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QJEEAdv07Ph4526: JEE Advanced 2007 Physics - Class 11 JEEAdv07Ph4526 Solution: A particle moves in the X - Y plane under the influence of a…

A particle moves in the X - Y plane under the influence of a force such that its linear momentum is p→(t)=A[i^cos⁡(kt)−j^sin⁡(kt)]\overrightarrow p \left( t \right) = A\left[ {\widehat i\cos (kt) - \widehat j\sin (kt)} \right] , where A and k are constants. The angle between the force and the momentum is

Multiple Choice Options

  • A 0∘0^\circ
  • B 30∘30^\circ
  • C 45∘45^\circ
  • D 90∘90^\circ

Step-by-Step Verified Solution

Given that,

P→(t)=A[i^cos⁡(kt)−j^sin⁡(kt)]\overrightarrow{\mathrm{P}}(t)=\mathrm{A}[\hat{i} \cos (k t)-\hat{j} \sin (k t)]

where, A\mathrm{A} and k=k= constant

 Force F→)amp;=dP→dt( from Newton’s 2nd  law) ⇒F→amp;=ddt[ A(i^cos⁡(kt)−j^sin⁡(kt))]amp;=Ak[i^cos⁡(kt)−j^sin⁡(kt)]\begin{aligned}\text { Force }\overrightarrow{\mathrm{F}}) & =\frac{d \overrightarrow{\mathrm{P}}}{d t}\left(\text { from Newton's } 2^{\text {nd }}\right. \text { law) } \\\\\Rightarrow \quad \overrightarrow{\mathrm{F}} & =\frac{d}{d t}[\mathrm{~A}(\hat{i} \cos (k t)-\hat{j} \sin (k t))] \\\\& =\mathrm{A} k[\hat{i} \cos (k t)-\hat{j} \sin (k t)]\end{aligned}

By applying dot product between F⃗&P⃗\vec{F} \& \vec{P} we have

amp;F→⋅P→=0amp;⇒∣F⃗∣∣P⃗∣cos⁡θ=0amp;⇒cos⁡θ=0amp;Hence,θ=90∘\begin{aligned}& \overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{P}}=0 \\\\& \Rightarrow|\vec{F}||\vec{P}| \cos \theta=0 \\\\& \Rightarrow \quad \cos \theta=0 \\\\& \text {Hence,} \theta=90^{\circ}\end{aligned}

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