Question

A bullet of mass $40g$ has its kinetic energy equal to $200J$. Then the speed of the bullet is___ $m{s}^{-1}$

Answer 1

given, K.E = 200 J

mass $m=40g=\frac{40}{1000}kg=0.04kg$

From the relation $K.E=\frac{1}{2}m{v}^2$,

Speed $v=\sqrt{\frac{2\times K.E}{m}}$ = $\sqrt{\frac{2\times 200}{0.04}}$ = 100 $m{s}^{-1}$