JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A particle moves in \(\mathrm{x}-\mathrm{y}\) plane under the influence of a force \(\vec{F}\) such that its linear momentum is \(\vec{P}(t)=\hat{i} \cos (k t)-\hat{j} \sin (k t)\). If \(k\) is constant, the angle between \(\overrightarrow{\mathrm{F}}\) and \(\overrightarrow{\mathrm{P}}\) will be _______.
- A \(\frac{\pi}{2}\)
- B \(\frac{\pi}{6}\)
- C \(\frac{\pi}{4}\)
- D \(\frac{\pi}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{P}}=\cos (\mathrm{kt}) \hat{\mathrm{i}}-\sin (\mathrm{kt}) \hat{\mathrm{j}} ;|\overrightarrow{\mathrm{P}}|=1\) \(\because \overrightarrow{\mathrm{P}}=\mathrm{m} \overrightarrow{\mathrm{v}}\) \(\therefore \hat{\mathrm{P}}=\hat{\mathrm{v}}\)…
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