The velocity of an object with a mass of #4 kg# is given by #v(t)= sin 3 t + cos 6 t #. What is the impulse applied to the object at #t= pi /3 #?

Question

The velocity of an object with a mass of #4   kg# is given by #v(t)= sin 3 t + cos  6 t  #.  What is the impulse applied to the object at #t=  pi /3  #?

Charlotte Aaron · Accepted Answer

The impulse is #-12# Newton seconds.

We know that impulse is change in momentum. Momentum is given by #p =mv#, therefore impulse is given by #J = mDeltav# So we want to find the rate of change, or the derivative of the velocity function, and evaluate it at time #pi/3#. #v'(t) = 3cos(3t) - 6sin(6t)# #v'(pi/3) = 3cos(3(pi/3)) - 6sin(6(pi/3))# #v'(pi/3) = -3# Then we have #J = mDelta v# #J= 4(-3)# #J = -12 kg" "Ns# Hopefully this helps!

Leo Abeyta · Answer

undefined To find the impulse applied to the object at $ t = \frac{\pi}{3} $, we first need to find the velocity at that time by plugging $ t = \frac{\pi}{3} $ into the velocity equation. Then, we can find the change in momentum, which is equal to the impulse.

The velocity at $ t = \frac{\pi}{3} $ is:
$$ v\left(\frac{\pi}{3}\right) = \sin(3 \cdot \frac{\pi}{3}) + \cos(6 \cdot \frac{\pi}{3}) $$
$$ = \sin(\pi) + \cos(2\pi) $$
$$ = 0 + 1 = 1 $$

The impulse applied to the object at $ t = \frac{\pi}{3} $ is equal to the change in momentum. Since impulse is the integral of force with respect to time, and momentum is the integral of velocity with respect to time, impulse equals the change in momentum.

Since impulse $ J = \Delta p $, and momentum $ p = mv $ where $ m $ is mass and $ v $ is velocity, the impulse is $ J = mv - mu $.

Thus, the impulse applied to the object at $ t = \frac{\pi}{3} $ is $ J = m \cdot v\left(\frac{\pi}{3}\right) - m \cdot v(0) $, where $ v(0) $ is the initial velocity. Given that $ m = 4 $ kg and $ v(0) = v(0) = \sin(0) + \cos(0) = 0 + 1 = 1 $, we can calculate:

$$ J = 4 \cdot 1 - 4 \cdot 1 = 0 $$

So, the impulse applied to the object at $ t = \frac{\pi}{3} $ is $ 0 $ Ns (Newtons seconds).