Question

When they are far apart, the momentum of a proton is $\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{21}}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{kg}\mathbf{.}\mathbf{m}\mathbf{/}\mathbf{s}$as it approaches another proton that is initially at rest. The two protons repel each other electrically, without coming close enough to touch. When they are once again far apart, one of the protons now has momentum$\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{21}}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{55}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{21}}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{kg}\mathbf{.}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{.}$ At that instant, what is the momentum of the other proton?

Short answer

The momentum of the other proton is $(1\times {10}^{-21},-1.55\times {10}^{-21},0)kg.m/s.$.

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The momentum of a proton is$(3.4\times {10}^{(}-21),0,0)\hspace{0.33em}kg.m/s.$ .
• At an instant, one of the protons has a momentum of$(2.4\times {10}^{-21},1.55\times {10}^{-21},0)\hspace{0.33em}kg.m/s.$ .

Significance of the law of conservation of energy for the proton

This law states that a body’s momentum remains constant before and after a collision if no external forces get involved.

The equation of the law of conservation of momentum gives the momentum of the other proton.

Significance of the law of conservation of momentum for the proton

From the law of conservation of momentum, the equation of the momentum of the other proton can be expressed as-

$p=p_1-p_2$

Here, p is the total momentum before the collision $p_{1}and{p}_2$, are the momenta of the first and the second object respectively.

Substituting the values in the above equation, we get-

$$\begin{gathered} p=\left(3.4\times {10}^{-21},0,0\right)\mathrm{kg}·\mathrm{m}/\mathrm{s}-\left(2.4\times {10}^{-21},1.55\times {10}^{-21},0\right)\mathrm{kg}·\mathrm{m}/\mathrm{s} \\ \mathrm{p}=\left(1\times {10}^{-21},-1.55\times {10}^{-21},0\right)\mathrm{kg}·\mathrm{m}/\mathrm{s}. \end{gathered}$$

Thus, the momentum of the other proton is$\left(1\times {10}^{-21},-1.55\times {10}^{-21},0\right)\mathrm{kg}·\mathrm{m}/\mathrm{s}.$