Question

The potential energy in joules of a particle of mass 1kg moving in a plane is given by U = 3x + 4y,the position coordinates of the point being x and y, measured in meters. If the particle is initially at rest at (6, 4) ,then

• A. Its acceleration is of magnitude $5m{s}^{-2}$
• B. Its speed when it crosses the y-axis is $10m{s}^{-1}$
• C. It crosses the y-axis (x=0) at y=-4
• D. It moves in a straight line passing through the origin (0,0).

Answer 1

Answer: (A), (B), (C)

The correct options are
A Its acceleration is of magnitude $5m{s}^{-2}$
B Its speed when it crosses the y-axis is $10m{s}^{-1}$
C It crosses the y-axis (x=0) at y=-4

w.r.t center of mass frame.
From law of conversion of momentum $mv=3m{v}^1$
$v^1=\frac{v}{3}(v^1-velocityofC.M)$
$\frac{mvL}{3}[\frac{\left(2m\right)L^2}{12}+2m{\left(\frac{L}{6}\right)}^2+m{\left(\frac{L}{6}\right)}^3]{\omega }^1\Rightarrow {\omega }^1=\frac{v}{L}$
Loss in $K.E=k_f-k_i=\frac{1}{2}m{v}^2-\left[\frac{1}{2}\right(3m)v^{1^2}+\frac{1}{2}{I}^1{\omega }^{1^2}]$
$\overrightarrow{F}=\frac{\delta U}{\delta x}\hat{i}-\frac{\delta U}{\delta y}\hat{j}=-3\hat{i}-4\hat{j}$
$a_x=-3m{s}^{-2},a_y=-4m{s}^{-2},$
$x=x_0+u_{x}t+\frac{1}{2}{a}_x{t}^{2}y=y_0+u_{y}t+\frac{1}{2}{a}_y{t}^2$
$U_x=U_y=0$;
$x_0=6;y_0=4$
$x=6-\frac{1}{2}\left(3\right)t^2$
$herex=0$
$t=2s$
$y=4-\frac{1}{2}\left(4\right)t^2$
$y=-4$