Question
A bomb of mass $$16kg$$ at rest explodes into two pieces of masses $$4kg$$ and $$12kg.$$ The velolcity of the $$12kg$$ mass is $$4\,ms^{ - 1}.$$ The kinetic energy of the other mass is
A.
$$144J$$
B.
$$288 J$$
C.
$$192 J$$
D.
$$96 J$$
Answer :
$$288 J$$
Solution :
Let the velocity and mass of $$4 kg$$ piece be $${v_1}$$ and $${m_1}$$ and that of $$12 kg$$ piece be $${v_2}$$ and $${m_2.}$$
Applying conservation of linear momentum
$$\eqalign{ & {m_2}{v_2} = {m_1}{v_1} \Rightarrow {v_1} = \frac{{12 \times 4}}{4} = 12m{s^{ - 1}} \cr & \therefore K.E{._1} = \frac{1}{2}{m_1}v_1^2 = \frac{1}{2} \times 4 \times 144 = 288J \cr} $$
Let the velocity and mass of $$4 kg$$ piece be $${v_1}$$ and $${m_1}$$ and that of $$12 kg$$ piece be $${v_2}$$ and $${m_2.}$$
Applying conservation of linear momentum
$$\eqalign{ & {m_2}{v_2} = {m_1}{v_1} \Rightarrow {v_1} = \frac{{12 \times 4}}{4} = 12m{s^{ - 1}} \cr & \therefore K.E{._1} = \frac{1}{2}{m_1}v_1^2 = \frac{1}{2} \times 4 \times 144 = 288J \cr} $$
