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JEE Advanced 2007Paper 2
Exam
JEE Advanced
Year
2007
Paper
Paper 2
Subject
Physics
Chapter
Mathematics in Physics

JEE Advanced 2007 · Paper 2 · Physics · Mathematics in Physics

A particle moves in the X-Y plane under the influence of a force...

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

Options

  1. A0∘0^{\circ}
  2. B30∘30^{\circ}
  3. C45∘45^{\circ}
  4. D90∘90^{\circ}

Correct answer

D. 90∘90^{\circ}

Step-by-step solution

F=dpdt=−kAsin⁡kti^−kAcos⁡ktj^p=Acos⁡kti^−Asin⁡kj^\begin{aligned}& \mathbf{F}=\frac{d \mathbf{p}}{d t}=- k A \sin k t \hat{ \mathbf{i} }-k A \cos k t \hat{\mathbf{j}} \\ & \mathbf{p}= A \cos k t \hat{\mathbf{i}}-A \sin k \hat{\mathbf{j}} \end{aligned}Since, F ⋅p=0∴\cdot \mathbf{p}=0\therefore Angle between F\mathbf{F} and pshould be 90∘90^{{}^{\circ}}. ∴\therefore Correct option is (d)(\mathrm{d}).
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