Question

A particle of mass 𝑚 moving eastward with a speed $v$ collides with another particle of the same mass moving northward with the same speed $v.$ The two particles coalesce on collision. The new particle of mass $2m$ will move in the north-easterly direction with a velocity

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

Answer 1

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

Hint: “Conservation of linear momentum”

Formula used:

$P_i=P_f$

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

Solution:

Momentum conservation:
$x:mv=2m{v}_0\cos \theta$
$\Rightarrow v_0\cos \theta =\frac{v}{2}$
$y:mv=2m{v}_0\sin \theta$
$\Rightarrow v_0\sin \theta =\frac{v}{2}$

$v_0=\sqrt{{\left(v_0\cos \theta \right)}^2+{\left(v_0\sin \theta \right)}^2}$
$\sqrt{2}\frac{v}{2}=\frac{v}{\sqrt{2}}$

Final Answer: (c).