SATHEE: Laws Of Motion Question 431

Laws Of Motion Question 431

Question: A particle moves in the $x - y$ plane under the action of a force $\vec{F}$ such that the value of its linear momentum $\vec{P}$ at any time $t$ is $P_{x} = 2 \cos t,$

$P_{y} = 2 \sin t$ . The angle $θ$ between $\vec{F}$ and $\vec{P}$ at a given time t will be

Options:

A) $90^{\circ}$

B) $0^{\circ}$

C) $180^{\circ}$

D) $30^{\circ}$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] $P = \sqrt{P_{x}^{2} + P_{y}^{2}} = \sqrt{{(2 \cos t)}^{2} + {(2 \sin t)}^{2}} = 2$

If m is the mass of the body, then kinetic energy $= \frac{P^{2}}{2 m} = \frac{{(2)}^{2}}{2 m} = \frac{2}{m}$

Since kinetic energy does not change with time, both work done and power are zero. Now, power $= F v \cos θ = 0$ As $F \neq 0, v \neq 0$

$∴$ $\cos θ = 0$ or $θ = 90^{\circ}$

As direction of $\vec{p}$ is same that of $\vec{v},$

$(∵ \vec{P} = m \vec{v}),$ hence angle between $\vec{F}$

and $\vec{P}$ is equal to $90^{\circ}$ .

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Laws Of Motion Question 431

Question: A particle moves in the x−y plane under the action of a force F→ such that the value of its linear momentum P→ at any time t is Px=2cos⁡t,

Options:

Answer:

Solution:

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Question: A particle moves in the $x - y$ plane under the action of a force $\vec{F}$ such that the value of its linear momentum $\vec{P}$ at any time $t$ is $P_{x} = 2 \cos t,$