Question

A particle of mass ${}^{\prime }{m}^{\prime }$ is moving with speed ${}^{\prime }2v^{\prime }$ and collides with a mass ${}^{\prime }2m^{\prime }$ moving with speed $v$ in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass ${}^{\prime }{m}^{\prime }$, which move at angle ${45}^{\circ }$ with respect to the original direction. The speed of each of the moving particle will be,

• A. $\frac{v}{\sqrt{2}}$
• B. $\frac{v}{2\sqrt{2}}$
• C. $2\sqrt{2v}$
• D. $\sqrt{2v}$
  

Answer 1

The correct option is C $2\sqrt{2v}$

Draw a diagram of given situation.

Find the speed of each of the moving particle.

Formula used: $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Given,
Before collision,

$m_1=m$
$u_1=2v$
$m_2=2m$
$u_2=v$

After collision,
$m_1=m$
$v_1=0$
$m_2=2m$
$v_2=v^{\prime }\cos \theta$

Conservation of linear momentum just before and after collision,
Total momentum before collision $=$ Total momentum after collision

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$m2v+2mv=m\times 0+m\frac{v^{\prime }}{\sqrt{2}}\times 2$

$v^{\prime }=2\sqrt{2}v$

Final answer: (b)