Question

Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial velocity of the ball A is 6 meters per second to the right, and the initial velocity of the balloon B is 12 meters per second to the left.

When the two balls hit each other, what will happen if it is a perfectly inelastic collision?

Answer 1

The two balls will be attach together and their will be one final velocity because in inelastic collisions the objects comes together. so
mAv0A+mBv0B=v(mA+mB)
v=(mAv0A+mBv0B)/(mA+mB)
v=(7kg * 6m/s+ 2kg * 12m/s)/(9kg)
v= 22/3 m/s. say thanks brainly. 

Answer 2

Answer :

The balls will stick to each other and move with a common velocity of 2 m/s.

Explanation :

It is given that,

Mass of ball A, $m_A=7\ kg$

Mass of ball B, $m_B=2\ kg$

Initial velocity of A,  $u_A=6\ m/s$

Initial velocity of B,  $u_B=-12\ m/s$ ( in left)

Let the final velocity after collision is v.

In an inelastic collision the momentum remains conserved. So,

$m_Au_A+m_Bu_B=(m_A+m_B)v$

$7\ kg\times 6\ m/s+2\ kg\times (-12\ m/s)=(7\ kg+2\ kg)\ v$

$42\ kgm/s-24\ kgm/s=9\ kg\times v$

$v=2\ m/s$

Since, it is a perfectly inelastic collision. So, after collision objects will stick to each other and move with a common velocity of 2 m/s.

Hence, this is the required solution.