Answer to Question #142696 in Mechanics | Relativity for Mageby

Question

A particle of mass m moves with momentum of magnitude p. (a) Show that the kinetic energy of the particle is K = p^2 / 2m. (b) An object has a kinetic energy of 275 J and a momentum of magnitude 25.0 kgm/s. Find the speed and mass of the object.

Answer 1

Since kinetic energy is given by $K. E. =\frac{1}{2}mv^2$

Momentum is given by, $P=mv$

$\implies v=P/m$

Then, I can write it as,

$K. E. =\frac{1}{2}mv^2=\frac{1}{2}m(p/m)^2= \frac{p^2}{2m}$

Given, kinetic energy, $K. E. =\frac{1}{2}mv^2 =275J$

Momentum, $P =mv=25.0 kgm/s$

Dividing both, $\frac{mv^2}{2mv}=\frac{275}{25}$

$v=2\times 11=22m/s$

Using the relationship which we have derived,

$K. E. = \frac{p^2}{2m}$

$275=\frac{25^2}{2m}$

$m =1.14 Kg$