Question

A bomb of $12\,kg$ explodes into two pieces of mass $4\,kg$ and $8\,kg$ . The velocity of $8\,kg$ mass is $6\,m{\sec ^{ - 1}}$ . The kinetic energy of another mass is?
A. $48$ Joules
B. $32$ Joules
C. $24$ Joules
D. $288$ Joule

Answer 1

Hint:In order to solve this question we need to understand conservation of momentum. So momentum is mathematically defined as a product of mass of body and acceleration. From Newton’ second law, Force is defined as the rate of change of momentum. So if net force on the body is zero, then the rate of change of momentum is also zero, so momentum is constant. In this way the momentum is always conserved during collision, because in collision net force on a body is zero.

Complete step by step answer:
According to problem, mass of one piece is, ${m_1} = 4kg$
And mass of other piece is, ${m_2} = 8kg$
Let us assume that mass one goes into positive x direction after explosion
So, the velocity of body two is, ${\vec v_2} = 6\,m{\sec ^{ - 1}}$
Also, let the velocity of mass one be, ${\vec v_1}$

Let us analyze the situation before the explosion. Before an explosion, A bomb of mass $M = 12kg$ does not explode so it is supposed to be at rest, that is $\vec V = 0$. So before collision, momentum of bomb is,
${\vec p_i} = M\vec V$
Putting values we get,
${\vec p_i} = (12)(0)$
$\Rightarrow {\vec p_i} = 0$
Now let us analyze after the explosion.So after explosion, momentum of mass one is given as,
${\vec p_1} = {m_1}{\vec v_1}$
Putting values we get,
${\vec p_1} = (4kg){\vec v_1}$

Momentum of mass two is given as,
${\vec p_2} = {m_2}{\vec v_2}$
Putting values we get,
${\vec p_2} = (8kg) \times (6m{s^{ - 1}})$
$\Rightarrow {\vec p_2} = 48kgm{s^{ - 1}}$
So total momentum of system after collision is,
${\vec p_f} = {\vec p_1} + {\vec p_2}$
Putting values we get,
${\vec p_f} = 4{\vec v_1} + 48$

Since momentum is conserved during collision, that is ${\vec p_i} = {\vec p_f}$.
Putting values we get,
$0 = 4{\vec v_1} + 48$
$\Rightarrow {\vec v_1} = \dfrac{{ - 48}}{4}m{\sec ^{ - 1}}$
$\Rightarrow {\vec v_1} = - 12m{\sec ^{ - 1}}$
So the velocity of mass one is equal to $12\,m{\sec ^{ - 1}}$ and it explodes in the opposite direction as that of the second mass.
So the kinetic energy of mass one is,
$K = \dfrac{1}{2}{m_1}{v_1}^2$
Putting values we get,
$K = \dfrac{1}{2}(4kg){(12)^2}$
$\therefore K = 288\,J$

So the correct option is D.

Note: It should be remembered that angular momentum of a body is also constant if the net torque on a body is zero, this is similar to the case of linear momentum. Basically collisions are of two types, perfectly elastic collision and inelastic collision. In a perfectly elastic collision, both energy and momentum of the body is conserved.