A particle moves in x-y under the action of the force F- Such that x and y components of linear momentum- P- at any time t are 2cost and 2 sint- Find the angle between F-P- at a given time

The correct option is D 90 ^0 F⃗=d P⃗dt #160;or, F⃗= 2sin t î +2cos t ĵ Now, F⃗. P⃗ = 0.So angle between them is 90^∘ . ...

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Standard XII

Physics

COM in Collision and Explosion

A particle mo...

Question

$A$ particle moves in x-y under the action of the force $\to F$ Such that $x$ and $y$ components of linear momentum, $\to P$ at any time $t$ are $2 cos t$ and $2 sin t . F$ ind the angle between $\to F & \to P$ at a given time

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Solution

The correct option is D 90 $^{0}$
$\to F = \frac{d \to P}{d t}$

or, $\to F = - 2 sin t^i + 2 cos t^j$
Now, $\to F . \to P$ = 0.
So angle between them is $90^{\circ}$ .

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A particle moves in x-y under the action of the force →F Such that x and y components of linear momentum, →P at any time t are 2cost and 2sint. Find the angle between →F&→P at a given time

The correct option is D 900→F=d→Pdt or,→F=−2sint^i+2cost^jNow, →F.→P = 0.So angle between them is 90∘.

$A$ particle moves in x-y under the action of the force $\to F$ Such that $x$ and $y$ components of linear momentum, $\to P$ at any time $t$ are $2 cos t$ and $2 sin t . F$ ind the angle between $\to F & \to P$ at a given time

The correct option is D 90 $^{0}$
$\to F = \frac{d \to P}{d t}$

or, $\to F = - 2 sin t^i + 2 cos t^j$
Now, $\to F . \to P$ = 0.
So angle between them is $90^{\circ}$ .