Answer:
impulse acting on it
Step-by-step explanation:
The impulse is defined as the product between the force applied to an object (F) and the time interval during which the force is applied (
):
![I=F\Delta t](https://img.qammunity.org/2020/formulas/physics/middle-school/lfdlnaamxvqpz93is4fg5bakjz3523dz03.png)
We can prove that this is equal to the change in momentum of the object. In fact, change in momentum is given by:
![\Delta p = m \Delta v](https://img.qammunity.org/2020/formulas/physics/middle-school/m7m9y1nkh3wackscg6a3gwh5ijb8vcwhg7.png)
where m is the mass and
is the change in velocity. Multiplying and dividing by
, we get
![\Delta p = m (\Delta v)/(\Delta t) \Delta t](https://img.qammunity.org/2020/formulas/physics/middle-school/6r5oifhmcgq7qrampghjzhifd101y9qote.png)
and since
is equal to the acceleration, a, we have
![\Delta p = ma \Delta t](https://img.qammunity.org/2020/formulas/physics/middle-school/i4aeqs513ktj7a6bst1iwzqibnwwrvxndi.png)
And since the product (ma) is equal to the force, we have
![\Delta p = F \Delta t](https://img.qammunity.org/2020/formulas/physics/middle-school/jxwgaglwhmgh8lr8e1md3a959d161rsq9n.png)
which corresponds to the impulse.