Question

An object moving to the right at 0.8c is struck head-on by a photon of wavelength $\mathit{\lambda }$ moving to the left. The object absorbs the photon (i.e., the photon disappears) and is afterward moving to the right at 0.6c. (a) Determine the ratio of the object’s mass after the collision to its mass before the collision. (Note: The object is not a “fundamental particle”, and its mass is, therefore, subject to change.) (b) Does Kinetic energy increase or decrease?

Short answer

1. The ratio of the two masses is $\frac{3}{2}=\frac{m_i}{m_f}$
2. Yes, increase.

Step-by-step solution

Step 1: Finding The kinetic energy

The energy of a photon E is inversely proportional to its wavelength as

$\mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{\lambda }}$

Hence h is the plank’s constant, and c is the speed of the light in vacuum.

De Broglie’s law will also be used, which states that the momentum of an object is inversely proportional to its wavelength as

$\mathit{P}\mathbf{=}\frac{\mathbf{h}}{\mathbf{\lambda }}$

Here h is the Plank’s constant

The relativistic momentum of an object with mass m and vel0city v is:

$\mathit{P}\mathbf{=}\frac{\mathbf{m}\mathbf{u}}{\sqrt{\mathbf{1}\mathbf{-}{\left({\displaystyle \frac{u}{c}}\right)}^{\mathbf{2}}}}$

Here c is the speed of light in the vacuum.

The energy E of an object with mass m and velocity u is given by:

$\mathit{E}\mathbf{=}\frac{\mathbf{m}{\mathbf{c}}^{\mathbf{2}}}{\sqrt{\mathbf{1}\mathbf{-}{\left({\displaystyle \frac{u}{c}}\right)}^{\mathbf{2}}}}$

The kinetic energy KE of an object with mass m and velocity u is,

role="math" localid="1659095182315" $\mathit{K}\mathit{E}\mathbf{=}\mathbf{(}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{=}{\left({\displaystyle \frac{u}{c}}\right)}^{\mathbf{2}}}}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{m}{\mathit{c}}^{}$