Question

A particle B of mass 0.6 kg slides down the smooth face PR of a wedge A of mass 1.7 kg which can move freely on a smooth horizontal surface. If the inclination of the face PR to the horizontal is ${45}^o$, then,

• A. the acceleration of A is $\frac{3g}{20}$
• B. the vertical component of the acceleration of B is $\frac{23g}{40}$
• C. the horizontal component of the acceleration of B is $\frac{17g}{40}$
• D. the vertical and horizontal components of the acceleration of B must be equal.

Answer 1

Answer: (A), (B), (C)

The correct options are
A the acceleration of A is $\frac{3g}{20}$
B the vertical component of the acceleration of B is $\frac{23g}{40}$
C the horizontal component of the acceleration of B is $\frac{17g}{40}$

Let acceleration of the wedge A is a .
$N+masin\theta =mgcos\theta$
$Nsin\theta =Ma$
Ma = ( mgcos\theta -ma sin\theta ) sin \theta
$a(M=msi{n}^2\theta =mgcos\theta sin\theta$
The acceleration of wedge A is
$a=\frac{mgsin\theta cos\theta }{M+msi{n}^2\theta }$
now we know ,$(\theta =45)$
$=\frac{\left(0.6\right)gsin{45}^{0}cos{45}^0}{1.7+\left(0.6\right)si{n}^2{45}^0}$
Therefore,
$=\frac{3g}{20}$
Acceleration of m relative to the wedge
$=acos\theta +gsin\theta$
Vertical component
$=(acos\theta +gsin\theta )sin\theta$
$=(\frac{3g}{20}+g)\frac{1}{2}$
$=\frac{23g}{40}$
$\Rightarrow$ Horizontal component$=$(Vertical component)-a
$=\frac{23g}{40}-\frac{3g}{20}$
$=\frac{17g}{40}$