Question

A small disc of mass $m$ is placed on a body of mass $M$ at rest. If the disc is set in horizontal motion with velocity $v$ as shown,

(Assume that disc breaks from surface of $M$ and all surfaces are smooth)
Find the height to which disc rises after breaking at point $A,$ from initial level.

• A. $\frac{M{v}^2}{2g(M+m)}$
• B. $\frac{M{v}^2}{2g\left(m\right)}$
• C. $\frac{m{v}^2}{2g(m+M)}$
• D. $\frac{m{v}^2}{2gM}$

Answer 1

The correct option is A $\frac{M{v}^2}{2g(M+m)}$

When the disk leaves the surface of $M$, its velocity towards right (along $x$ axis) equals the velocity of the body $M$.
Let it be $v_x$
Let $v_y$ be the disk velocity in vertical direction (along $y$ axis) at the same moment (disc leaving surface).
There is no net external force in horizontal direction, so linear momentum will be conserved for the system
$mv=(M+m)v_x$
$v_x=\frac{mv}{M+m}...\left(i\right)$
Applying law of conservation of energy for the system,
$\frac{1}{2}m{v}^2=\frac{1}{2}(M+m)v_x^2+\frac{1}{2}m{v}_y^2+mg{h}_1....\left(ii\right)$
Substituting the value of $v_x$ from equation $\left(i\right)$ in equation $\left(ii\right)$
Where $h_1$ is the height of break off point from initial level
$v_y^2=v^2-\frac{m{v}^2}{M+m}-2g{h}_1$
Let $h_2$ be the height raised after the break off point, then
$v_y^2=2g{h}_2....\left(iii\right)$
From equation $\left(ii\right)$ and $\left(iii\right)$
$h_1+h_2=\frac{M{v}^2}{2g(M+m)}$