The momentum of a particle is p→=2costi^+2sintj^. The angle between the force F→ acting on the particle and the momentum p→ is x°. What will be the value of x if F→ = dP→dt?

As force F→=dp→dt=(−2sint)i^+(2cost)j^Now we get: cosθ=F→.P→Fp F→.P→=(−2sinti^+2costj^).(2costj^+2sintj^) = 0 ⇒cosθ=0∴ θ=90°x° = 90°

The momentum of a particle is p→=2costi^+2sintj^. The angle between the force F→ acting on the particle and the momentum p→ is x°. What will be the value of x if F→ = dP→dt?

As force F→=dp→dt=(−2sint)i^+(2cost)j^Now we get: cosθ=F→.P→Fp F→.P→=(−2sinti^+2costj^).(2costj^+2sintj^) = 0 ⇒cosθ=0∴ θ=90°x° = 90°

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The momentum of a particle is $\vec{p} = 2 cost \hat{i} + 2 sint \hat{j}$ . The angle between the force $\vec{F}$ acting on the particle and the momentum $\vec{p}$ is x°. What will be the value of x if $\vec{F} = \frac{d \vec{P}}{dt}$ ?

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As force $\vec{F} = \frac{d \vec{p}}{dt} = (- 2 sint) \hat{i} + (2 cost) \hat{j}$

Now we get:

$cosθ = \frac{\vec{F} . \vec{P}}{Fp} \vec{F} . \vec{P} = (- 2 sint \hat{i} + 2 cost \hat{j}) . (2 cost \hat{j} + 2 sint \hat{j}) = 0 \Rightarrow cosθ = 0$

$∴ θ = 90^{°}$

x° = 90°

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The momentum of a particle is p→=2costi^+2sintj^. The angle between the force F→ acting on the particle and the momentum p→ is x°. What will be the value of x if F→ = dP→dt?

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The momentum of a particle is $\vec{p} = 2 cost \hat{i} + 2 sint \hat{j}$ . The angle between the force $\vec{F}$ acting on the particle and the momentum $\vec{p}$ is x°. What will be the value of x if $\vec{F} = \frac{d \vec{P}}{dt}$ ?