Question

A bomb of mass $16\,{\text{kg}}$ at rest explodes into two pieces of mass $4\,{\text{kg}}$ and $12\,{\text{kg}}$ . The velocity of the $12\,{\text{kg}}$ is $4\,{\text{m}}{{\text{s}}^{ - 1}}$ . The kinetic energy of the other mass is-
A. $192\,{\text{J}}$
B. $96\,{\text{J}}$
C. $144\,{\text{J}}$
D. $288\,{\text{J}}$

Answer 1

First of all, we will find out velocity of the other body by applying the principle of conservation of momentum. After that, we will use formula for kinetic energy and substitute the values and manipulate accordingly.

Stepwise solution:

In the given question:
We have a bomb with overall mass of $16\,{\text{kg}}$ , which explodes into two pieces of mass $4\,{\text{kg}}$ and $12\,{\text{kg}}$ .
Let us say that the:
$m = 16\,{\text{kg}}$
${m_1} = 4\,{\text{kg}}$
${m_2} = 12\,{\text{kg}}$
${v_2} = 4\,{\text{m}}{{\text{s}}^{ - 1}}$

We are asked to find the kinetic energy of the mass $4\,{\text{kg}}$ , while the velocity of the mass $12\,{\text{kg}}$ is given as $4\,{\text{m}}{{\text{s}}^{ - 1}}$ . This problem is truly based on the principle of conservation of linear momentum, that is we can alternatively say that momentum of the system is conserved before and after the explosion.

First, we will try to find the velocity of the object with mass $4\,{\text{kg}}$ , then we try to find the amount of kinetic energy associated with that said mass.

Now, we apply the principle of conservation of linear momentum, which can be illustrated as follows:
$mv = {m_1}{v_1} + {m_2}{v_2}$ …… (1)
In the formula,
$m$ indicates the mass of the bomb.
$v$ indicates velocity of the bomb.
${m_1}$ indicates the mass of the first piece.
${v_1}$ indicates velocity of the first piece.
${m_2}$ indicates mass of the second piece.
${v_2}$ indicates the velocity of the second piece.

Since, the bomb was at rest initially, then the value of velocity of the bomb is zero.
Substituting the required values in equation (1), we get:
mv = {m_1}{v_1} + {m_2}{v_2} \\\ 16 \times 0 = 4 \times {v_1} + 12 \times 4 \\\ 0 = 4 \times {v_1} + 48 \\\ {v_1} = - 12\,{\text{m}}{{\text{s}}^{ - 1}} \\\
The negative sign indicates that the two pieces are moving opposite to its other.
The velocity of the other mass comes out to be $12\,{\text{m}}{{\text{s}}^{ - 1}}$ .

Now, we will try to find the kinetic energy associated to it, which is given by the formula:
${E_K} = \dfrac{1}{2}{m_1}v_1^2$ …… (2)
Substituting the required values in the equation (2), we get:
{E_K} = \dfrac{1}{2} \times 4 \times {12^2} \\\ {E_K} = 2 \times 144 \\\ {E_K} = 288\,{\text{J}} \\\
Hence, the kinetic energy associated with it is $288\,{\text{J}}$ .
The correct option is D.

Note: In this problem, you will need the concept of conservation of linear momentum. It is important to note that, if it would have been more than two pieces also, in that condition too, linear momentum would have been conserved. Since, velocity is a vector quantity, hence it will depend on direction too. Positive sign indicates that the bodies are moving in the same direction, while negative indicates that they are moving in the opposite direction.