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

`v=21.36\,\,(m)/(s)\\`

`m=1.2357\,\,kg`

Step-by-step 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=(1)/(2) m\,v^2=(1)/(2) (m^2\,v^2)/(m) =(1)/(2)((m\,v)^2)/(m) =(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=(p^2)/(2\,m)\\282=(26.4^2)/(2\,m)\\m=(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=(26.4)/(1.2357) \,(m)/(s) \\v=21.36\,\,(m)/(s)`