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 #?

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Answer 1 · 2016-06-04T22:04:26

It is #-6.69# #kgm/s#.

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#.