Question

3. (a) Derive the formula for kinetic energy.
(b) Derive the relation between momentum and kinetic energy. ​

Answer 1

e=FS
F=ma
by 3rd eqn.of motion =2as=v^2-u^2
s=v^2-u^2/2a
ma×v^2-u^2/2a
therefore, 1/2×m×v^2-u^2

Answer 2

$\mathfrak{Derivation \: of \:Kinetic \: Energy}$

$\mathsf{Things \: to \: know \: before\: Derivation}$

$\mathsf{\implies Work done = Fs}$

$\mathsf{\implies v² = u² + 2as}$

$\mathsf{\implies s = \frac{v² - u²}{2a}}$

$\mathsf{\implies u = 0 m/s \: for \:a \: body \: starting \: from \: rest}$

$\mathsf{\implies Work \: Done = \: Energy}$

$\mathsf{\implies Kinetic \: Energy \: is \: also \: written \: as \: E_k}$

$\mathsf{\implies Force = mass \: × \: acceleration \: = ma}$

$\mathsf{Derivation}$

$\mathsf{\hookrightarrow E_k = Work done = Fs }$

$\mathsf{\hookrightarrow \: = \: Fs \: = ma × s }$

$\mathsf{\hookrightarrow E_k = m × \frac{v² - u²}{2a} × a}$

$\mathsf{\hookrightarrow E_k = \frac{1}{2}mv²}$