 degrees)
The tortoise has more momentum (larger magnitude).

4. Calculating Total Momentum
It is very useful to be able to calculate the total momentum of a system. It will help us use momentum conservation.
What’s the total momentum of the bee and the tortoise from before?
pbee = (0.02kgm/s, 90 degrees)
pbee = (0.05kgm/s, 180 degrees)
Momentum is a vector so we need to do vector addition.
ptotal = Σp = p1 + p2 + p3 + …
0.05kgm/s
0.02kgm/s + =
0.02
0.05
Pythagorean Theorem:
( )0.5 = kgm/s
Use arctan to find the angle:
Tan-1(0.02/0.05) = degrees
180 – = degrees
Answer: (0.0538kgm/s, degrees)

5. Momentum Conservation
In energy we saw that if no energy was added or lost, the energy would be conserved, Ei + W = Ef
The same is true for momentum. pi + J = pf
J is momentum added or taken away from the system.
If there are no outside forces acting on an system, the momentum of that system remains constant; it is conserved.
A 10 kg object moves at a constant velocity 2 m/s to the right and collides with a 6 kg object moving at a velocity 5 m/s to the left. Ignore outside forces.
What are the initial and final momentum of the system?
Step 1: Givens/Unknowns
m1 = 10kg, v1 = 2m/s, m1 = 6kg, v1 = -5m/s, J = 0 because there are no outside forces
Step 2: Equations
p = mv
pi + J = pf
Step 3: Solve
pi = m1v1i + m2v2i
pi = 10*2 + 6*(-5) = -10kgm/s = (10kgm/s, 180 degrees)
pi + 0 = pf
pf = (10kgm/s, 180 degrees)

6. Momentum Conservation Example:
Train Car Collision
A freight car A with a mass of 24,000 kg moves at a constant velocity of 8 m/s on a horizontal railroad track and collides with an empty stationary car B with a mass of 8,000 kg. After the collision the car A is moving at 3 m/s in the same direction. What is the velocity of car B after the collision?
Step 1: Givens/Unknowns
m1 = 24,000kg
v1i = 8m/s
v1f = 5m/s
m2 = 10,000kg
v2i = 0m/s
v2f = ?
Step 2: Equations
p = mv
pi + J = pf
Step 3: Solve
p1i + p2i + J = p1f +p2f
(Note J = 0 because there are no outside forces)
m1v1i + m2v2i = m1v1f + m2v2f
24,000*8 + 10,000*0 = 24,000*5 + 10,000*v2f
v2f = 7.2 m/s
Step 4: Solution Statement
7.2m/s seems like a reasonable speed.
Cart A lost momentum, and cart B received it. However, because Cart A is more massive, a change of 3m/s for Cart A resulted in a much larger change for cart B.

7. Collision Types Momentum is conserved in all collisions.
Collisions are categorized by change in energy of the system due to the collision.
Elastic Collisions:
This is a “bouncy” collision, one where objects bounce off each other.
Energy is conserved: KEi = KEf
Example: A bowling ball hits a bowling pin.
Inelastic Collisions:
This is a “non-bouncy” collision.
Energy is lost: KEi > KEf
When two objects collide and stick together it is a perfectly inelastic collision.
Otherwise the collision is a partially inelastic collision.
Example: An arrow hits and gets stuck in a flying target (perfectly inelastic).
Example: A beanbag hits a cardboard box (partially inelastic).
Superelastic “Collisions”:
These are explosive situations.
Energy is gained: KEi < KEf
Example: A gun fires a bullet.

8. Collision Example
A loaded freight car A with a mass of 24,000 kg moves at a constant velocity of 8 m/s on a horizontal railroad track and collides with car B (v = 2m/s) with a mass of 8,000 kg. After the collision the cars stick to each other and move like one object.
What type of collision is this?
What is the velocity of two cars after the collision?
This is an perfectly inelastic collision because the two cars stick together after colliding.
Step 1: Givens/Unknowns
m1 = 24,000kg
v1i = 8m/s
m2 = 8,000kg
v2i = 2m/s
v1f = v2f = vf = ?
The two cars move together so
they have the same final v
Step 2: Equations
p = mv
pi + J = pf (No outside forces, J = 0)
Step 3: Solve
p1i + p2i = p12f
m1v1i + m2v2i = (m1 + m2) vf
24,000*8 + 8,000*2 = (24,000+8,000)vf
vf = 6.5m/s
Step 4: Solution Statement
When two objects collide and stick together, we expect their velocities to be somewhere between that of the faster one and that of the slower one. 6.5 m/s seems like a reasonable speed.

9. Momentum Conservation in 2D
Momentum is conserved separately in the x and y directions.
The ball (1kg) on the left is travelling right at 6m/s when it collides with a ball(1.5kg) travelling upward at 2m/s. After the collision, the two balls stick together
Determine the direction and speed of the combined mass.
Step 3: Solve
First solve for vf in the x, then in the y, then combine them
p1ix + p2ix = p12fx
m1v1ix + m2v2ix = (m1 + m2) vfx
1* *0 = ( )vfx
vfx = 2.4 m/s
p1iy + p2iy = p12fy
m1v1iy + m2v2iy = (m1 + m2) vfy
1* *2 = 2.5 vfy
vfy = 1.2m/s
vf2 = vfx2 + vfy2 θ = tan-1(vfy/vfx)
vf2 = θ = tan-1(1.2/2.4)
vf = 2.683m/s θ = degrees
Step 4: Solution Statement
The result is up to the right, which makes sense, since those were the directions of the original objects. The speed is also smaller than the sum of the two original speeds, so that makes sense.
Step 1: Givens/Unknowns
m1 = 1kg
v1ix = 6m/s
v1iy = 0m/s
m2 = 1.5kg
v2ix = 0m/s
v2iy = 2m/s
v12f = ?
Step 2: Equations
p = mv
pi + J = pf (No outside forces, J = 0)

10. Impulse: Change in Momentum
We saw that energy could be added or removed from the system by doing positive or negative work. W = Fdcos(θ)
Similarly, momentum can added or removed by applying an impulse (J or I) to the system. Don’t confuse the variable J with the unit for energy Joules.
The equation for impulse is J = Ft
Note that impulse is a vector in the same direction as the force that created it.
A shopper applies a force of 7.5N to a shopping cart for 3 seconds, causing it to speed up from 1.2m/s to 1.8m/s.
What is the mass of the cart?
Step 1: Givens/Unknowns
F = 7.5N
t = 3s
vi = 1.2m/s
vf = 1.8m/s
m = ?
Step 2: Equations
p = mv
pi + J = pf
J = Ft
Step 3: Solve
mvi + Ft = mvf
m(vf – vi) = Ft
m( ) = 7.5*3
m = 37.5kg

11. Impulse: Change in Momentum
J = Δp = mΔv = Ft
A bowling ball (5kg) moving at 5m/s bowls through a set of bowling pins. Right before the ball falls down the shoot it is going 3m/s
What is the impulse the bowling pins provide to a bowling ball?
J = mΔv
J = m*(vfinal – vinitial)‏
J = 5*(3-5)=-10kgm/s
This shows that the force was in the negative direction and slowed down the ball
In football, a field goal is kicked with the football initially at rest. The football (300g) is kicked at 25m/s.  The player's foot was in contact with the ball for 0.1 seconds.
What was the impulse on the ball?
What is the average force during the time of contact?
300g=.3kg J = mΔv = m*(Vfinal-Vinitial) = .3*(25-0) = 7.5kgm/s J = FΔt = F * F= 75 N

12. Energy and Momentum Conservation
Energy and Momentum conservation are powerful tools to solve problems, but it can be difficult to know when to use which.
In general, you want to use momentum conservation for collisions where there are no major outside forces on the objects involved. Energy is only conserved in elastic collisions which are rare; typically you would need to be told that energy was conserved.
In general, you want to use energy conservation to solve problems with changes in height. Momentum will not be conserved in these cases because gravity applies an external force on the system.
Examples:
A car rolls down a hill without friction.
Energy is conserved. There is no work done or energy lost to heat. The normal force is always at 90 degrees to the distance, and the force of gravity is already accounted for with PEg.
Potential energy is completely converted to kinetic energy since no energy is lost to friction or heat.
Momentum is not conserved. There was an impulse due to the outside force gravity. Fgx is an unbalanced force that is applied for the time it takes the car to go down the hill.
It’s easy to see that the total momentum changes. The car, which is the only object in the system, speeds up as it rolls down the hill.

13. Energy and Momentum Conservation Examples continued
A bullet imbeds itself in a block which is free to move.
Energy is not conserved. This is an inelastic collision and energy is converted into heat and sound. The kinetic energy of the system decreased without an equal increase in potential energy.
Momentum in conserved. Both objects in the collision are free from outside forces. As with all collisions where the objects are not attached to things outside of the system, the initial momentum will equal the final momentum.
A ball hits the wall bouncing back with a slower speed.
Energy is not conserved. We can see that kinetic energy has decreased without an increase in potential energy. This energy was lost to heat.
Momentum is not conserved. The initial momentum was positive, but the final momentum was negative. There was an outside impulse caused by the normal force applied by the wall.

14. Energy and Momentum Combined
Sometimes different sections of a problem will use different conservation laws for different parts of the problem.
Ballistic pendulums were devices used to measure the speed of bullets.
A bullet would collide with a pendulum(AB) and the pendulum would swing upward (BC).
Momentum is conserved in the collision (AB) but energy is not.
Energy is conserved as the pendulum swings (BC) but momentum is not.
By combining the two principles, we can work backward to determine the velocity of the bullet.

15. Energy and Momentum Combined
A 10g bullet strikes a 3.090kg block-pendulum, causing it to swing up to a height of 25cm.
How fast was the bullet travelling when it hit the block?
Step 1: Givens/Unknowns
m1 = 10g = 0.01kg
m2 = 3.09kg
hC = 25cm = 0.25m
v1A = ?
Note: v1A denotes the velocity of object 1 in situation A
Step 2: Equations
KE = ½ mv2
PE = mgh
Ei + W = Ef
p =mv
pi + J = pf
Step 3: Solve
We want to work backwards from C to A
Use energy conservation from B to C to find vB
Then use momentum conservation from A to B to find v1A
EB + W = EC
KEB = PEC
½ mvB2 = mghC
vB = (2*9.8*0.25)0.5
vB = 2.214m/s
pA + J = pB
m1v1A + m2v2A = (m1 +m2) vB
0.01*v1A + 0 = ( )*2.214
v1A = m/s
This is very fast, but bullets are also very fast.