Question

What is the percentage increase IN kinetic energy(K.E) ,If the momentum(p) of a moving body is increase by 10%?And How? (If K.E= p^2/2m)

Answer 1

$KE = \frac{p^2}{2m}$ (start kinetic energy)
$KE = \frac{(1.1p)^2}{2m} =1.21 \times \frac{p^2}{2m}$
$KE_n_e_w=1.21\times KE_o_l_d$
Therefore increasing the momentum by 10% increases KE by 21%.