Question

Two particles $A$ and $B$ each of mass $m$ are attached by a light in-extensible string of length $2l$. The whole system lies on a smooth horizontal table with $B$ initially at a distance $l$ from $A$. The particle at end $B$ is projected across the table with speed $u$ perpendicular to $AB$. Velocity of ball $A$ just after the jerk, is :

• A. $v_A=\frac{u\sqrt{3}}{4}$
• B. $v_B=\frac{u\sqrt{3}}{4}$
• C. component of $v_B$ along $AB$ is $\frac{u\sqrt{3}}{4}$
• D. component of $v_B$ perpendicular to $AB=\frac{u}{2}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $v_A=\frac{u\sqrt{3}}{4}$
C component of $v_B$ along $AB$ is $\frac{u\sqrt{3}}{4}$
D component of $v_B$ perpendicular to $AB=\frac{u}{2}$

The moment just before jerk:

$u_B=u\hat{j}$ position vector of B w.r.t A: $l\hat{i}+l\sqrt{3}\hat{j}$

The velocity of B along this vector = $ucos\theta =u\sqrt{3}/2$

Since conservation of momentum, $m_B{u}_{line}=(m_A+m_B)v_{line}$

$v_{line}=u_{line}/2=u\sqrt{3}/4$