Question 270: Laws of Motion

Question

A particle of mass $m$ is moving in the $x y$-plane such that its velocity at a point $(x, y)$ is given as $\overrightarrow{\mathrm{v}}=\alpha(y \hat{x}+2 x \hat{y})$, where $\alpha$ is a non-zero constant. What is the force $\vec{F}$ acting on the particle?

Accepted Answer

Given :
$ \vec{v} = \alpha(y \hat{x} + 2x \hat{y}) $
For the x-component, $ v_x = \alpha y $ :
$ \frac{dv_x}{dt} = \alpha \frac{dy}{dt} = \alpha v_y $
Given $ v_y = 2\alpha x $ :
$ \frac{dv_x}{dt} = \alpha (2\alpha x) = 2\alpha^2 x $
For the y-component, $ v_y = 2\alpha x $ :
$ \frac{dv_y}{dt} = 2\alpha \frac{dx}{dt} = 2\alpha v_x $
Given $ v_x = \alpha y $ :
$ \frac{dv_y}{dt} = 2\alpha (\alpha y) = 2\alpha^2 y $
Thus, combining the components :
$ \vec{a} = 2\alpha^2 x \hat{x} + 2\alpha^2 y \hat{y} $
Now, the force acting on the particle is given by :
$ \vec{F} = m\vec{a} $
$ \vec{F} = m\alpha(2 \alpha x \hat{x} + 2 \alpha y \hat{y}) $
$ \vec{F} = m\alpha^2(2x \hat{x} + 2y \hat{y}) $
This is equivalent to :
$ \vec{F} = 2m\alpha^2(x \hat{x} + y \hat{y}) $
The correct answer is Option A :
$ \vec{F} = 2m\alpha^2(x \hat{x} + y \hat{y}) $

Answer

$\vec{F}=2 m \alpha^2(x \hat{x}+y \hat{y})$

Answer

$\vec{F}=m \alpha^2(y \hat{x}+2 x \hat{y})$

Answer

$\vec{F}=2 m \alpha^2(y \hat{x}+x \hat{y})$

Answer

$\vec{F}=m \alpha^2(x \hat{x}+2 y \hat{y})$

Laws of Motion

A particle of mass m is moving in the x y-plane such that its velocity at a point (x, y) is given as \overrightarrow{\mathrm{v}}=\alpha(y \hat{x}+2 x \hat{y}), where \alpha is a non-zero constant. What is the force \vec{F} acting on the particle?

Choose the correct answer: