Key Terminology
Momentum: Momentum is a vector quantity in physics for the "mass in motion". Momentum (p) = mass*velocity. It can also be understood as "how difficult it is to stop an object". Momentum is conserved in isolated systems where there is no external net force (no impulse). In non-isolated systems, there must be an impulse and thus final momentum is different from initial momentum.
Impulse: Impulse (J) is a force acting over some time, it is given by J = F*Δt. We know that F = ma, so J = ma*Δt, and a = Δv/t, so J = mΔv or J = Δp, the change in momentum. Thus F*Δt=Δp
Explosion: The rapid release of potential energy into kinetic energy
Impulse: Impulse (J) is a force acting over some time, it is given by J = F*Δt. We know that F = ma, so J = ma*Δt, and a = Δv/t, so J = mΔv or J = Δp, the change in momentum. Thus F*Δt=Δp
Explosion: The rapid release of potential energy into kinetic energy
Representing Momentum: LIL chart
We can represent momentum with momentum bar charts or the LIL chart. The first L shows the initial momentum of all the objects in the system, the I shows the impulse (negative/positive) and the final L shows the final momentum of all the objects in the system. In this example, a big fish of mass 3m with velocity 2m/s eats a small fish of mass m at rest. There is no external net force acted on this system of 2 fish so momentum is conserved. We can use this to find the final velocity of the big fish after it eats the small fish if we know this:
pi = pf
3m*2 + m*0 = (3m+m)*v
6m = (4m)v
v = 1.5 m/s
pi = pf
3m*2 + m*0 = (3m+m)*v
6m = (4m)v
v = 1.5 m/s
Force x Time graphs
Recall that Impulse = Force * time, so the area under a force time graph = Impulse or Δp
In this example, the two cars feel the same Δp because they experience the same force (because of Newton's Third Law) and their contact time is the same. Momentum is transferred from the blue car to the red car.
In this example, the two cars feel the same Δp because they experience the same force (because of Newton's Third Law) and their contact time is the same. Momentum is transferred from the blue car to the red car.
Sometimes it is hard to find the area under the curve. So instead we can find the area under the curve by multiplying the average force and time.
Collisions
There are 4 types of collisions: perfectly inelastic, inelastic, elastic and superelastic (explosions)
Perfectly inelastic (stick together):
Perfectly inelastic collisions are when the two objects merge into one object after colliding. Deformation occurs, thus kinetic energy is lost
E.g. a bullet digs into the flesh
Inelastic:
Deformation occurs but the two objects does not become one. Kinetic energy is lost
E.g. A car crash
Elastic:
No deformation occurs so kinetic energy conserved
E.g. snooker balls collide
Explosions:
Release of potential energy into kinetic energy, kinetic energy increases.
However, in all of those four cases, momentum is conserved. In explosions, the momentum of the CENTER of MASS remains unchanged.
Perfectly inelastic (stick together):
Perfectly inelastic collisions are when the two objects merge into one object after colliding. Deformation occurs, thus kinetic energy is lost
E.g. a bullet digs into the flesh
Inelastic:
Deformation occurs but the two objects does not become one. Kinetic energy is lost
E.g. A car crash
Elastic:
No deformation occurs so kinetic energy conserved
E.g. snooker balls collide
Explosions:
Release of potential energy into kinetic energy, kinetic energy increases.
However, in all of those four cases, momentum is conserved. In explosions, the momentum of the CENTER of MASS remains unchanged.
Collisions in Two Dimensions
- Momentum in the x-direction is conserved
- Total px (before) = Total px (after)
- Momentum in the y-direction is conserved
- Total py (before) = Total py (after)
- Treat x and y coordinates independently
- Ignore x when calculating y
- Ignore y when calculating x
- and Ignore x and y while calculating z (momentum in z-direction also conserved)
Momentum Problem Solving and relating Momentum, Energy, Forces, and Kinematics
This is an example of an elastic collision:
In collisions, momentum is conserved, so pi = pf.
This collision has two dimensions, x and y, and each momentum are conserved.
So pix = pfxw (white ball) + pfxb (black ball) and piy = pfyw + pfyb.
First look at the two balls individually after the collision and find the x and y components.
The black ball
vxb = cos(45)*2 = 2/√2
vyb = sin(45)*2 = 2/√2
The white ball
vxw = cos(Θ)*vw
vyw = sin(Θ)*vw
pix = 4*0.5 = 2 kgm/s = 0.5*vxb + 0.5*vxw = 0.5*2/√2 + 0.5*cos(Θ)*vw
piy = zero (since the cue ball is only moving in the x direction initially) = 0.5*2/√2 + 0.5*sin(Θ)*vw
Dividing by common factor 0.5,
4 = 2/√2 + cos(Θ)*vw and 4-(2/√2) = cos(Θ)*vw
0 = 2/√2 + sin(Θ)*vw and -2/√2 = sin(Θ)*vw
Dividing equation 2 by 1, vw cancels out:
tan(Θ) = (-2/√2)/(4-(2/√2))
Θ = 28.7°
This collision has two dimensions, x and y, and each momentum are conserved.
So pix = pfxw (white ball) + pfxb (black ball) and piy = pfyw + pfyb.
First look at the two balls individually after the collision and find the x and y components.
The black ball
vxb = cos(45)*2 = 2/√2
vyb = sin(45)*2 = 2/√2
The white ball
vxw = cos(Θ)*vw
vyw = sin(Θ)*vw
pix = 4*0.5 = 2 kgm/s = 0.5*vxb + 0.5*vxw = 0.5*2/√2 + 0.5*cos(Θ)*vw
piy = zero (since the cue ball is only moving in the x direction initially) = 0.5*2/√2 + 0.5*sin(Θ)*vw
Dividing by common factor 0.5,
4 = 2/√2 + cos(Θ)*vw and 4-(2/√2) = cos(Θ)*vw
0 = 2/√2 + sin(Θ)*vw and -2/√2 = sin(Θ)*vw
Dividing equation 2 by 1, vw cancels out:
tan(Θ) = (-2/√2)/(4-(2/√2))
Θ = 28.7°
This is an example of an perfectly inelastic collision
In a ballistic pendulum an object of mass mmm is fired with an initial speed v0 at a pendulum bob. The bob has a mass M, which is suspended by a rod of length L and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement θ as shown. Find an expression for v0, the initial speed of the fired object.
There are two distinct physical processes at work in the ballistic pendulum. Momentum is conserved in the collision and mechanical energy is conserved in the following swing.
The collision:
mv0 = (M+m)v
v0 = (M+m)v/m
The swing:
KE = Ug
0.5 (M+m)v^2 = (M+m)gh
The collision:
mv0 = (M+m)v
v0 = (M+m)v/m
The swing:
KE = Ug
0.5 (M+m)v^2 = (M+m)gh
0.5 (M+m)v^2 = (M+m)g(L-Lcosθ)
v = √2g(L-Lcosθ)
So v0 = (M+m)√2g(L-Lcosθ)/m
v = √2g(L-Lcosθ)
So v0 = (M+m)√2g(L-Lcosθ)/m
Collision involving force
A 70-kg astronaut pushes against the inside back wall of a 2000-kg spaceship and moves toward the front. Her speed increases from 0 to 1.2 m/s.
If the astronaut is the system, there is an external force acting on him and momentum is not conserved, we can write that:
pi + J = pf
0 + Favg * t = 1.2*70
If her push lasts 0.30 s, what is the average force that the astronaut exerts on the wall of the spaceship?
Favg = 1.2*70/0.30 = 280N
If the astronaut is the system, there is an external force acting on him and momentum is not conserved, we can write that:
pi + J = pf
0 + Favg * t = 1.2*70
If her push lasts 0.30 s, what is the average force that the astronaut exerts on the wall of the spaceship?
Favg = 1.2*70/0.30 = 280N
Collision and kinematics
You can use kinematic equations to connect kinematics (distance, acceleration) to velocity, which is useful since momentum depends on velocity. For example a 0.4kg apple falls from 4 meters to the ground and you want to find the momentum right before it hits the ground. From the kinematic equation vf^2 = vi^2 + 2ax, we can find that
vf^2 = 0 + 2g(4)
vf = √8g
And momentum = 0.4 * √8g = 3.49 kg*m/s
vf^2 = 0 + 2g(4)
vf = √8g
And momentum = 0.4 * √8g = 3.49 kg*m/s
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