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

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

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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 px=2cost and py=2sint. The angle between →F and →p at the time t will be:

The correct option is C 90oFx=dpx/dt=−2sint and Fy=dpy/dt=2costso angle between →F and →p will be 900 because we see that their dot product is zero.

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:

The correct option is C $90^{o}$
$F_{x} = d p_{x} / d t = - 2 s i n t$ and $F_{y} = d p_{y} / d t = 2 c o s t$
so angle between $\to F$ and $\to p$ will be $90^{0}$ because we see that their $d o t$ product is $z e r o$ .