JEE Advanced · Physics · 1. Math in Physics
A particle moves in the \(X-Y\) plane under the influence of a force such that its linear momentum is \(\mathbf{p}(t)=A[\hat{\mathbf{i}} \cos k t-\hat{\mathbf{j}} \sin k t]\), where \(A\) and \(k\) are constants. The angle between the force and the momentum is
- A \(0^{\circ}\)
- B \(30^{\circ}\)
- C \(45^{\circ}\)
- D \(90^{\circ}\)
Answer & Solution
Correct Answer
(D) \(90^{\circ}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}& \mathbf{F}=\frac{d \mathbf{p}}{d t}=-k A \sin k t \hat{\mathbf{i}}-k A \cos k t \hat{\mathbf{j}} \\& \mathbf{p}=A \cos k t \hat{\mathbf{i}}-A \sin k \hat{\mathbf{j}}\end{aligned}\)
Since,
F \(\cdot \mathbf{p}=0\)
\(\therefore\) Angle between \(\mathbf{F}\) and pshould be \(90^{\circ}\).
\(\therefore\) Correct option is \((\mathrm{d})\).
Since,
F \(\cdot \mathbf{p}=0\)
\(\therefore\) Angle between \(\mathbf{F}\) and pshould be \(90^{\circ}\).
\(\therefore\) Correct option is \((\mathrm{d})\).
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