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 ?

1 Answer
Jun 4, 2016

It is -6.69 kgm/s.

Explanation:

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

p=mv

In your case

p=4*[sin(3t)+cos(7t)]

and this has to be evaluated when t=(5pi)/12

then

p=4*[sin(3*(5pi)/12)+cos(7*(5pi)/12)]

=4*[sin(5/4pi)+cos(35/12pi)]

we have that sin(5/4pi)=-sin(pi/4) because the angle is pi+pi/4 and cos(35/12pi)=cos(11/12pi) because the angle is 2pi+11/12pi.
So we have

p=4*[-sin(pi/4)+cos(11/12pi)]

=4*(-1/sqrt(2)-(1+sqrt(3))/(2sqrt(2)))

=4*(-(3+sqrt(3))/(2sqrt(2)))

=-(6+2sqrt(3))/(sqrt(2))

=-(6sqrt(2)+2sqrt(6))/2

=-3sqrt(2)-sqrt(6)=-sqrt(2)(3+sqrt(3))\approx-6.69 kgm/s.