Question

The Atwood machine in figure has a third mass attached to it by a limp string. After being released, mass $2m$ falls a distance $x$ before the limp string becomes taut. Thereafter both the masses on the left rise at the same speed. Assume that pulley is ideal. Choose the correct options.

• A. Speed of mass $2m$ after falling through a distance $x$, just before string is taut, is $\sqrt{2gx}$.
• B. Speed of mass $2m$ after falling through a distance $x$, just before string is taut, is $\sqrt{\frac{2gx}{3}}$.
• C. Final speed of all the masses after the string is taut, is $\sqrt{\frac{8gx}{3}}$.
• D. Final speed of all the masses after the string is taut, is $\sqrt{\frac{3gx}{8}}$.

Answer 1

Answer: (B), (D)

The correct options are
B Speed of mass $2m$ after falling through a distance $x$, just before string is taut, is $\sqrt{\frac{2gx}{3}}$.
D Final speed of all the masses after the string is taut, is $\sqrt{\frac{3gx}{8}}$.

For block $B$: $2mg-T=2ma$
For block $A$: $T-mg=ma$
Solving we get $a=\frac{g}{3}$

Velocity of $m$ and $2m$ after falling through a distance before the string is taut
$v_0=\sqrt{2ax}=\sqrt{\frac{2gx}{3}}$

Impulse equation, for block $B$
$\Delta P=T\Delta t=2m(v-\sqrt{\frac{2gx}{3}})$
$T\Delta t-T^{{}^{\prime }}\Delta t=m(v-\sqrt{\frac{2gx}{3}})$
For block $C$
$T^{{}^{\prime }}\Delta t=m(v-0)$

Solving we get,
$v=\sqrt{\frac{3gx}{8}}$