Question

A particle is projected up from a point at an angle $\theta$ with the horizontal direction. At any time t, if P is the linear momentum, y is the vertical displacement, x is horizontal displacement, the graph among the following which does not represent the variation of kinetic energy KE of the particle is

• A. graph (A)
• B. graph (B)
• C. graph (C)
• D. graph (D)

Answer 1

The correct option is A graph (A)

Let the speed with which the particle was thrown be $u$

So,

$v_x=u\cos \theta$

$v_y=u\sin \theta -gt$

$x=u\cos \theta t$

$y=u\sin \theta t-\frac{1}{2}g{t}^2$

Let $v$ be the resultant velocity at any time $t$

$\Rightarrow v^2=v_x^2+v_y^2$

$\Rightarrow v^2=u^2-2ugt\sin \theta +g^2{t}^2$

$\Rightarrow v^2=u^2-2gy$

Now, $KE\alpha v^2$ [$\because$ mass of particle is constant]

$\Rightarrow KE\alpha u^2-2gy$

This means the graph between $KE$ and $y$ will be a straight line with a negative slope.

Hence the graph shown in figure $A$ is incorrect.

Note- Equation for all other graphs can be easily obtained by substituting the equations developed above

Correct answer is option $A$