1. m . a F = F = m . v P = m . v t F.t = m . v P = m . v
New Definition…
F =
m . a
Momentum = mass . velocity
F = m . v
P = m . v t
(Unit = kg . m/s)
F.t =
m . v
P = m . v
“Change in” Momentum =
(This is called Impulse)
 F.t = P
* Impulse is the unbalanced force multiplied by the time of interaction, that causes an object to change its Momentum

2. Momentum
– is a vector!
Which means that momentum can have a Positive OR Negative Direction!

3. Conservation of Momentum
If no outside unbalanced forces (F) act on an object, its momentum remains constant.
Consider a system of 2 or more objects…without any outside forces acting on the system, momentum is constant.

4. Conservation of Momentum
Consider a system where 2 objects collide with each other. If there is no friction or air resistance, the Combined Total Momentum of the objects will be the same before and after the collision

5. Conservation of Momentum
Equation:
Pbefore = Pafter (P = P’)
(mv)A + (mv)B = (mv)’A + (mv)’B A B v v

6. Example: Cart A and B are each 2 kg
Example: Cart A and B are each 2 kg. VA = 6 m/s; VB = 4 m/s in the opposite direction. If Cart A has a velocity of 0.5 m/s after collision, what is the velocity of cart B after the collision? A B v v

7. (mv)A + (mv)B = (mv)’A + (mv)’B
(2 kg)(6 m/s)A + (2 kg)(-4 m/s)B
= (2 kg)(0.5m/s) + (2 kg)(v’B) A B v v

8. (mv)A + (mv)B = (mv)’A + (mv)’B
12 kg m/s - 8 kg m/s
= (2 kg)(0.5m/s) + (2 kg)(v’B) A B v v

9. (mv)A + (mv)B = (mv)’A + (mv)’B
12 kg m/s - 8 kg m/s
= 1 kg m/s + (2kg)(v’B) A B v v

10. (mv)A + (mv)B = (mv)’A + (mv)’B
4 kg m/s
= 1 kg m/s + (2 kg)(v’B)
- 1 kgm/s
3 kgm/s = kg v’B
2 kg
2 kg
v’B = +1.5 m/s A B v v