Question

26. A particle moves in the x - y plane under the
influence of a force such that its linear momentum
is p(t) = [i cos(at) - sin(at)] B. where a and B
are constants. The angle between the force and
the momentum is
(1) 0°
(2) 45°
(3) 60°
(4) 90°​

Answer 1

${\mathfrak{\red{\underline{\underline{Answer:-}}}}}$

$\bf{The\;angle\;between\;force\;and\;momentum\;is\;90^{\circ}.}$

${\mathfrak{\red{\underline{\underline{Explanation:-}}}}}$

The given expression is,

$\bf{\overrightarrow{p(t)}=A \Bigg[\hat{i}\;cos \Bigg(kt \Bigg)-\hat{j}\;sin\;\Bigg(kt \Bigg) \Bigg]}$

$\bf{\overrightarrow{F}=\dfrac{d\overrightarrow{p}}{dt}}$

$\bf{=\dfrac{d}{dt}\;\Bigg[A\;\bigg[ \hat{i}\;cos \bigg(kt \bigg)- \hat{j}\;sin \bigg(kt \bigg)\bigg] \; \Bigg]}$

$\bf{= Ak\; \Bigg[- \hat{i}\;sin \bigg(kt \bigg)- \hat{j}\;cos\;\bigg(kt \bigg) \Bigg]}$

$\bf{Now,}$

$\bf{\overrightarrow{F}\;.\;\overrightarrow{p}=Ak\;\;\Bigg[-\hat{i}\;sin\bigg(kt \bigg)- \hat{j}\;cos\bigg(kt \bigg) \Bigg].A.} \\ \\ \\ \;\bf{\Bigg[\hat{i}\;cos\bigg(kt \bigg)-\hat{j}\;sin\bigg(kt \bigg)\Bigg]}$

$\bf{=A^{2}k\; \Bigg[-sin\;\bigg(kt \bigg)\;cos \;\bigg(kt \bigg)+cos\bigg(kt \bigg)\;sin\;\bigg(kt \bigg) \Bigg]}$

$\bf{\Bigg[Since\; \hat{i}\;.\; \hat{i}=\hat{j}\;.\; \hat{j}=0 \Bigg]}$

${\boxed{\boxed{\bf{Thus,\;angle\;between\;force\;and\;momentum\;is\;90^{\circ}.}}}}$