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A bomb explodes in the air when it has a horizontal speed of $100\,km{h^{ - 1}}$. It breaks into two pieces A, B of mass ratio $1:2$. If A goes vertically at speed $400\,km{h^{ - 1}}$, the speed of B will be:
A. $200\,km{h^{ - 1}}$
B. $250\,km{h^{ - 1}}$
C. $300\,km{h^{ - 1}}$
D. $500\,km{h^{ - 1}}$

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Last updated date: 23rd Jul 2024
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Answer
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Hint:Speed can be explained as the ratio of distance travelled to the time it took to travel that distance. Since speed has only direction and no magnitude, it is a scalar quantity.Momentum is a vector quantity in the sense that it has both a magnitude and a direction.

Complete step by step answer:
We can solve this question using the law of conservation of momentum. If there is no additional force acting on the colliding objects, the law of conservation of momentum tells that if two objects collide, their overall momentum before and after the collision will be the same. Since the bomb breaks into two pieces, A and B having the mass ratio $1:2$ which means that: the mass of piece A can be written as $1\,m$ and the mass of piece B can be written as $2\,m$. The total mass of the bomb can be written as $3m$.

When we apply the law of conservation of momentum in the vertical direction, we get,
$m \times 400 = 2m \times {v_y}$
where $400$ is the speed of piece A travelling vertically
${v_y} = 200\,km{h^{ - 1}}$
By applying the law of conservation of momentum in the horizontal direction, we get,
$3m \times 100 = 2m \times {v_x}$
where $100$ is the speed of bomb travelling horizontally
${v_x} = 150\,km{h^{ - 1}}$

The speed of piece B can be calculated using the formula ${v_B} = \sqrt {v_x^2 + v_y^2} $.
$\Rightarrow {v_B} = \sqrt {v_x^2 + v_y^2} \\
\Rightarrow {v_B}= \sqrt {{{150}^2} + {{200}^2}} \\
\Rightarrow {v_B}= \sqrt {22500 + 40000} \\
\Rightarrow {v_B}= \sqrt {100\left( {225 + 400} \right)} \\
\Rightarrow {v_B}= 10 \times \sqrt {625} \\
\Rightarrow {v_B}= 10 \times 25 \\
\therefore {v_B}= 250\,km{h^{ - 1}} \\ $
Hence, the correct option is B.

Note: As there are no external forces in an isolated system (such as the universe), momentum is still conserved. Since momentum is conserved, its components will be conserved in every direction. In order to solve collision problems, the law of conservation of momentum must be applied.