An electron travelling at 7.72 x 107 m/s [E] enters a force field that reduces its velocity to 2.46 x 107 m/s [E]. The acceleration is const

Question

An electron travelling at 7.72 x 107 m/s [E] enters a force field that reduces its velocity to 2.46 x 107 m/s [E]. The acceleration is constant. The displacement during the acceleration is 0.478 m [E]. Determine (a) the electron’s acceleration (b) the time interval over which the acceleration occurs.

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Verity 5 years 2021-07-12T09:02:12+00:00 1 Answers 352 views 1

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  1. binhan
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    2021-07-12T09:03:59+00:00
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    Answer:

    Acceleration of this electron: -5.60 \times 10^{15}\; \rm m \cdot s^{-2}.

    Time taken: approximately 9.39 \times 10^{-9}\; \rm s.

    Explanation:

    • Let u denote the velocity of this electron before the change.
    • Let v denote the velocity of this electron after the change.
    • Let x denote the displacement.
    • Let a denote the acceleration.
    • Let t denote the time taken.

    Apply the SUVAT equation that does not involve time:

    v^{2} - u^{2} = 2\, a \, x.

    Equivalently:

    \begin{aligned}a &= \frac{v^{2} - u^{2}}{2\, x}\end{aligned}.

    By this equation, the acceleration of this electron would be:

    \begin{aligned}a &= \frac{v^{2} - u^{2}}{2\, x} \\ &= \frac{(7.72 \times 10^{7}\; \rm m \cdot s^{-1})^{2} - (2.46 \times 10^{7} \; \rm m \cdot s^{-1})^{2}}{2 \times 0.478\; \rm m} \\ &\approx -5.60 \times 10^{15}\; \rm m \cdot s^{-2}\end{aligned}.

    The speed of this electron has changed from u = 7.72 \times 10^{7}\; \rm m\cdot s^{-1} to v = 2.46 \times 10^{7}\; \rm m \cdot s^{-1}. Calculate the time required to achieve this change at a rate of a \approx -5.60 \times 10^{15}\; \rm m\cdot s^{-2}:

    \begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{2.46\times 10^{7}\; \rm m \cdot s^{-1} - 7.72 \times 10^{7}\; \rm m\cdot s^{-1}}{-5.60 \times 10^{15}\; \rm m\cdot s^{-2}} \\ &\approx 9.39 \times 10^{-9}\; \rm s\end{aligned}.

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