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= 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= sin t . The angle between F and p at time t is (A) 90° (B) 0° (C) 180° (D) 30°

Tardigrade Admin · Accepted Answer

px=2 cos t, py=2 sin t ∴ p =px i +py j =2 cos t i +2 sin t j Force = rate of change of momentum =(d p /d t) ⇒ F =-2 sin t i +2 cos t j ∴ cos θ=( F ⋅ p /| F | ⋅| p |) =((-2 sin t i +2 cos t j )(2 cos t i +2 sin t j )/√4 sin 2 t+4 cos 2 t √4 cos 2 t+4 sin 2 t) ⇒ cos θ =0 ⇒ θ =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 $p_{x} = 2 cos t, p_{y} = sin t .$ The angle between $F$ and $p$ at time $t$ is

$9 0^{\circ}$

$0^{\circ}$

$18 0^{\circ}$

$3 0^{\circ}$

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​=sint. The angle between F and p at time t is

90∘

0∘

180∘

30∘

px​=2cost,py​=2sint ∴p=px​i+py​j =2costi+2sintj Force = rate of change of momentum =dtdp​ ⇒F=−2sinti+2costj ∴cosθ=∣F∣⋅∣p∣F⋅p​ =4sin2t+4cos2t​4cos2t+4sin2t​(−2sinti+2costj)(2costi+2sintj)​ ⇒cosθ=0 ⇒θ=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 $p_{x} = 2 cos t, p_{y} = sin t .$ The angle between $F$ and $p$ at time $t$ is

$9 0^{\circ}$

$0^{\circ}$

$18 0^{\circ}$

$3 0^{\circ}$