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Newton’s Laws of Motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsNewton’s Laws of Motion1 Mark
Question
Two blocks A and B of mass $m_A$ and $m_B$ respectively are kept in contact on a frictionless table. The experimenter pushes the block A from behind so that the blocks accelerate. If the block A exerts a force F on the block B, what is the force exerted by the experimenter on A?
✓

Answer

Let, F → contact force between $m_A$ & $m_B$. And, f → force exerted by experimenter.

$\left.F+m_A a-f=0 m_B a-f=0 \Rightarrow F=f-m_A a \ldots \text {... } i\right) \Rightarrow F=m_B a \ldots \text {...(ii) From eqn. (i) and eqn. (ii) } \Rightarrow f-m_A a=m_B a \Rightarrow f=$
$m_B a+m_A a \Rightarrow f=a\left(m_A+m_B\right)$
$\Rightarrow\text{f}=\frac{\text{F}}{\text{m}_{\text{B}}}(\text{m}_{\text{B}}+\text{m}_{\text{A}})=\text{F}\Big(1+\frac{\text{m}_{\text{A}}}{\text{m}_{\text{B}}}\Big)$
$\Big[\text{because }\text{a}=\frac{\text{F}}{\text{m}_{\text{B}}}\Big]$
$\therefore$ The force exerted by the experimenter is $\text{F}\Big(1+\frac{\text{m}_{\text{A}}}{\text{m}_{\text{B}}}\Big)$

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