Blog: At what speed relative to the lab will a 0.641-kg object have the same momentum as a 1.30-kg object that is moving at 0.515c relative to the lab?

Hello!

Denote the given masses as $\displaystyle{{m}_{{1}}^{{0}}}$ and $\displaystyle{{m}_{{2}}^{{0}}},$ and the speeds as $\displaystyle{v}_{{1}}$ and $\displaystyle{v}_{{2}}.$ I suppose the given masses are the rest masses, and they'll change with the changing of the speeds.

Relativistic momentum is defined the same way as non-relativistic momentum: $\displaystyle{p}={m}{v},$ where $\displaystyle{m}$ is the current (observed) mass depending of a frame of reference. Also we know that the mass $\displaystyle{m}$ depends on the speed $\displaystyle{v}$ as $\displaystyle{m}=\frac{{m}_{{0}}}{\sqrt{{{1}-\frac{{{v}}^{{2}}}{{{c}}^{{2}}}}}},$ so

$\displaystyle{p}_{{1}}=\ldots$

Hello!

$\displaystyle{p}_{{1}}={m}_{{1}}{v}_{{1}}=\frac{{{{m}_{{1}}^{{0}}}{v}_{{1}}}}{\sqrt{{{1}-\frac{{{v}_{{1}}^{{2}}}}{{{c}}^{{2}}}}}},$ $\displaystyle{p}_{{2}}={m}_{{2}}{v}_{{2}}=\frac{{{{m}_{{2}}^{{0}}}{v}_{{2}}}}{\sqrt{{{1}-\frac{{{v}_{{2}}^{{2}}}}{{{c}}^{{2}}}}}}.$

It is given that $\displaystyle{p}_{{1}}={p}_{{2}}.$ The only unknown quantity here is $\displaystyle{v}_{{1}},$ and we can find it from this equation. It is more convenient to express $\displaystyle{v}_{{1}}$ in terms of $\displaystyle{c},$ i.e. to find $\displaystyle{k}_{{1}}=\frac{{v}_{{1}}}{{c}}$ ( $\displaystyle{k}_{{2}}=\frac{{v}_{{2}}}{{c}}$ is given). The equation is then equivalent to

$\displaystyle\frac{{{{m}_{{1}}^{{0}}}{k}_{{1}}}}{\sqrt{{{1}-{{k}_{{1}}^{{2}}}}}}=\frac{{{{m}_{{2}}^{{0}}}{k}_{{2}}}}{\sqrt{{{1}-{{k}_{{2}}^{{2}}}}}},$ or squared $\displaystyle\frac{{{{\left({{m}_{{1}}^{{0}}}\right)}}^{{2}}{{k}_{{1}}^{{2}}}}}{{{1}-{{k}_{{1}}^{{2}}}}}=\frac{{{{\left({{m}_{{2}}^{{0}}}\right)}}^{{2}}{{k}_{{2}}^{{2}}}}}{{{1}-{{k}_{{2}}^{{2}}}}}.$

Multiply by $\displaystyle{1}-{{k}_{{1}}^{{2}}},$ collect the terms with $\displaystyle{{k}_{{1}}^{{2}}}$ and it becomes

$\displaystyle{{k}_{{1}}^{{2}}}{\left({{\left({{m}_{{1}}^{{0}}}\right)}}^{{2}}+\frac{{{{\left({{m}_{{2}}^{{0}}}\right)}}^{{2}}{{k}_{{2}}^{{2}}}}}{{{1}-{{k}_{{2}}^{{2}}}}}\right)}=\frac{{{{\left({{m}_{{2}}^{{0}}}\right)}}^{{2}}{{k}_{{2}}^{{2}}}}}{{{1}-{{k}_{{2}}^{{2}}}}},$

and finally

$\displaystyle{{k}_{{1}}^{{2}}}=\frac{{\frac{{{{\left({{m}_{{2}}^{{0}}}\right)}}^{{2}}{{k}_{{2}}^{{2}}}}}{{{1}-{{k}_{{2}}^{{2}}}}}}}{{{{\left({{m}_{{1}}^{{0}}}\right)}}^{{2}}+\frac{{{{\left({{m}_{{2}}^{{0}}}\right)}}^{{2}}{{k}_{{2}}^{{2}}}}}{{{1}-{{k}_{{2}}^{{2}}}}}}}.$

To find the number, first compute $\displaystyle\frac{{{{\left({{m}_{{2}}^{{0}}}\right)}}^{{2}}{{k}_{{2}}^{{2}}}}}{{{1}-{{k}_{{2}}^{{2}}}}}={{\left({1.30}\right)}}^{{2}}\cdot\frac{{{\left({0.515}\right)}}^{{2}}}{{{1}-{{\left({0.515}\right)}}^{{2}}}}\approx{0.61}.$ Then $\displaystyle{{k}_{{1}}^{{2}}}\approx\frac{{0.61}}{{{{\left({0.641}\right)}}^{{2}}+{0.61}}}\approx{0.60}.$ And $\displaystyle{k}_{{1}}\approx\sqrt{{{0.60}}}\approx{0.77}.$ So the answer is: the first object must have a speed of about 0.77c.