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of inspiration for Jaan Kalda’s handouts and to some NBPhO problems. You may find problems from old IPhO
$2.2.34.$ 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)
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="description" content="A website with solutions to physics problems from Savchenko Textbook">
<meta name="description" content="The largest dataset of solutions of 'Savchenko. Problems in Physics'. Savchenko’s Problems in General Physics is widely used to prepare for olympiads and it is a useful tool to
master and sharpen your skills and techniques in comptetitive problem solving. Some of these problems were a source
of inspiration for Jaan Kalda’s handouts and to some NBPhO problems. You may find problems from old IPhO
$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)
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
