A particle moves in the x-y plane under the action of a force F such that the value of its linear momentum $(\to P)$ at any time 't' is $P_{x} = 2 c o s t, P_{y} = 2 s i n t$ .The angle $θ$ between $\to F$ and $\to P$ at a given time t. will be

Q. A particle moves in X-Y plane under the action of forces

F

such that the values of linear momentum

p

at any times is

p_{x} = 2 cos t

and

p_{y} = 2 sin t

. The angle between

\to F

and

\to p

at the time

t

will be:

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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A particle moves in the x-y plane under the action of a force →F=−2k sin kt(^i+^j) at any time t its linear momentum is given by PX = 2cos (kt) and PY = 2cos (kt)The angle between →Fand→P at time t is

The correct option is D 1800 →P=2coskt(^i+^j)→F=−2k sin kt(^i+^j)Angle between→Fand→P=1800

A particle moves in the x-y plane under the action of a force $\to F = - 2 k s i n k t (^i +^j)$ at any time t its linear

momentum is given by P_X = 2cos (kt) and P_Y = 2cos (kt)The angle between $\to F a n d \to P$ at time t is

The correct option is D
180⁰

$\to P = 2 c o s k t (^i +^j) \to F = - 2 k s i n k t (^i +^j) Angle between \to F a n d \to P = 180^{0}$