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Question: A \(30\,kg\) shell is flying at \(48\,m/s\) , when it explodes its one part is \(18\,kg\) stops whil...

A 30kg30\,kg shell is flying at 48m/s48\,m/s , when it explodes its one part is 18kg18\,kg stops while the remaining part flies on. Find the velocity of the latter.

Explanation

Solution

In order to this question, to know the velocity of the remaining part which is flying, we will first rewrite the given facts and then we will find the mass of the remaining part of the shell. And then we will apply the law of conservation of momentum to find the velocity of the remaining part of the shell.

Complete step by step answer:
We will rewrite facts of the given question-
Mass of the shell, M=30kgM = 30\,kg
Velocity of the shell, V=48m/sV = 48\,m/s
After explosion-
Mass of the first part, m1=18kg{m_1} = 18\,kg
As its stops, so the velocity of the first part, v1=0m/s{v_1} = 0\,m/s
So, the mass of remaining part is, m2=30kg18kg=12kg{m_2} = 30\,kg - 18\,kg = 12\,kg
We have to find the velocity of the remaining part or v2{v_2}
Now, according to the law of conservation of momentum:-
MV=m2v2 30×48=12v2 v2=30×4812 v2=120m/s MV = {m_2}{v_2} \\\ \Rightarrow 30 \times 48 = 12{v_2} \\\ \Rightarrow {v_2} = \dfrac{{30 \times 48}}{{12}} \\\ \therefore {v_2} = 120\,m/s \\\
Hence, the velocity of the required part is 120m/s120\,m/s.

Note: The general law of physics states that in an isolated set of objects, the quantity called momentum, which characterises motion, never changes; that is, the total momentum of a system remains constant.