Question

A 30 kg shell is flying at 48 m/sec.when it exposed,it's one part of 18 kg stops, while the remaining part files on. Find the velocity of the latter

Answer 1

Given,

Mass of shell = 30 kg

Velocity of shell = 48 m/sec

Mass of one part of shell = 18 kg

To Find,

The velocity of the latter.

Solution,

By using the law of conservation of momentum,

MV = m₁v₁+m₂v₂

where, M = 30, V = 48, m₁ = 18, v₁=0, m₂=12

Now,

30(48) = 18(0)+12(v₂)

1440 = 12v

v = 120 m/sec

Hence, the velocity of the latter is 120 m/sec.

Answer 2

Concept:
The law of conservation of momentum states that " The total momentum of a system remains constant. " Which means momentum after the explosion and before the explosion is the same.
Given:
Mass of the shell before the explosion $= M = 30kg$
Velocity before the explosion $= V= 48ms^{-1}$
After the explosion, the shell is divided into two parts i.e Part A and Part B
Mass of Part A $= m_1= 18kg$
The Velocity of Part A $V' = 0$
Mass of Part B $= m_2 = M - m_1= 30-18=12kg$
The Velocity of Part B $= V''$
To Find:
The Velocity of the other part of the explosive
Solution:
According to the law of conservation of momentum,
$M \times V = m_1\times V' + m_2 \times V''$
$\Rightarrow 30 \times 48= 18 \times 0 + 12 \times V''\\ V''= \frac{30 \times 48 }{12}\\V''= 120ms^{-1}$

$\therefore$ The Velocity of the latter $= 120 ms^{-1}$