Choosing the Right Force: Bug Survival Odds on a Frictionless Table

[MarkFL]

Here is this week's POTW:

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A small bug is placed between two blocks of masses $m_1$ and $m_2$ ($m_1>m_2$) on a frictionless table. A horizontal force $\bf{F}$ can be applied to either $m_1$ or $m_2$. For which of these cases does the bug have a greater chance of surviving?

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[MarkFL]

Congratulations to the following members for their correct submission:

• kaliprasad

My solution is as follows:

Spoiler

For the entire system (neglecting the mass of the bug), we find:

$$\sum F_x=F=\left(m_1+m_2\right)a\implies a=\frac{F}{m_1+m_2}$$

Let the force $P_1$ represent the contact force between the blocks when $F$ is applied to $m_1$:

$$P_1=m_2a=\frac{m_2}{m_1+m_2}F$$

Likewise, we find:

$$P_2=m_1a=\frac{m_1}{m_1+m_2}F$$

Thus:

$$P_2>P_1$$

From this we may conclude that when $F$ is applied to $m_1$ the contact force is smaller giving the bug a greater chance of surviving.