Question

A block $B$ of mass $0.6\text{kg}$ slides down the smooth face $PR$ of a wedge $A$ of mass $1.7\text{kg}$ which can move freely on a smooth horizontal surface. The inclination of the face $PR$ from the horizontal is ${45}^{\circ }$, then

• A. The acceleration of wedge $A$ is $\frac{3g}{20}$ relative to the ground.
• B. The vertical component of the acceleration of $B$ is $\frac{23g}{40}$ with respect to the ground.
• C. The horizontal component of the acceleration of $B$ is $\frac{17g}{40}$ with respect to the ground.
• D. None of these

Answer 1

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

The correct options are
A The acceleration of wedge $A$ is $\frac{3g}{20}$ relative to the ground.
B The vertical component of the acceleration of $B$ is $\frac{23g}{40}$ with respect to the ground.
C The horizontal component of the acceleration of $B$ is $\frac{17g}{40}$ with respect to the ground.

Assume $a$ is the acceleration of the wedge $A$ towards left.
F.B.D of wedge $A$:


F.B.D of block $B$: (from frame of reference of $A$)


Applying equilibrium condition perpendicular to the wedge surface on block $B$,
$mg\cos \theta =N+ma\cos ({90}^{\circ }-\theta )$
$\Rightarrow N=mg\cos \theta -ma\sin \theta ...\left(1\right)$

Let $M=$ mass of wedge $A$
$m=$ mass of block $B$
For wedge $A$, applying newton's $2^{nd}$ law in the direction of acceleration;
$N\sin \theta =Ma$
Multiply eq. $\left(1\right)$ by $\sin \theta$ and substitute
$(mg\cos \theta -ma\sin \theta )\sin \theta =Ma$
$mg\cos \theta \sin \theta =a(M+m{\sin }^2\theta )$
$\Rightarrow a=\frac{mg\cos \theta \sin \theta }{M+m{\sin }^2\theta }$

As given, $\theta ={45}^{\circ }$, $m=0.6\text{kg},M=1.7\text{kg}$
$\Rightarrow a=\frac{0.6\times g\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}}{1.7+0.6\times {\left(\frac{1}{\sqrt{2}}\right)}^2}$
$\therefore a=\frac{3g}{20}$
Hence the acceleration of wedge $A$ towards left$(-vex\text{axis})$ is $\frac{3g}{20}$

Let acceleration of block $B$ relative to wedge is $a^{\prime }$ down the incline. Applying Newton's $2^{nd}$ law in the direction of acceleration:
$mg\sin \theta +ma\sin ({90}^{\circ }-\theta )=m{a}^{\prime }$
$\Rightarrow a^{\prime }=a\cos \theta +g\sin \theta$
$\Rightarrow a^{\prime }=\frac{3g}{20}\times \frac{1}{\sqrt{2}}+\frac{g}{\sqrt{2}}$
$\therefore a^{\prime }=\frac{23g}{20\sqrt{2}}$

$\Rightarrow$Vertical component of acceleration of block $B$ w.r.t ground
$a_y^{\prime }=a^{\prime }\sin \theta =\frac{23g}{20\times \sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{23g}{40}$
and Horizontal component of acceleration of block $B$ w.r.t ground
$a_x^{\prime }=a^{\prime }\cos \theta -a$
($\because -a$ is the accelerarion of wedge w.r.t ground and $a^{\prime }\cos \theta$ is the acceleration of block w.r.t wedge in horizontal direction)
$\Rightarrow a_x^{\prime }=\frac{23g}{20\sqrt{2}}\times \frac{1}{\sqrt{2}}-\frac{3g}{20}$
$\therefore a_x^{\prime }=\frac{17g}{40}$
Hence, option A, B and C are correct.