Question

A mass m moving horizontally along x axis with velocity v collides and sticks to a mass of 3m moving vertically upwards with its velocity 2v. The final velocity of combination is

Answer 1

mass = m
velocity = v
second mass= 3m
second velocity = 2v

final velocity = massxvelocity + 2nd mass x 2nd velocity/ sum of masses

=mvi + 3m₂vj/m + 3m
=v/4 i + 3v/2 j
 

Answer 2

The final velocity is $\frac{v}{4} \hat{\imath}+\frac{3 v}{2} \hat{\jmath}$

Given:

Mass in horizontal direction = m  

Velocity in horizontal direction = v

Mass of stick = 3m  

Velocity of stick = 2v

Solution:

From the conservation of momentum, we get,

$m1 u1 + m2 u2 = m1 v1 + m2 v2$

For this case, the formula becomes

$m1 u1 + m2 u2 = v (m1 + m2)$

$v=\frac{\mathrm{m} 1 \mathrm{u} 1+\mathrm{m} 2 \mathrm{u} 2}{\mathrm{m} 1+\mathrm{m} 2}$

$v=\frac{(m v)+(3 m \times 2 v)}{m+3 m}$

$v=\frac{m v+6 m v}{4 m}$

$v=\frac{m v}{4 m}+\frac{6 m v}{4 m}$

$\Rightarrow v=\frac{m v}{4 m}+\frac{6 m v}{4 m}(\hat{\imath}+\hat{\jmath})$

$\therefore v=\frac{v}{4} \hat{\imath}+\frac{3 v}{2} \hat{\jmath}$

The above obtained equation is the velocity of the final combination.