Question

A particle $A$ of mass $1\text{kg}$ moves along the line $3x-4y=0$ with a speed of $10{\text{ms}}^{-1}$ and another particle $B$ of mass $2\text{kg}$ moves along the line $12y-5x=0$ with a speed of $v{\text{ms}}^{-1}$. Both of them start simultaneously from the origin, with $A$ and $B$ moving in first and fourth quadrants respectively, such that their center of mass is always on $x-$axis. The value of $v$ is

• A. $\frac{39}{5}$
• B. $\frac{13}{4}$
• C. $\frac{78}{5}$
• D. $\text{None}$

Answer 1

Answer: (A)

$\frac{39}{5}$

Line $3x-4y=0$ has slope

$tan{\theta }_1=\frac{3}{4}$

Line $12y-5x=0$ has slope

$tan{\theta }_2=+\frac{5}{12}$

for A, $y_A=\left(10sin{\theta }_1\right)t$;

for B, $y_B=\left(vsin{\theta }_2\right)t$

Now according to question

$y_{cm}=\frac{m_A{y}_A+m_B{y}_B}{m_A+m_B}=0$

$\Rightarrow \left(1\right)\left(10sin{\theta }_1\right)t-\left(2\right)\left(vsin{\theta }_2\right)t=0$

$\Rightarrow v=\frac{10}{2}\frac{sin{\theta }_1}{sin{\theta }_2}=\left(5\right)(\frac{3}{5}\times \frac{13}{5})=\frac{39}{5}$