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

The Correct Option is A

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.