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Found in: Page 170

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# A proton $\left(\mathrm{mass}m=1.67×{10}^{-27}\mathrm{kg}\right)$ is being accelerated along a straight line at ${\mathbf{3}}{\mathbf{.}}{\mathbf{6}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{15}}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{{\mathbf{s}}}^{{\mathbf{2}}}$ in a machine. If the proton has an initial speed of ${\mathbf{2}}{\mathbf{.}}{\mathbf{4}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{7}}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}$ and travels 3.5 cm, what then is (a) its speed and (b) the increase in its kinetic energy?

1. The final speed of the proton is $2.9×{10}^{7}\mathrm{m}/\mathrm{s}$.
2. The increase in the kinetic energy of the proton is $2.1×{10}^{-13}\mathrm{J}$.
See the step by step solution

## Step 1: Given data

1. Mass of the proton is, $\mathrm{m}=1.67×{10}^{-27}\mathrm{kg}$.
2. Acceleration is, $\mathrm{a}=3.6×{10}^{15}\mathrm{m}/{\mathrm{s}}^{2}$.
3. The initial speed is, ${\mathrm{v}}_{\mathrm{i}}=2.4×{10}^{7}\mathrm{m}/\mathrm{s}$.
4. Distance traveled is, $\mathrm{d}=3.5\mathrm{cm}=0.035\mathrm{m}$.

## Step 2: Understanding the concept

Using the third kinematic equation, we can find the final speed of the proton from a given mass, acceleration, and initial speed of the proton. Also, we can use this final velocity and given initial velocity to calculate the increase in its kinetic energy.

## Step 3: a) Calculate the speed of the proton

From newtons law of motion equations, the final velocity can be calculated as,

${\mathrm{v}}_{\mathrm{f}}^{2}={\mathrm{v}}_{\mathrm{i}}^{2}+2\mathrm{ad}$

Substitute the given values in the above equation, and we get,

${\mathrm{v}}_{\mathrm{f}}^{2}={\left(2.4×{10}^{7}\frac{\mathrm{m}}{\mathrm{s}}\right)}^{2}+2\left(3.6×{10}^{15}\frac{\mathrm{m}}{{\mathrm{s}}^{2}}\right)\left(0.035\mathrm{m}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{v}}_{\mathrm{f}}^{2}=5.76×{10}^{14}{\mathrm{m}}^{2}/{\mathrm{s}}^{2}+2.52×{10}^{14}{\mathrm{m}}^{2}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{v}}_{\mathrm{f}}^{2}=8.41×{10}^{14}{\mathrm{m}}^{2}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{v}}_{\mathrm{f}}=2.9×{10}^{7}\mathrm{m}/\mathrm{s}$

Therefore, the final speed of the proton is $2.9×{10}^{7}\mathrm{m}/\mathrm{s}$.

## Step 4: (b) Calculate the increase in the proton’s kinetic energy

The expression for kinetic energy is,

$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^{2}$

Initial kinetic energy can be calculated as,

${\mathrm{K}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^{2}$

Substitute the given values in the above equation, and we get,

Final kinetic energy can be calculated as,

${\mathrm{K}}_{\mathrm{f}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^{2}$

Substitute the given values in the above equation, and we get,

Therefore, an increase in the kinetic energy of the proton can be calculated as,

$∆\mathrm{K}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}$

Substitute the given values in the above equation, and we get,

$∆\mathrm{K}=6.9×{10}^{-13}\mathrm{J}-4.8×{10}^{-13}\mathrm{J}\phantom{\rule{0ex}{0ex}}∆\mathrm{K}=2.1×{10}^{-13}\mathrm{J}$

Therefore, an increase in the kinetic energy of the proton is $2.1×{10}^{-13}\mathrm{J}$.