Answer:
The average force times the time of interaction equals the change in momentum
Step-by-step explanation:
The impulse-momentum theorem states that impulse (equal to the average force times the time of interaction) is equal to the change in momentum:
![I=F \Delta t = \Delta p](https://img.qammunity.org/2020/formulas/physics/middle-school/m0dotavr9f00r343p4xprtisyfjcjp80kt.png)
where
I is the impulse
F is the average force
is the time of the interaction
is the change in momentum
We can demonstrate this theorem, by re-writing the force as mass (m) times acceleration (a):
![F \Delta t = ma \Delta t](https://img.qammunity.org/2020/formulas/physics/middle-school/d2evn2batqd9q1g9jrxndgd3e16854ifnj.png)
The acceleration is equal to the change in velocity divided by the time interval:
![ma \Delta t = m (\Delta v)/(\Delta t) \Delta t](https://img.qammunity.org/2020/formulas/physics/middle-school/4tlvlpaj03r49xl2rwvpvi04hugaziem4l.png)
And by simplifying
, we get:
![= m \Delta v = \Delta p](https://img.qammunity.org/2020/formulas/physics/middle-school/v7bl6naqsu1ayo69s1267pj8x1n9kd22hw.png)
which is equal to the change in momentum.