Question

A completely inelastic collision occurs between two balls of wet putty that move directly towards each other along a vertical axis. Just before the collision, one ball of mass $3.0kg$ is moving upward at $18m{s}^{-1}$ and the other ball of mass $1.9kg$ is moving downward at $13m{s}^{-1}$ How high do the combined two balls of putty rise above the collision point? (Neglect air drag)

Answer 1

Hint: In order to solve this question, we will first balance the linear momentum using the law of conservation of linear momentum for inelastic collision and then by using Newton's equation of motion we will solve for distance balls that will cover together vertically.

Complete answer:
According to the question, we have given that
$m_1=3kg$ mass of first body
$v_1=18m{s}^{-1}$ velocity of the first body which is moving upwards.
$m_2=1.9kg$ mass of second body
$v_2=-13m{s}^{-1}$ velocity of the second body which is moving downward, it’s negative as it’s moving in the opposite direction to that of the first body.
Let $m=m_1+m_2=4.9kg$ be the combined mass of balls after inelastic collision and v be the velocity with which both balls together rise upward.
So, Initial momentum of system P can be written as
$P=m_1{v}_1+m_2{v}_2$ on putting the values we get,
$P=54-24.7$
$P=29.3kgm{s}^{-1}\to (i)$
Now, final momentum of system P’ can be written as
$P^{\prime }=(m_1+m_2)v$ on putting the values we get,
$P^{\prime }=4.9v\to (ii)$
now, according to law of conservation of momentum we know that,
initial momentum is equal to final momentum so,
$P=P^{\prime }$ so from equations (i) and (ii) we get,
$29.3=4.9v$
$\Rightarrow v=5.98m{s}^{-1}$
Now, both balls stick together and moves upwards against the force of gravity with initial velocity of $v=5.98m{s}^{-1}$ and acceleration due to gravity acts downward against the motion of balls so, $a=g=-9.8m{s}^{-2}$ and let h be the height to which ball moves before coming to rest so, final velocity will be zero.
Now, using newton’s equation of motion we have,
$${\left(v_{final}\right)}^2-{\left(v_{initial}\right)}^2=2as$$.
Now, we know that $v_{initial}=v=5.98m{s}^{-1}$, $v_{final}=0$, $s=h$, and $a=g=-9.8m{s}^{-2}$.
Now, substituting these values, we get,
$0-(5.98{)}^2=-2(9.8)h$
Simplifying the calculations, we get,
$35.76=19.6h$
$\Rightarrow h=1.35m$
Hence, balls moved to a height of $1.35m$ above the collision point.

Note:
It should be remembered that, in a perfectly inelastic collision both bodies stick together after collision and move together with same velocity and here it was mentioned to ignore air drag. air drag is the resistance to motion due to air particles which slows down the velocity of bodies.