Question

$n$ blocks of different masses are placed on a frictionless inclined plane such that they are just in contact with each other. If they are released simultaneously, then find the force of interaction between $(n-1{)}^{th}$ block and $n^{th}$ block

• A. $(m_{n-1}-m_n)g\sin \theta$
• B. zero
• C. $mng\cos \theta$
• D. None of these

Answer 1

The correct option is B zero

Since friction is absent on the surface, therefore each block will have acceleration $a=g\sin \theta$ which is independent of their masses.
Consider the FBD of $(n-1{)}^{th}$ block with mass, $m_{n-1}$ and let $N$ be the interaction force between body with mass $m_{n-1}$ and $m_n$

$m_{n-1}g\sin \theta -N=m_{n-1}a$
$\Rightarrow m_{n-1}g\sin \theta -N=m_{n-1}g\sin \theta$
$\Rightarrow N=0$
Hence, the force of interaction between ant two bodies will be zero.