Question

Two particles of masses $m_1$ and $m_2$ are released from rest on a smooth wedge of mass $M$, then

• A. Reaction force on the wedge due to ground is equal to $(m_1+m_2+M)g$
• B. The wedge remains at rest if $m_1=m_2$
• C. The wedge remains stationary if $\frac{(m_1}{m_2}=\tan \theta$
• D. if $(m_1>m_2$, the wedge will accelerate towards right

Answer 1

Answer: (A), (D)

The correct options are
A Reaction force on the wedge due to ground is equal to $(m_1+m_2+M)g$
D if $(m_1>m_2$, the wedge will accelerate towards right

$A)$Here normal reaction between wedge and earth will be equals to the (total mass that wedge contains)$\times g$.

$\therefore N=(m_1+m_2+M)g$

$B)$Wedge experience normal force from both masses which is $m_{1}g\cos \theta$ and $m_{2}g\sin \theta$.

Horizontal components of these forces helps in moving wedge.

i.e. $m_{1}g\cos \theta \times \sin \theta$ and $m_{2}g\sin \theta \times \cos \theta$.

$\therefore$Wedge in equilibrium when $m_1=m_2$.

$C)$ false.

$D)$if $m_1>m_2$ then, $m_{1}g\cos \theta \sin \theta >m_{2}g\sin \theta \cos \theta$

$\therefore$rightward force > leftward force on wedge.and wedge moves rightwards.