Question

Two blocks $A$ and $B$ shown in figure have masses $M_A=5kg;M_B=4kg$. The system is released from rest. The speed of $B$ after $A$ has travelled a distance $1m$ along the incline is (do not consider dimensional analysis in the options)

• A. $\frac{\sqrt{3}}{2}\sqrt{g}$
• B. $\frac{\sqrt{3}}{4}\sqrt{g}$
• C. $\frac{\sqrt{g}}{2\sqrt{3}}$
• D. $\frac{\sqrt{g}}{2}$

Answer 1

The correct option is C $\frac{\sqrt{g}}{2\sqrt{3}}$

Given:

$M_A=5kg$

$M_B=4kg$

Displacement of B when A move $d_A=5m$

From FBD of 5 kg mass:

$M_{A}gsin37-T=M_A{a}_A$

$5\times \frac{3g}{5}-T=5a_A$

$3g-T=5a_A$.................(1)

From FBD of 4 kg mass:

$2T-M_{B}g=M_B{a}_B$

$2T-4g=4a_B$ ....................(2)

From constrain equation:

$2T{a}_B-T{a}_A=0$

$a_A=2a_B$.................(3)

From equation 1,2 and 3

$a_A=\frac{2g}{12}{a}_B=\frac{g}{12}$

$S_a=ut+\frac{1}{2}g{t}^2$

$1=0+\frac{1}{2}\times \frac{2g}{12}\times t^2$

$\frac{g}{12}{t}^2=1$

$t=\sqrt{\frac{12}{g}}$

Speed of B $=u_B+a_{B}t$

$V_B=\frac{g}{12}\sqrt{\frac{12}{g}}$

$V_B=\sqrt{\frac{g}{12}}$

$V_B=\frac{\sqrt{g}}{2\sqrt{3}}$