Question

A ball A of mass $m$ is moving with a velocity $v$ along north. It collides with another, ball $B$ of same moving with velocity $v$ along east. After the collision both balls stick together and move along northeast. The velocity of the combination is :

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

Answer 1

The correct option is A $\frac{v}{\sqrt{2}}$

Given
$m_1=m_2=m$
$u_1=u_2=v$
The components of the momentum before collision is $($in component form $)$
$p^2=p_x^2+p_y^2$
$p^2=(mv{)}^2+(mv{)}^2$
$p^2=2m^2{v}^2$
$p=\sqrt{2}mv$ ................(1)
Now, if $v^{{}^{\prime }}$ is the velocity of combination after the collision $($when both balls stick together and move along northeast $)$, the momentum of the combined system is
$p=(m+m)v^{\prime }=2m{v}^{\prime }$ ...............(2)
From (1) and (2), we get
$\sqrt{2}mv=2m{v}^{\prime }$
since, momentum will remain conserved hence,
$v^{\prime }=\frac{\sqrt{2}v}{2}$
$v^{\prime }=\frac{v}{\sqrt{2}}$