Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A bomb at rest explodes into 3 parts of the same mass. The momentum of two parts is $- 3P\widehat i$ and $2P\widehat j$ respectively. The magnitude of the momentum of the third part is.A) PB) $\sqrt 5 P$C) $\sqrt {11} P$D) $\sqrt {13} P$

Last updated date: 14th Sep 2024
Total views: 79.8k
Views today: 0.79k
Verified
79.8k+ views
Hint: Since, the bomb was initially at rest state, hence after it explodes, the net momentum of it’s centre of mass should be zero, because the rest bomb had momentum zero before it exploded. So, we just have to conserve linear momentum of the bomb to find the momentum of the third particle.

Question states that bomb explodes into three parts of same mass, so all particles should have mass
${m_1} + {m_2} + {m_3} = \dfrac{m}{3}$,
Now initial momentum of bomb ($\overrightarrow {{P_{initial}}}$)$= 0$,
Momentum of first particle $\left( {\overrightarrow {{P_1}} } \right)$$= - 3P\widehat i Momentum of second particle \left( {\overrightarrow {{P_2}} } \right)$$ = 2P\widehat j,$
Let momentum of third particle be $\overrightarrow {{P_3}}$
Using law of conservation of momentum,
$\overrightarrow {{P_{initial}}} = \overrightarrow {{P_{final}}}$
So, we get,
$\overrightarrow {{P_{initial}}} = \overrightarrow {{P_1}} + \overrightarrow {{P_2}} + \overrightarrow {{P_3}}$
Initial momentum is 0
$\overrightarrow {{P_1}} + \overrightarrow {{P_2}} + \overrightarrow {{P_3}} = 0$
Putting all values, we get,
$- 3P\widehat i + 2P\widehat j + \overrightarrow {{P_3}} = 0 \\ \overrightarrow {{P_3}} = 3P\widehat i - 2P\widehat j \\$
Now, we want to find its magnitude, so,
$\overrightarrow {{P_3}} = \sqrt {{{\left( {3P} \right)}^2} + {{\left( { - 2P} \right)}^2}} \\ \overrightarrow {{P_3}} = \sqrt {13{P^2}} \\ \overrightarrow {{P_3}} = \sqrt {13} P \\$
Hence magnitude of momentum of third particle will be $\sqrt {13} P$
So our answer is option (D)

Note: We were able to apply the law of conservation of linear momentum in the bomb because there was no external force acting on the bomb before it exploded. Now, the initial linear momentum of the bomb was 0 because rest objects always have linear momentum 0 because their velocity is 0.