Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

starting at time t = 0 from the origin with speed 1 m/s , a particle follow a two dimensional trajectory in the x - y plane so that co ordinates are related by the equation y = x²/2. The X and Y components of its acceleration are denoted by ax and ay, respectively Then ..

solution : equation of path is y = x²/2

at t = 0, x = 0 and y = 0 , u = 1 m/s

now differentiating equation with respect to t,

dy/dt = 2x/2 dx/dt

⇒dy/dt = x dx/dt

⇒vy = x vx ........(1)

again differentiating with respect to time,

⇒dvy/dt = x dvx/dt + dx/dt vx

⇒ay = x ax + vx²

if ax = 1 and particle at origin (x = 0, y = 0)

ay = 0 × 1 + 1² = 1 m/s²

Therefore option (A) is correct choice.

option (B) , ax = 0 ⇒ay = vx²

but vx = constant = 1 ( all time )

so, ay = 1 at all times.

so option (B) is also correct choice.

at t = 0, x = 0, vy = x vx

here vx = 1 ⇒vy = 0 × 1 = 0

here it is clear that particle's velocity points x - direction at t = 0

so option (C) is also correct choice.

ay = vx² + x ax

if ax = 0 implies that ay = vx² = 1² = 1 m/s² at t = 1 sec

vy = uy + ay × t

⇒vy = 0 + 1 × 1 = 1 m/s

also from equation (1) we get,

tanθ = vy/vx = xvx/vx = x

i.e., θ is the angle with x - axis.

now tanθ = vy/vx = 1/1 = 1 = tan45°

θ = 45°

so option (D) is also correct choice

Therefore options (A), (B) , (C) and (D) are correct choices.

Answer 2

Given :

• Here correct options are suggested and we've to give exact answer for it. That which options are correct as given in quest.'

To find :

• Correct option for the question is wht.

Step - By - Step Solution :

Hence, => Eq = y = x^2/2 (when t = 0) (x = 0) & (we get y = 0) including that (v = 1m/s)

=] dy/dt=x(dvx/dt)+(DX/dt) vx

=] dvy/dt=x(dvx/dt) + (DX/dt)vx

=> ay = x × ax/1 + vx^2

=] ay=XAX+V×2

=> X=0;y=0){and)ay= 0×1+1+1x^2

→ Hence, Option A is correct.. Same as Options are also correct. ✅✔️☑️