A particle of mass m is moving in the x y -plane such that its velocity at a point (x, y) is given as \overrightarrowv=\alpha(y \hatx+2 x \haty) , where \alpha is a non-zero constant. What is the force \vecF acting on the particle?

\vecF=m \alpha^2(y \hatx+2 x \haty)

\vecF=2 m \alpha^2(y \hatx+x \haty)

\vecF=m \alpha^2(x \hatx+2 y \haty)

Laws of Motion PYQ — JEE Advanced 2023 · Shift 2 · Q37 with Solution

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Correct answer: A

The velocity of the particle at position $(x,y)$ is

\vec v = \alpha(y\hat x + 2x\hat y).

So the components are

v_x = \alpha y, \qquad v_y = 2\alpha x.

Since velocity depends on position, acceleration is

\vec a = \frac{d\vec v}{dt} = \left(\frac{dv_x}{dt}\right)\hat x + \left(\frac{dv_y}{dt}\right)\hat y.

Now,

\frac{dv_x}{dt} = \frac{d(\alpha y)}{dt} = \alpha \frac{dy}{dt} = \alpha v_y.

But

v_y = 2\alpha x,

a_x = \alpha(2\alpha x) = 2\alpha^2 x.

Similarly,

\frac{dv_y}{dt} = \frac{d(2\alpha x)}{dt} = 2\alpha \frac{dx}{dt} = 2\alpha v_x.

But

v_x = \alpha y,

a_y = 2\alpha(\alpha y) = 2\alpha^2 y.

Thus,

\vec a = 2\alpha^2 x\,\hat x + 2\alpha^2 y\,\hat y = 2\alpha^2(x\hat x + y\hat y).

\vec F = m\vec a = 2m\alpha^2(x\hat x + y\hat y).

This matches:

\boxed{\text{A: } \vec F = 2m\alpha^2(x\hat x + y\hat y)}

Stored correct answer: A

Our derived answer is also A, so they agree.