Momentum!

Change in Momentum Review Question 1: One of the smallest planes ever flown was the Bumble Bee II, which had a mass of 1.80 x 10^2kg. If the pilot's mass was 7.0 x 10^1kg, what was the momentum of the plane and pilot while it moves west at 3.00 x 10^2km/hr?

Since we want to find the momentum of the pilot and plane combined, we must add their weights.

Because momentum is measured in kgm/s, we must have velocity in terms of meters per second.  To do this, we use dimensional analysis.

We then get v = 83.33m/sec.

Now that we have mass and velocity, we can use the formula: momentum (p) equals mass (m) times velocity (v). 

p = m * v

p = 250kg * 83.33m/s

p = 20833kgm/s

We then find that momentum equals 20833kgm/s.

Change in Momentum Review Recoil 1: A rocket sits on a launch pad with fuel. The igniter causes the rocket to rise off the pad, oxidizing 100g of fuel and ejecting it out the back of the rocket at -650m/s. After launch, the rocket has a mass of 3kg. What is its speed after launch?

To begin, we find the momentum of the fuel. We are given the mass of the fuel (100g) and velocity of the fuel (-650m/s). In order to plug these in to the formula momentum equals mass times velocity, we must convert grams of fuel to kg (1,000g = 1kg, so 100g = .100kg). 

p = m * v

p = .100kg * -650m/s

p = -65.0kgm/s

When we plug it in, we get momentum equals -65.0kgm/s.

Pfuel = -Procket

-65.9=0kgm/s = -Procket

Procket = 65.0kgm/s

This number gives us the momentum of the fuel. However, since the fuel powers the rocet, its momentum has the same value as the rocket's momentum (but it is positive since it moves in the opposite direction of the fuel). Therefore, the momentum of the rocket also equals 65.0kgm/s. This is exemplified by the Law of Conservation of Momentum.

p = m * v

65.0kgm/s = 3kg * v

v = 21.67m/s

Since we have the rocket's momentum and mass, we can find its velocity using momentum equals mass times velocity (p = m * v) again. When we plug in 65.0kgm/s and 3kg, we find that velocity of the rocket after launch is 21.67m/s.




Impulse Example: Egg Drop Lab

For this lab, we constructed an egg catcher using paper and tape. The egg was then dropped from 1 meter above the ground and lined up to land in the structure.

To find the momentum, you must have the mass and velocity of the egg. We were given the mass (.057kg). In order to find the velocity, you must find how long it took, so we use the y-displacement formula. We use -9.8m/s^2 for acceleration since that is the force of gravity which is pulling the egg down to the Earth. 

Δy = 1/2 * a * t^2 + Voy * t

-1m = 1/2 * -9.8m/s^2 * t^2 + O * t

-1m = -4.9m/s^2 * t^2

-1/4.9 = t^2

t = .452s

Once you solve, you get t = .452s. We then find the velocity using velocity equals change in position over time. The drop was 1m, and t = .452s, so we got v = 2.21m/s. To find the momentum, we use momentum equals mass times velocity.

p = m * v

p = .057kg * 2.21m/s

p = .126kgm/s

We then know that momentum equals .126kgm/s.


In order to find how much impulse was exerted by the floor to stop the egg, we will use the equation force times change in time equsl mass times change in velocity (F * Δt = m *  Δv). We found time and velocity previously (.452s and 2.21m/s, respectively) and were mass (.057kg), so we simply plug them into the equation and solve using algebra.

F Δt = m *  Δv

F * .452s = .057kg * 2.21m/s

F = .28N

We now know that the floor would exert .28N of impulse to stop the egg.

Our apparatus exerts this impulse safely by slowing down the egg as it goes further down into the cone and touches the sides.

To find the amount of average force that our apparatus exerted, we will yet again use that equation, however we will solve using .125s (the average length of time for the drop) as our time.

F *  Δt = m *  Δv

F = (m *  Δv) /  Δt

F = (.057kg * 2.21m/s) / .125s

F = 1.01N

Therefore, our apparatus exerted 1.01N of average force.