Updated spacing between @ latex expressions

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			<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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			Solutions of Savchenko Problems in Physics <br>
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			<h3 id="back-link"><a href="../../#1.4">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$1.4.14.$ A nucleus traveling at speed $v$ splits into two identical fragments. Determine the maximum possible angle $\alpha$ between the velocities of one of the fragments and the vector $v$, if the fragments have a velocity $u < v$ during the decay of a resting nucleus.
			</p>

			<h3>Solution</h3>
			<p>
			<p>The velocity and one of the accelerations of the nucleus are</p>
			<p class="exp">$$ \vec{u} = \vec{v}_{0} + \vec{v} $$</p>

			<p>where $ \vec{v}_{0} $ is its velocity before the decay of the nucleus, $ \vec{v} $ is the velocity of the fragment relative to the resting nucleus (that is, in the coordinate system in which the nucleus was at rest before the decay).</p>

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		@@ -71,14 +71,14 @@
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			<p>Let's arrange the vector $ \vec{v}_{0} $ as shown in the figure. The end of the vector can lie on a circle of radius $\| \vec{v} \|$. </p>
		−	<p>It is clear from the figure that the vector and forms the maximum angle $ \alpha $ with the vector $ \vec{v}_{0} $ if the angle between the vectors $ \vec{u} $ and $ \vec{v} $ is $ \pi / 2 $. In this case</p>
		+	<p>It is clear from the figure that the vector and forms the maximum angle $ \alpha$ with the vector $ \vec{v}_{0} $ if the angle between the vectors $ \vec{u} $ and $ \vec{v} $ is $ \pi / 2 $. In this case</p>

			<p class="exp">$$\fbox{$ \alpha = \arcsin \frac{v}{v_0} $}$$</p>
			</p>

			<h4>Answer</h4>
			<p>
		−	$$\sin \alpha = u/v$$
		+	$$\sin\alpha = u/v$$
			</p>


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	<h3 id="back-link"><a href="../../#1.4">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#1.4">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<p>
	$1.4.14.$ A nucleus traveling at speed $v$ splits into two identical fragments. Determine the maximum possible angle $\alpha$ between the velocities of one of the fragments and the vector $v$, if the fragments have a velocity $u < v$ during the decay of a resting nucleus.		$1.4.14.$ A nucleus traveling at speed $v$ splits into two identical fragments. Determine the maximum possible angle $\alpha$ between the velocities of one of the fragments and the vector $v$, if the fragments have a velocity $u < v$ during the decay of a resting nucleus.
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	<p>The velocity and one of the accelerations of the nucleus are</p>		<p>The velocity and one of the accelerations of the nucleus are</p>
	<p class="exp">$$ \vec{u} = \vec{v}_{0} + \vec{v} $$</p>		<p class="exp">$$ \vec{u} = \vec{v}_{0} + \vec{v} $$</p>

	<p>where $ \vec{v}_{0} $ is its velocity before the decay of the nucleus, $ \vec{v} $ is the velocity of the fragment relative to the resting nucleus (that is, in the coordinate system in which the nucleus was at rest before the decay).</p>		<p>where $ \vec{v}_{0} $ is its velocity before the decay of the nucleus, $ \vec{v} $ is the velocity of the fragment relative to the resting nucleus (that is, in the coordinate system in which the nucleus was at rest before the decay).</p>

	<center>		<center>
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	Vector image		Vector image
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	@@ -71,14 +71,14 @@
	</center>		</center>

	<p>Let's arrange the vector $ \vec{v}_{0} $ as shown in the figure. The end of the vector can lie on a circle of radius $\| \vec{v} \|$. </p>		<p>Let's arrange the vector $ \vec{v}_{0} $ as shown in the figure. The end of the vector can lie on a circle of radius $\| \vec{v} \|$. </p>
	<p>It is clear from the figure that the vector and forms the maximum angle $ \alpha $ with the vector $ \vec{v}_{0} $ if the angle between the vectors $ \vec{u} $ and $ \vec{v} $ is $ \pi / 2 $. In this case</p>		<p>It is clear from the figure that the vector and forms the maximum angle $ \alpha$ with the vector $ \vec{v}_{0} $ if the angle between the vectors $ \vec{u} $ and $ \vec{v} $ is $ \pi / 2 $. In this case</p>

	<p class="exp">$$\fbox{$ \alpha = \arcsin \frac{v}{v_0} $}$$</p>		<p class="exp">$$\fbox{$ \alpha = \arcsin \frac{v}{v_0} $}$$</p>
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$\sin \alpha = u/v$$		$$\sin\alpha = u/v$$
	</p>		</p>


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