Question

A particle is projected up from a point at an angle 9, with the horizontal direction. At any time t, if p is the linear momentum, y is the vertical displacement, x is the horizontal displacement, the graph among the following, which does not represent the variation of kinetic energy K of the projectile is

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

Answer 1

Answer: (A)

graph (A)

Answer 2

In above figure, a block is projected at an angle $\theta$ in X-Y plane. After some time particle reaches at point P, at this point, its momentum becomes p, let its mass is m. $\therefore$ $p=mv$ $\frac{p}{m}=v$ (velocity at point P) Kinetic energy, $K=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m\left( \frac{{{p}^{2}}}{{{m}^{2}}} \right)$ $K=\frac{1}{2}\left( \frac{{{p}^{2}}}{m} \right)$ $K\propto {{p}^{2}}$ Graph between K and ${{p}^{2}}$ will be as shown below:Velocity at point P, ${{v}^{2}}={{u}^{2}}+2gy$ ${{v}^{2}}=2gy$ $\therefore$ $K=\frac{1}{2}m(2gy)$ or $K=mgy$ or $K\propto y$ Graph between K and y will be as shown below:Now, ${{u}_{x}}=\left( \frac{x}{t} \right)$ ${{K}_{x}}=\frac{1}{2}mu_{x}^{2}=\frac{1}{2}m\left( \frac{{{x}^{2}}}{{{t}^{2}}} \right)$ $=\frac{1}{2}\frac{m}{{{t}^{2}}}({{x}^{2}})$ or ${{K}_{x}}\propto {{x}^{2}}$ Therefore graph between K and $x$ will be as show below:Also, ${{K}_{y}}=mgy=mg({{u}_{x}}t+\frac{1}{2}g{{t}^{2}})$ Therefore, required graph will be as shown: ${{K}_{y}}\propto {{t}^{2}}$ Hence, graph A is wrong.