Question

A proton moving with a speed of 1.0x10e7 m/sec passes through a 0.020 cm thick sheet of paper and emerges with a speed of 2.0x10e6 m/sec. Assuming uniform deceleration, find retardation and time taken to pass through the paper?

Answer 1

Answer on Question 64278, Physics, Mechanics, Relativity

Question:

A proton moving with a speed of $1.0 \cdot 10^{7} \, \text{m/s}$ passes through a $0.020 \, \text{cm}$ thick sheet of paper and emerges with a speed of $2.0 \cdot 10^{6} \, \text{m/s}$. Assuming uniform deceleration, find retardation and time taken to pass through the paper?

Solution:

1) We can find the retardation of the proton from the kinematic equation:

$$v_{f}^{2} = v_{i}^{2} + 2 a s,$$

here, $v_{i}$ is the initial speed of the proton, $v_{f}$ is the final speed of the proton (after it passes through the sheet of paper), $a$ is the retardation of the proton, $s$ is the distance that proton passes when moving through the sheet of paper.

Then, we can find the retardation of the proton:

$$a = \frac{v_{f}^{2} - v_{i}^{2}}{2 s} = \frac{\left(2.0 \cdot 10^{6} \, \frac{\text{m}}{\text{s}}\right)^{2} - \left(1.0 \cdot 10^{7} \, \frac{\text{m}}{\text{s}}\right)^{2}}{2 \cdot 2 \cdot 10^{-4} \, \text{m}} = -2.4 \cdot 10^{17} \, \frac{\text{m}}{\text{s}^{2}}.$$

The sign minus indicates that the proton decelerates.

b) We can find the time taken to pass through the paper from another kinematic equation:

$$v_{f} = v_{i} + a t,$$ $$t = \frac{v_{f} - v_{i}}{a} = \frac{2.0 \cdot 10^{6} \, \frac{\text{m}}{\text{s}} - 1.0 \cdot 10^{7} \, \frac{\text{m}}{\text{s}}}{-2.4 \cdot 10^{17} \, \frac{\text{m}}{\text{s}^{2}}} = 3.3 \cdot 10^{-11} \, \text{s}.$$

Answer:

a) $a = -2.4 \cdot 10^{17} \, \frac{\text{m}}{\text{s}^{2}}$.

b) $t = 3.3 \cdot 10^{-11} \, \text{s}$.

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