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			<h2>Solutions of Savchenko Physics Textbook</h2>
			<p class="author">
			Aliaksandr Melnichenka <br/>
			October 2023
			</p>
			</header>

			<h3 id="back-link"><a href="../#2.2">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$2.2.34.$ Two trolleys of mass $M$ each move in parallel with the initial speeds $v_1$ and $v_2$ ($\vec{v}_2 > \vec{v}_1$). A load of mass $m$, which initially lay on the first trolley, is transferred to the second trolley with almost zero speed relative to this trolley. Then, with almost zero speed relative to the second trolley, it is transferred back to the first one. What will be the speed difference of the trolleys after $N$ such transfers of cargo back and forth? Try to explain qualitatively the viscous friction that occurs when gas layers slip relative to each other.

			</p>
			<center>
			<figure>
			<img src="statement.png"
			loading="lazy" width="250" />
			<figcaption>
			For problem $2.2.34$
			</figcaption>
			</figure>
			</center>
			<p>
			</p>

			<h3>Solution</h3>
			<p>
			From first to second:

			$$m v_1 + M v_2 = (M + m) u_2$$

			$$
			u_2 = \frac{m v_1 + M v_2}{M + m}
			$$

			$$P_x = \text{const!}$$

			Second to first:

			$$
			M v_1 + m u_2 = (M + m) u_1
			$$

			$$
			u_1 = \frac{M v_1 + m u_2}{M + m}
			$$

			About the change:

			$$
			\Delta u = u_1 - u_2
			$$

			$$
			u_1 > u_2
			$$

			$$
			\Delta u = \left(\frac{M}{M + m}\right)^2 (v_1 - v_2)
			$$

			$$
			\kappa = \left(\frac{M}{M + m}\right)^2
			$$

			$$
			\Delta v = v_1 - v_2
			$$

			$$
			\Delta u_N = \kappa^N \Delta v
			$$

			$$
			\Delta u_N = \left(\frac{M}{M + m}\right)^{2N} (v_1 - v_2)
			$$

			$$
			\boxed{\|\Delta u_N\| = \left(\frac{M}{M + m}\right)^{2N} (v_2 - v_1)}
			$$
			</p>

			<h4>Answer</h4>
		@@ -120,6 +120,10 @@
			<p>
			$$\left(\frac{M}{M+m}\right)^{2N}(v_2-v_1).$$
			</p>
		+
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	Almaskhan Arsen<br>
		+	</p>


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	<body style="">		<body style="">
	<header style="text-align:center;">		<header style="text-align:center;">
	<h2>Solutions of Savchenko Physics Textbook</h2>		<h2>Solutions of Savchenko Physics Textbook</h2>
	<p class="author">		<p class="author">
	Aliaksandr Melnichenka <br/>		Aliaksandr Melnichenka <br/>
	October 2023		October 2023
	</p>		</p>
	</header>		</header>

	<h3 id="back-link"><a href="../#2.2">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../#2.2">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<p>
	$2.2.34.$ Two trolleys of mass $M$ each move in parallel with the initial speeds $v_1$ and $v_2$ ($\vec{v}_2 > \vec{v}_1$). A load of mass $m$, which initially lay on the first trolley, is transferred to the second trolley with almost zero speed relative to this trolley. Then, with almost zero speed relative to the second trolley, it is transferred back to the first one. What will be the speed difference of the trolleys after $N$ such transfers of cargo back and forth? Try to explain qualitatively the viscous friction that occurs when gas layers slip relative to each other.		$2.2.34.$ Two trolleys of mass $M$ each move in parallel with the initial speeds $v_1$ and $v_2$ ($\vec{v}_2 > \vec{v}_1$). A load of mass $m$, which initially lay on the first trolley, is transferred to the second trolley with almost zero speed relative to this trolley. Then, with almost zero speed relative to the second trolley, it is transferred back to the first one. What will be the speed difference of the trolleys after $N$ such transfers of cargo back and forth? Try to explain qualitatively the viscous friction that occurs when gas layers slip relative to each other.

	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="statement.png"		<img src="statement.png"
	loading="lazy" width="250" />		loading="lazy" width="250" />
	<figcaption>		<figcaption>
	For problem $2.2.34$		For problem $2.2.34$
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	From first to second:		From first to second:

	$$m v_1 + M v_2 = (M + m) u_2$$		$$m v_1 + M v_2 = (M + m) u_2$$

	$$		$$
	u_2 = \frac{m v_1 + M v_2}{M + m}		u_2 = \frac{m v_1 + M v_2}{M + m}
	$$		$$

	$$P_x = \text{const!}$$		$$P_x = \text{const!}$$

	Second to first:		Second to first:

	$$		$$
	M v_1 + m u_2 = (M + m) u_1		M v_1 + m u_2 = (M + m) u_1
	$$		$$

	$$		$$
	u_1 = \frac{M v_1 + m u_2}{M + m}		u_1 = \frac{M v_1 + m u_2}{M + m}
	$$		$$

	About the change:		About the change:

	$$		$$
	\Delta u = u_1 - u_2		\Delta u = u_1 - u_2
	$$		$$

	$$		$$
	u_1 > u_2		u_1 > u_2
	$$		$$

	$$		$$
	\Delta u = \left(\frac{M}{M + m}\right)^2 (v_1 - v_2)		\Delta u = \left(\frac{M}{M + m}\right)^2 (v_1 - v_2)
	$$		$$

	$$		$$
	\kappa = \left(\frac{M}{M + m}\right)^2		\kappa = \left(\frac{M}{M + m}\right)^2
	$$		$$

	$$		$$
	\Delta v = v_1 - v_2		\Delta v = v_1 - v_2
	$$		$$

	$$		$$
	\Delta u_N = \kappa^N \Delta v		\Delta u_N = \kappa^N \Delta v
	$$		$$

	$$		$$
	\Delta u_N = \left(\frac{M}{M + m}\right)^{2N} (v_1 - v_2)		\Delta u_N = \left(\frac{M}{M + m}\right)^{2N} (v_1 - v_2)
	$$		$$

	$$		$$
	\boxed{\|\Delta u_N\| = \left(\frac{M}{M + m}\right)^{2N} (v_2 - v_1)}		\boxed{\|\Delta u_N\| = \left(\frac{M}{M + m}\right)^{2N} (v_2 - v_1)}
	$$		$$
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	@@ -120,6 +120,10 @@
	<p>		<p>
	$$\left(\frac{M}{M+m}\right)^{2N}(v_2-v_1).$$		$$\left(\frac{M}{M+m}\right)^{2N}(v_2-v_1).$$
	</p>		</p>

			<p style="text-align: right; font-style: italic; font-size: 14;">
			Almaskhan Arsen<br>
			</p>


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