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 block B, what is the force exerted by the experimenter on A ?

Answer 1

Let , F = contact force between $m_A$ and $m_B$

and f = force exerted by experimenter

$F=m_{A}a-f=0$

$\Rightarrow F=f-m_{A}a$ .....(i)

Now, $m_B.a-F=0$

$\Rightarrow F=m_{B}a$ .....(ii)

From equation (i) and equation (ii), we have

$\Rightarrow f-m_{A}a=m_{B}a$

$\Rightarrow f=m_{B}a+m_{A}a$

$\Rightarrow =a(m_B+m_A)$

$\Rightarrow f=\frac{F}{m_b}(m_B+m_A)$

$=F(1+\frac{m_A}{m_B})$

[since $a=\frac{F}{m_B}$]

$\therefore$ Force exerted by the experimenter

= $=F(1+\frac{m_A}{m_B})$