updated: Nov. 2014
Course: sph 4U1
Unit: momentum
Lesson 1: Title: Conservation of momentum
Demos: full pop cans rolling at each other; also pop cans and bottles of water/pop
Lesson:
Where have you heard of momentum? – like inertia – something keeps on going. Inertia is actually the same as mass.
The most important thing about momentum is that it is conserved.
What does ‘conserved’ mean in physics? (What does the law of conservation of energy tell us?)
We can state the conservation of momentum as:
Momentum cannot be created or destroyed (in any closed system)
the total momentum in a closed system is constant
the initial momentum = the final momentum (in any closed system)
When you find a conservation law in physics, it is like finding a diamond. You can use it to solve so many problems.
What other quantities are conserved? mass, energy, charge
Symbol: (I don’t know why it’s p)
Formula:this is a vector equation
Units: there are no special units so …
Units: kg m/s
Momentum problems almost always involve collisions. The problems generally look like this:
crazy stuff happens with forces, acceleration, deformations. Normally some energy is ‘lost’.
We ignore this stage / the object(s) move at constant velocity again after the collision is over
initial momentumfinal momentum
The initial momentum is always equal to the final momentum, regardless of what happens during the collision. pi = pf
Common mistake: thinking that velocities are conserved. They’re not!
You have to multiply velocity by mass.
Momentum in 1 dimension (linear)
As always, in 1 dimension we can leave off the vector symbols and just make sure we use + and – when appropriate.
Examples
1. A 1000 kg car travelling at 80 km/hr collides with a stationary 5000 kg truck. What is their final speed if they stick together? (will the final speed be greater or lesser than 80 km/hr?)
pi = pf
m1v1 + m2v2 = (m1+m2)vf
(1000kg)(80 km/hr) + (5000 kg)(0) = (6000kg)vf
vf = 80000/6000 km/hr
= 13.3 km/hr
<This is the first simple type of problem that you can solve with momentum. Learn to recognize the types of problems. >
2. A 300g ball moving at +3 m/s collides head-on with a 400 g ball moving towards it at 5 m/s. If the first ball rebounds at -2 m/s, what is the final velocity of the second ball?
(head-on = a one dimensional or linear collision)
How to diagram:
before
m1 v1 v2 m2
after
v1' m1m2
pi = pf
m1v1 + m2v2 = m1v1' + m2v2'NOTE: we use ' (prime) to mean after the collision.
(300g)(3m/s) + (400g)(-5m/s) = (300g)(-2m/s) + 400g v2'
…
v2' = – 1.25 m/s
Homework:
Nelson p 243 #1, 2, 3, 5,6,7,9
(rest of class – finish up torque lab, etc. and hand them in)
Giancoli p 187 #1,2,4,5,7
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Problem:
You can’t move a car from the inside – pushing on it. Right? There must be an external unbalanced force.
Consider an astronaut in space. He throws a 20 kg hammer away from him at 15 m/s. He weighs 100 kg. What is his speed?
0 = m1v1 + m2v2
= 20(-15) + 100(v2)
v2 = 3 m/s
Now the astronaut is in a large, light rectangular box, weighing 50 kg. He is right up against the end of the box and throws the hammer towards the other end, which is 100 m away. What happens as the hammer floats though the vacuum to the other end? Calculate the speed of all objects. Will the box move? Yes, but how is it possible?
0 = m1v1 + m2v2
= 20(-15) + 150(v2)
v2 = 2 m/s An external observer sees the box start moving at 2 m/s to the right!
How long does it take the hammer to reach the other end? Less than 10 seconds, because the end of the box is moving closer!
What speed does the astronaut see? Hammer moves away at 10 m/s and he moves away at 2 m/s 12m/s.
How far does the box move? d = vt. We need to find t!
What speed does the hammer see? It is moving left at 10 m/s w.r.t. external observer, and the box is moving right at 2m/s. The hammer’s speed relative to the box is also 12 m/s.
t = 100 m / (8 m/s) = 12.5 sec. X this is more than 10 sec. it must be wrong.
t = 100 m / (12 m/s) = 8.33 sec
The box will move a distance of 2 m/s * 8.33 s = 16.7 m before the hammer hits the other end.
Now what happens when the hammer hits the other end?
pi (moving) = pf(hammer hits other end) * from external view point.
m1v1 + m2v2 = mtotal (vfinal)* mtotal, because once the hammer hits the other end,
20(-15) + 150(2) = 170 (vf) the hammer, box, and astronaut are all connected
-300 + 300 = 170 vf and touching each other.
0 = 170 vf
.: vf = 0
What does the astronaut see? What does an external observer see?
Has the box moved? Yes. How can this be? ** The centre of mass is still in the same place.
So you can move a box or a boat from the inside – e.g. walking to the front of a canoe. As long as friction is extremely low and the masses are small.