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 #?

Question

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  #?

William Audy · Accepted Answer

No answer to this

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.

Layla Ahmed · Answer

undefined To find the impulse applied to the object at $ t = \frac{5\pi}{12} $, we first need to find the velocity of the object at that time by substituting $ t = \frac{5\pi}{12} $ into the velocity equation:

$ v\left(\frac{5\pi}{12}\right) = \sin(2 \cdot \frac{5\pi}{12}) + \cos(4 \cdot \frac{5\pi}{12}) $

$ v\left(\frac{5\pi}{12}\right) = \sin\left(\frac{5\pi}{6}\right) + \cos\left(\frac{5\pi}{3}\right) $

$ v\left(\frac{5\pi}{12}\right) = \sin\left(\frac{\pi}{6}\right) + \cos\left(\frac{\pi}{3}\right) $

$ v\left(\frac{5\pi}{12}\right) = \frac{1}{2} + \frac{1}{2} $

$ v\left(\frac{5\pi}{12}\right) = 1 $

The impulse applied to the object at $ t = \frac{5\pi}{12} $ is given by the change in momentum, which is the product of mass and change in velocity. Since impulse is a vector quantity, we need to consider the direction as well.

$ \text{Impulse} = \Delta p = m \cdot \Delta v $

Since the object's mass $ m $ is given as 6 kg and the change in velocity $ \Delta v $ is $ 1 - 0 = 1 $, the impulse is $ 6 \times 1 = 6 \, \text{kg m/s} $.