Question

A proton has mass$\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{-27}\mathbf{kg}$. What is the magnitude of the impulse required to increase its speed from 0.990c to 0.994c?

Short answer

The magnitude of the impulse required to increase its speed from 0.990c to 0.994c is $2.004\times {10}^{-21}\mathrm{N}\mathrm{s}$.

Step-by-step solution

Identification of the given data

• The mass of a proton is $\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{-27}\mathbf{kg}$.
• The initial speed of a proton is 0.990c
• The final speed of a proton is 0.994c

Definition of Impulse

$\begin{array}{l}m,\text{mass\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}proton}=1.7\times {10}^{-27}\mathrm{kg}\\ v,\text{final\hspace{0.33em}velocity}=0.994\text{c}=0.994\times 3\times {10}^8=2.982\times {10}^8\text{\hspace{0.33em}m/s}\\ u,\text{initial\hspace{0.33em}velocity}=0.990\text{c}=0.990\times 3\times {10}^8=2.97\times {10}^8\text{\hspace{0.33em}m/s}\end{array}$

Impulse is the equivalent to the force as the change in momentum,

$\mathrm{I}\text{mpulse}=mv-mu\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Substitute these values in Equation (1),

localid="1657118294846" $\begin{array}{c}\mathrm{Im}pluse=\left(1.67\times {10}^{-27}\text{\hspace{0.33em}kg}\times 2.982\times {10}^8\text{\hspace{0.33em}m/s}\right)-\left(1.67\times {10}^{-27}\text{\hspace{0.33em}kg}\times 2.97\times {10}^8\text{\hspace{0.33em}m/s}\right)\\ =\left[\left(1.67\times {10}^{-27}\times 2.982\times {10}^8\right)-\left(1.67\times {10}^{-27}\times 2.97\times {10}^8\right)\right]\left(\frac{1\text{\hspace{0.33em}kg}\cdot 1\text{\hspace{0.33em}m}}{1\text{\hspace{0.33em}s}}\right)\\ =2.004\times {10}^{-21}\mathrm{N}\mathrm{s}\end{array}$

Hence,the magnitude of the impulse required to increase its speed from 0.990c to 0.994c is $\mathbf{2}\mathbf{.}\mathbf{004}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{21}}\mathbf{}\mathbf{N}\mathbf{}\mathbf{s}$.