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 #?

1 Answer
Jan 3, 2016

Answer:

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.