A particle of mass m moves along a curve y = x2. When particle has x -co-ordinate as 1/2 and x-component of velocity as 4m/s. Then pick the incorrect statements:

Question

A particle of mass m moves along a curve y = x2. When particle has x -co-ordinate as 1/2 and x-component of velocity as 4m/s. Then pick the incorrect statements: (A) The position coordinate of particle are (1/2, 1/4). (B) The velocity of particle will be along the line 4x -4y -1 =0. (C) The magnitude of velocity at that instant is 4√2 m/s (D) The magnitude of angular momentum of particle about origin at that position is 0

Tardigrade Admin · Accepted Answer

y=x2 at x=(1/2), y=(1/4) so position is ((1/2), (1/4)) Vx=4, (d y/d t)=2 x (d x/d t) V y =2 xV x at Vx=4 Vy=2((1/2)) 4 Vy=4 ((1/2), (1/4))

Q. A particle of mass m moves along a curve $y = x^{2}$ . When particle has $x - co$ -ordinate as $1/2$ and $x$ -component of velocity as $4 m / s$ . Then pick the incorrect statements:

The position coordinate of particle are $(1/2, 1/4)$ .

The velocity of particle will be along the line $4 x - 4 y - 1 = 0$ .

The magnitude of velocity at that instant is $42 m / s$

The magnitude of angular momentum of particle about origin at that position is 0

$y = x^{2}$
at $x = \frac{1}{2}, y = \frac{1}{4}$
so position is $(\frac{1}{2}, \frac{1}{4})$
$V_{x} = 4, \frac{d y}{d t} = 2 x \frac{d x}{d t}$
$V_{y} = 2 x V_{x}$
at $V_{x} = 4$
$V_{y} = 2 (\frac{1}{2}) 4$
$V_{y} = 4$
$(\frac{1}{2}, \frac{1}{4})$

Q. A particle of mass m moves along a curve y=x2. When particle has x−co-ordinate as 1/2 and x-component of velocity as 4m/s. Then pick the incorrect statements:

The position coordinate of particle are (1/2,1/4).

The velocity of particle will be along the line 4x−4y−1=0.

The magnitude of velocity at that instant is 42​m/s