Question

In the arrangement shown in figure wedge of mass M moves towards left with an acceleration a. All surfaces are smooth. The acceleration of mass m relative to wedge is –

• A. a/2
• B. $\frac { 2 \mathrm { Ma } } { \mathrm { m } }$
• C. $\frac { 2 ( M + m ) a } { m }$
• D. $\frac { ( \mathrm { M } + \mathrm { m } ) \mathrm { a } } { \mathrm { m } }$

Answer 1

Answer: (C)

$\frac { 2 ( M + m ) a } { m }$

Answer 2

Let acceleration of mass m relative to wedge down the plane is a_r. Its absolute acceleration in horizontal direction is a_r cos 60° – a(towards right). Hence, let N be the normal reaction between the mass and the wedge. Then

N sin q = Ma = m(a_r cos 60° – a)

or ar = $\frac { ( M + m ) a } { m \cos 60 ^ { \circ } }$ =