A particle moves in the x y-plane under the action of a force F such that the components of its linear momentum p at any time t are px=2 cos t, py=2 sin t. The angle between F and p at time t is

Question

A particle moves in the x y-plane under the action of a force F such that the components of its linear momentum p at any time t are px=2 cos t, py=2 sin t. The angle between F and p at time t is (A) 90° (B) 0° (C) 180° (D) 30°

Tardigrade Admin · Accepted Answer

Given that vecp=px hati+py hatj =2 cos t hati+2 sin t hatj ∴ vecF=(d vecp/d t)=-2 sin t hati+2 cos t hatj Now, vecF ⋅ vecp=0 i.e. angle between vecF and vecp is 90°.

Q. A particle moves in the $x y$ -plane under the action of a force $F$ such that the components of its linear momentum $p$ at any time $t$ are $p_{x} = 2 cos t, p_{y} = 2 sin t$ . The angle between $F$ and $p$ at time $t$ is

$9 0^{\circ}$

$0^{\circ}$

$18 0^{\circ}$

$3 0^{\circ}$

Given that $p = p_{x} \hat{i} + p_{y} \hat{j}$
$= 2 cos t \hat{i} + 2 sin t \hat{j}$
$∴ F = \frac{d p}{d t} = - 2 sin t \hat{i} + 2 cos t \hat{j}$
Now, $F \cdot p = 0$ i.e. angle between $F$ and $p$ is $9 0^{\circ}$ .

Q. A particle moves in the xy-plane under the action of a force F such that the components of its linear momentum p at any time t are px​=2cost,py​=2sint. The angle between F and p at time t is

90∘

0∘

180∘

30∘