Question

A particle is dropped from a tower in a uniform gravitational field at $t=0$. The particle is blown over by a horizontal wind with constant velocity.
Slope (m) of trajectory of the particle with horizontal and its kinetic energy vary according to the curves.

• A.
• B.
• C.
• D.

Answer 1

Answer: (B), (C)

The correct options are
B
C

Since the particle is dropped, it means that the initial velocity of the particle is equal to zero. But the particle is blown over by a wind with a constant velocity along horizontal direction, therefore, the particle has a horizontal component of velocity.
Let this component be $v_0.$ Then it may be assumed that the particle is projected horizontally from the top of the tower with velocity $v_0$.
Hence, for the particle, initial velocity $u=v_0$ and angle of projection $\theta =0^{\circ }$.
We know equation of trajectory is:
$y=xtan\theta -\frac{g{x}^2}{2u^{2}co{s}^2\theta }$
Here, $y=-\frac{g{x}^2}{2v_0^2}$
The slope of the trajectory of the particle
$\frac{dy}{dx}=-\frac{2gx}{2v_0^2}=-\frac{g}{v_0^2}x$
Hence, the curve between the slope and horizontal will be a straight line passing through the origin and will have a negative slope. It means that option (b) is correct.
Since, horizontal velocity of the particle remains constant, therefore $x=v_{0}t.$ We get $\frac{dy}{dx}=-\frac{gt}{v_0}$
So the graph between $m$ and time $t$ will have the same shape as the graph between $m$ and $x$.
Hence, option (a) is wrong.
The vertical component of velocity of the particle at time $t$ is equal to $gt$. Hence, at time $t$, KE of the particle, KE = $\frac{1}{2}m\left[\right(gt{)}^2+\left(v_0{)}^2\right]$
It means, the graph between KE and time t should be a parabola having value $\frac{1}{2}m{v}_0^{2}att=0.$ Therefore, option (c) is correct.
As the particle falls, its height decreases and KE increases. The KE increases linearly with height of its fall or the graph between KE and height of the particle will be a straight line having negative slope. Hence, option (d) is wrong