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Laws of Motion question

2024 · 5 Apr · Shift 2 · Q63
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  5. /2024 · 5 Apr · Shift 2 · Q63

Laws of Motion question

2024 · 5 Apr · Shift 2 · Q63

JEE MainPhysicsLaws of MotionMCQ+4 / −1
A particle moves in xxx-yyy plane under the influence of a force F⃗\vec{F}F such that its linear momentum is p→(t)=i^cos⁡(kt)−j^sin⁡(kt)\overrightarrow{\mathrm{p}}(\mathrm{t})=\hat{i} \cos (\mathrm{kt})-\hat{j} \sin (\mathrm{kt})p​(t)=i^cos(kt)−j^​sin(kt). If k\mathrm{k}k is constant, the angle between F→\overrightarrow{\mathrm{F}}F and p→\overrightarrow{\mathrm{p}}p​ will be :
  1. A
    π2\frac{\pi}{2}2π​
  2. B
    π3\frac{\pi}{3}3π​
  3. C
    π4\frac{\pi}{4}4π​
  4. D
    π6\frac{\pi}{6}6π​
View written solutionFree

Correct answer: A

  1. The given linear momentum is
p⃗(t)=i^cos⁡(kt)−j^sin⁡(kt)\vec p(t)=\hat i\cos(kt)-\hat j\sin(kt)p​(t)=i^cos(kt)−j^​sin(kt)

So its components are

px=cos⁡(kt),py=−sin⁡(kt)p_x=\cos(kt),\qquad p_y=-\sin(kt)px​=cos(kt),py​=−sin(kt)
  1. Force is the time rate of change of momentum:
F⃗=dp⃗dt\vec F=\frac{d\vec p}{dt}F=dtdp​​

Differentiate component-wise:

ddt[cos⁡(kt)]=−ksin⁡(kt)\frac{d}{dt}[\cos(kt)]=-k\sin(kt)dtd​[cos(kt)]=−ksin(kt)

and

ddt[−sin⁡(kt)]=−kcos⁡(kt)\frac{d}{dt}[-\sin(kt)]=-k\cos(kt)dtd​[−sin(kt)]=−kcos(kt)

Hence,

F⃗=−ksin⁡(kt) i^−kcos⁡(kt) j^\vec F=-k\sin(kt)\,\hat i-k\cos(kt)\,\hat jF=−ksin(kt)i^−kcos(kt)j^​
  1. To find the angle between F⃗\vec FF and p⃗\vec pp​, compute their dot product:
p⃗⋅F⃗=(cos⁡kt)(−ksin⁡kt)+(−sin⁡kt)(−kcos⁡kt)\vec p\cdot \vec F=(\cos kt)(-k\sin kt)+(-\sin kt)(-k\cos kt)p​⋅F=(coskt)(−ksinkt)+(−sinkt)(−kcoskt) p⃗⋅F⃗=−ksin⁡ktcos⁡kt+ksin⁡ktcos⁡kt=0\vec p\cdot \vec F=-k\sin kt\cos kt+k\sin kt\cos kt=0p​⋅F=−ksinktcoskt+ksinktcoskt=0
  1. Since
p⃗⋅F⃗=∣p⃗∣ ∣F⃗∣cos⁡θ=0\vec p\cdot \vec F=|\vec p|\,|\vec F|\cos\theta=0p​⋅F=∣p​∣∣F∣cosθ=0

we get

cos⁡θ=0\cos\theta=0cosθ=0

Therefore,

θ=π2\theta=\frac{\pi}{2}θ=2π​
  1. So the correct option is:
A π2\boxed{\text{A }\frac{\pi}{2}}A 2π​​
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