Momentum for AP Physics B & C
Practice problems for these concepts can be found at: Momentum Practice Problems for AP Physics B & C
Momentum and Impulse
The units of momentum are kg.m/s which is the same as N·s. Momentum is a vector quantity, and it is often abbreviated with a p.
Impulse (designated as I )is an object's change in momentum. It is also equal to the force acting on an object multiplied by the time interval over which that force was applied. The above equation is often referred to as the "impulse–momentum theorem."
The FΔt definition of impulse explains why airbags are used in cars and why hitting someone with a pillow is less dangerous than hitting him or her with a cement block. The key is the Δt term. An example will help illustrate this point.
This is a multistep problem. We start by calculating the man's velocity the instant before he hits the ground. That's a kinematics problem, so we start by choosing a positive direction—we'll choose "down" to be positive—and by writing out our table of variables.
We have three variables with known values, so we can solve for the other two. We don't care about time, t, so we will just solve for vf.
v2f= v20 + 2a(x – x0).
vf= 7.7 m/s.
Now we can solve for the man's momentum the instant before he hits the ground.
p = mv = (70)(7.7) = 540 kg.m/s.
Once he hits the ground, the man quickly comes to rest. That is, his momentum changes from 540 kg. m/s to 0.
I = Δp = Pf – p0.
I = –540 N·s = FΔt.
If the man does not bend his knees, then
–540 = F(0.003 s).
F = –180,000 N.
The negative sign in our answer just means that the force exerted on the man is directed in the negative direction: up.
Now, what if he had bent his knees?
I = –540 = FΔt.
–540 = F(0.10 seconds).
F = –5400 N.
If he bends his knees, he allows for his momentum to change more slowly, and as a result, the ground exerts a lot less force on him than had he landed stiff-legged. More to the point, hundreds of thousands of newtons applied to a person's legs will cause major damage—this is the equivalent of almost 20 tons sitting on his legs. So we would assume that the man would bend his knees upon landing, reducing the force on his legs by a factor of 30.
Conservation of Momentum
Momentum in an isolated system, where no net external forces act, is always conserved. A rough approximation of a closed system is a billiard table covered with hard tile instead of felt. When the billiard balls collide, they transfer momentum to one another, but the total momentum of all the balls remains constant.
The key to solving conservation of momentum problems is remembering that momentum is a vector.
Let's begin by drawing a picture.
Momentum is conserved, so we write
The tick marks on the right side of the equation mean "after the collision." We know the momentum of each space traveler before the collision, and we know the UFO's final momentum. So we solve for the satellite's final velocity. (Note that we must define a positive direction; because the UFO is moving to the left, its velocity is plugged in as negative.)
Now, what if the satellite and the UFO had stuck together upon colliding? We can solve for their final velocity easily:
Practice problems for these concepts can be found at: Momentum Practice Problems for AP Physics
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