Question

A particle $A$ of mass $100g$ moving along +ve x-axis with $10$ m/sec, collides at origin, with particle $B$ of mass $200gm$ moving along +ve y-axis with $10$ m/sec. After collision the particle $B$ moves along line $4x-3y=0$ with speed $5$ m/sec. The equation of line along which $A$ moves after collision.

• A. $y-3x=0$
• B. $3y-x=0$
• C. $4y-3x=0$
• D. None

Answer 1

The correct option is A $y-3x=0$

Situation is shown in figure, from figure

$\tan \theta =\frac{4}{3}⟹\theta ={53}^o$

Final velocity of B in x ditection = $5\cos {53}^o=3m/s$

Final velocity of B in y ditection = $5\sin {53}^o=4m/s$

let A maxes angle $\varphi$ with horizontal after collision and its final velocity be $v$

Final velocity of A in x ditection = $v\cos \varphi$

Final velocity of A in y ditection = -$v\sin \varphi$

Since there is no external force, so momentum would remain conserved.

applying momentum conservation in x direction

$m_A{u}_A=m_B{v}_{Bx}+m_A{v}_{Ax}$

$0.1\times 10=0.2\times 3+0.1v\cos \varphi$

$4=v\cos \varphi$ ....(1)

applying momentum conservation in y direction

$m_B{u}_B=m_B{v}_{By}+m_A{v}_{Ay}$

$0.2\times 10=0.2\times 4-0.1v\sin \varphi$

$12=v\sin \varphi$ ...(2)

dividing 1 and 2.

$\frac{12}{4}=\tan \varphi$

so slope of line = 3

Hence particle moves along $y=3x$.