Question

The velocity of an object with a mass of `3 kg` is given by `v(t)= - t^2 +4 t`. What is the impulse applied to the object at `t= 5`?

Answer 1

The impulse of an object is associated to a change on its linear momentum, `J = Delta p`.
Let us calculate it for `t=0` and `t=5`.

Explanation:

Let us suppose that the object starts its motion at `t=0`, and we want to calculate its impulse at `t=5`, i.e. the change of linear momentum it has experienced.

Linear momentum is given by: `p = m cdot v`.

• At `t=0`, linear momentum is:
  `p(0) = m cdot v(0) = 3 cdot (-0^2 + 4 cdot 0) = 0`

• At `t=5`, linear momentum is:
  `p(5) = m cdot v(5) = 3 cdot (-5^2 + 4 cdot 5) = -15 " kg" cdot "m/s"`

So impulse finally is given by:

`J = Delta p = p(5) - p(0) = (-15) - (0) = -15 " kg" cdot "m/s"`

The negative sign just means that the object is moving backwards.

P.S.: the vectorial expression is `vec J = Delta vec p`, but we have assumed that object moves only on one direction, and we just take in account the modulus of the magnitudes.