# The velocity of an object with a mass of 4 kg is given by v(t)= sin 3 t + cos 7 t . What is the impulse applied to the object at t= (5pi)/12 ?

Jun 4, 2016

It is $- 6.69$ $k g \frac{m}{s}$.

#### Explanation:

In classical mechanics and with constant mass, the instantaneous impulse is defined as the momentum

$p = m v$

$p = 4 \cdot \left[\sin \left(3 t\right) + \cos \left(7 t\right)\right]$

and this has to be evaluated when $t = \frac{5 \pi}{12}$

then

$p = 4 \cdot \left[\sin \left(3 \cdot \frac{5 \pi}{12}\right) + \cos \left(7 \cdot \frac{5 \pi}{12}\right)\right]$

$= 4 \cdot \left[\sin \left(\frac{5}{4} \pi\right) + \cos \left(\frac{35}{12} \pi\right)\right]$

we have that $\sin \left(\frac{5}{4} \pi\right) = - \sin \left(\frac{\pi}{4}\right)$ because the angle is $\pi + \frac{\pi}{4}$ and $\cos \left(\frac{35}{12} \pi\right) = \cos \left(\frac{11}{12} \pi\right)$ because the angle is $2 \pi + \frac{11}{12} \pi$.
So we have

$p = 4 \cdot \left[- \sin \left(\frac{\pi}{4}\right) + \cos \left(\frac{11}{12} \pi\right)\right]$

$= 4 \cdot \left(- \frac{1}{\sqrt{2}} - \frac{1 + \sqrt{3}}{2 \sqrt{2}}\right)$

$= 4 \cdot \left(- \frac{3 + \sqrt{3}}{2 \sqrt{2}}\right)$

$= - \frac{6 + 2 \sqrt{3}}{\sqrt{2}}$

$= - \frac{6 \sqrt{2} + 2 \sqrt{6}}{2}$

$= - 3 \sqrt{2} - \sqrt{6} = - \sqrt{2} \left(3 + \sqrt{3}\right) \setminus \approx - 6.69$ $k g \frac{m}{s}$.