The velocity of an object with a mass of #6 kg# is given by #v(t)= sin 2 t + cos 4 t #. What is the impulse applied to the object at #t= (5pi)/12 #?

1 Answer
Oct 11, 2016

Answer:

No answer to this

Explanation:

Impulse is #vec J = int_a^b vec F dt #

#= int_(t_1)^(t_2) (d vec p)/(dt) dt #

#= vec p(t_2) - vec p(t_1) #

So we need a time period for there to be an impulse within the definition provided, and the Impulse is the change of momentum over that time period.

We can calculate the momentum of the particle at #t= (5pi)/12# as

#v =6 ( sin (10pi)/12 + cos (20pi)/12 ) = 6 \ kg \ m \ s^(-1)#

But that is the instantaneous momentum.

We can try

#\vec J = lim_(Delta t = 0) vec p(t + Delta t) - vec p(t) #

#= 6 lim_(Delta t = 0) sin 2(t + Delta t) + cos 4(t + Delta t) -sin 2t - cos 4t #

#= 6 lim_(Delta t = 0) sin 2t cos 2 Delta t + cos 2t sin 2 Delta t + cos 4t cos 4 Delta t - sin 4t sin 4 Delta t -sin 2t - cos 4t = 0#

No luck :-(

The next port of call might be the Dirac delta function but I'm not sure where that might lead as it's been a while.