Question

A mass $m$ moves with a velocity $v$ and collides inelastically with another identical mass. After collision, the $1^{st}$ mass moves with velocity $3v$ in a direction perpendicular to the initial direction of motion. Find the speed of the $2^{nd}$ mass after collision.

• A. $v$
• B. $\sqrt{3}v$
• C. $\frac{v}{\sqrt{3}}$
• D. $\frac{2v}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{2v}{\sqrt{3}}$

Hint: “Conservation of linear momentum”

Formula used:

$P_i=P_f$

$v=\sqrt{v_x^2+v_y^2}$

Solution:

In the collision, the momentum is conserved. Before the collision momentum is along the $x-axis,$ after the collision, the resultant momentum should be along the $x-axis,$ for this the second particle should move at some angle with the $x-axis.$ Here, the momentum should be conserved in the $x-$ and $y-$directions separately.

Momentum conservation:
$x-axis:mv=mV\cos \theta$
$\Rightarrow V\cos \theta =v$
$y-axis:0=m\frac{v}{\sqrt{3}}-mV\sin \theta$
$\Rightarrow V\sin \theta =\frac{v}{\sqrt{3}}$
$V=\sqrt{{\left(V\sin \theta \right)}^2+{\left(V\cos \theta \right)}^2}=\frac{2v}{\sqrt{3}}$

Final Answer: (a).