A wedge Q of mass M is placed on a horizontal frictionless surface A B and a block P of mass m is released on its frictionless slope. As P slides by a length L on this slope of inclination θ, Q would slide by a distance of

Question

A wedge Q of mass M is placed on a horizontal frictionless surface A B and a block P of mass m is released on its frictionless slope. As P slides by a length L on this slope of inclination θ, Q would slide by a distance of (A) ((m/M)) L cos θ (B) (m/L)(M+m) (C) ((M+m)/(m L cos θ)) (D) ((m L cos θ)/(m+M))

Tardigrade Admin · Accepted Answer

Let Q slides by a distance x towards left, then net horizontal displacement of P w.r.t. ground =L cos θ-x Now apply m1 x1=m2 x2 ⇒ M x=m(L cos θ-x) ⇒ x=(m L cos θ/M+m)

Q. A wedge $Q$ of mass $M$ is placed on a horizontal frictionless surface $A B$ and a block $P$ of mass $m$ is released on its frictionless slope. As $P$ slides by a length $L$ on this slope of inclination $θ, Q$ would slide by a distance of

$(\frac{m}{M}) L cos θ$

$\frac{m}{L} (M + m)$

$\frac{( M + m )}{( m L c o s θ )}$

$\frac{( m L c o s θ )}{( m + M )}$

Let $Q$ slides by a distance $x$ towards left, then net horizontal displacement of $P$ w.r.t. ground
$= L cos θ - x$
Now apply $m_{1} x_{1} = m_{2} x_{2}$
$\Rightarrow M x = m (L cos θ - x)$
$\Rightarrow x = \frac{m L c o s θ}{M + m}$

Q. A wedge Q of mass M is placed on a horizontal frictionless surface AB and a block P of mass m is released on its frictionless slope. As P slides by a length L on this slope of inclination θ,Q would slide by a distance of

(Mm​)Lcosθ

Lm​(M+m)

(mLcosθ)(M+m)​

(m+M)(mLcosθ)​