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3 years ago

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An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit

y was the alpha particle released ?

1 answer:

lora16 [44]3 years ago

4 0

<u>Answer:</u> The velocity of released alpha particle is $1.127\times 10^7m/s$

<u>Explanation:</u>

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

$m_1v_1=m_2v_2$

where,

$m_1\text{ and }v_1$ = Initial mass and velocity

$m_2\text{ and }v_2$ = Final mass and velocity

We are given:

$m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s$

Putting values in above equation, we get:

$238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s$

Hence, the velocity of released alpha particle is $1.127\times 10^7m/s$

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Answer:

Approximately $\displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right]$ .

Explanation:

Consider this $45^{\circ}$ slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin $(0, 0)$ .

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at $45^{\circ}$ to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as $y = -x$ .

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$x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}$ .

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$\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}$ .

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Alternatively, equate the two expressions of $y$ (right-hand side of the equation, the part where $y$ is expressed as a function of $x$ .)

$-0.0052535\;x^{2} = -x$ ,

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The value of $y$ can be found by evaluating either equation at this particular $x$ -value: $x = 190.35$ .

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