Question

A particle with kinetic energy equal to 282 J has a momentum of magnitude 26.4 kg · m/s. Calculate the speed (in m/s) and the mass (in kg) of the particle.

Answer 1

Answer:

$v=21.36\,\,\frac{m}{s}$

$m=1.2357\,\,kg$

Explanation:

Recall the formula for linear momentum (p):

$p = m\,v$  which in our case equals 26.4 kg m/s

and notice that the kinetic energy can be written in terms of the linear momentum (p) as shown below:

$K=\frac{1}{2} m\,v^2=\frac{1}{2} \frac{m^2\,v^2}{m} =\frac{1}{2}\frac{(m\,v)^2}{m} =\frac{p^2}{2\,m}$

Then, we can solve for the mass (m) given the information we have on the kinetic energy and momentum of the particle:

$K=\frac{p^2}{2\,m}\\282=\frac{26.4^2}{2\,m}\\m=\frac{26.4^2}{2\,(282)}\,kg\\m=1.2357\,\,kg$

Now by knowing the particle's mass, we use the momentum formula to find its speed:

$p=m\,v\\26.4=1.2357\,v\\v=\frac{26.4}{1.2357} \,\frac{m}{s} \\v=21.36\,\,\frac{m}{s}$