Question

A bomb of mass 16 kg at rest explodes into two
pieces of masses 4 kg and 12 kg. The ratio of
kinetic energy of heavier piece to lighter piece is
(1) 1:3
(2) 1:4
(3) 4:1
(4) 1:1​

Answer 1

☆Momentum conservation:

Since the bomb explodes due to result of its internal forces, internal force don't change the Momentum of system.

Momentum will remain conserved.

$\implies \rm \vec{F}_{ext} = \dfrac{dp}{dt} = 0$

$\implies \rm \vec{P}_{initial} = \vec{P}_{final}$

☆Solution:

Refer the attached figure!

1st particle:

• $\ mass = m_1 = 4kg$

2nd particle:

• $\ mass = m_2 = 12kg$

By Momentum conservation

$\longrightarrow \rm \vec{P}_{bomb} = \vec{P}_1 + \vec{P}_2$

$\longrightarrow \rm 0 = \vec{P}_1 + \vec{P}_2$-----(due to rest initially)

$\longrightarrow \rm \vec{P}_1 = - \vec{P}_2$

• We know that

$\implies \rm KE = \dfrac{1}{2}mv^2 = \dfrac{1\times m}{2\times m}mv^2 =\dfrac{m^2v^2}{2m}$

$\implies \bf KE = \dfrac{P^2}{2m}$

•Now the according to ques, ratio of KE of heavier(2) to lighter(1):

$\longrightarrow \displaystyle \rm \dfrac{KE_2}{KE_1} = \dfrac{\dfrac{\vec{P}_2^2}{2m_2}}{\dfrac{\vec{P}_1^2}{2m_1}}$

$\longrightarrow \displaystyle \rm \dfrac{KE_2}{KE_1} = \dfrac{\vec{P}_2^2 \times m_1}{(-\vec{P}_2)^2 \times m_2}$

$\longrightarrow \displaystyle \rm \dfrac{KE_2}{KE_1} = \dfrac{P_2^2 \times m_1}{P_2^2 \times m_2}$

$\longrightarrow \rm \dfrac{KE_2}{KE_1} = \dfrac{m_1}{m_2}$

$\longrightarrow \rm \dfrac{KE_2}{KE_1} = \dfrac{4\: kg}{12\: kg}$

$\longrightarrow \bf \dfrac{KE_2}{KE_1} = \dfrac{1}{3}$

☆The ratio of kinetic energy of heavier piece to lighter piece is 1:3.