
A body at rest breaks up into $3$ parts. If $2$ parts having equal masses fly off perpendicularly each after with a velocity of $12\,m{s^{ - 1}}$ then the velocity of the third part which has $3$ times mass of each part is
A. $4\sqrt 2 m{s^{ - 1}}$ at angle of ${45^o}$ from each body
B. $24\sqrt 2 m{s^{ - 1}}$ at angle of ${135^o}$ from each body
C. $6\sqrt 2 m{s^{ - 1}}$ at angle of ${135^o}$ from each body
D. $4\sqrt 2 m{s^{ - 1}}$ at angle of ${135^o}$ from each body
Answer
162.3k+ views
Hint:In order to solve this question, we will apply the law of conservation of linear momentum and then using this principle and equations we will determine the velocity of the third part.
Formula used:
The principle of conservation of linear momentum says that Initial momentum of a system is always equal to final momentum of a system.
${P_i} = {P_f}$
where P denotes the momentum of a body defined as the product of the mass of the body and the velocity of the body. $P = mv$
Complete step by step solution:
Initial body was at rest and after the explosion, two bodies of equal mass let’s say m moves perpendicular to each other with velocity of $u = 12\,m{s^{ - 1}}$. Now, to conserve the linear momentum of the system the third body having mass of $3m$ must fly exactly opposite to the resultant momentum direction of two perpendicular bodies.
The rough diagram of all the three particles are shown as,

So, Resultant momentum of bodies having equal mass m and velocity $u = 12\,m{s^{ - 1}}$ can be find as,
$R = m\sqrt {{u^2} + {u^2}} \\
\Rightarrow R = 12\sqrt 2\,m\, kgm{s^{ - 1}}$
And due to equal velocities this resultant will make an angle of ${45^o}$ from each of equal mass bodies.
Now, momentum of third body having mass $3m$ is $R' = 3mv$ where v is its velocity must be equal to Resultant R so, we have
$12\sqrt 2\,m = 3mv \\
\Rightarrow v = 4\sqrt 2 m{s^{ - 1}} $
Now, since R and R' form a straight line because they are opposite to each other, R’ will have an angle of $= 180 - 45 = {135^o}$ with each of the other two bodies.
Hence, the correct answer is option D.
Note: It should be remembered that, always make sure to take negative sign velocity as in opposite direction to that of positive one as velocity is a vector quantity therefore momentum is also a vector quantity and always draw rough diagram of such kind of problems to solve easily.
Formula used:
The principle of conservation of linear momentum says that Initial momentum of a system is always equal to final momentum of a system.
${P_i} = {P_f}$
where P denotes the momentum of a body defined as the product of the mass of the body and the velocity of the body. $P = mv$
Complete step by step solution:
Initial body was at rest and after the explosion, two bodies of equal mass let’s say m moves perpendicular to each other with velocity of $u = 12\,m{s^{ - 1}}$. Now, to conserve the linear momentum of the system the third body having mass of $3m$ must fly exactly opposite to the resultant momentum direction of two perpendicular bodies.
The rough diagram of all the three particles are shown as,

So, Resultant momentum of bodies having equal mass m and velocity $u = 12\,m{s^{ - 1}}$ can be find as,
$R = m\sqrt {{u^2} + {u^2}} \\
\Rightarrow R = 12\sqrt 2\,m\, kgm{s^{ - 1}}$
And due to equal velocities this resultant will make an angle of ${45^o}$ from each of equal mass bodies.
Now, momentum of third body having mass $3m$ is $R' = 3mv$ where v is its velocity must be equal to Resultant R so, we have
$12\sqrt 2\,m = 3mv \\
\Rightarrow v = 4\sqrt 2 m{s^{ - 1}} $
Now, since R and R' form a straight line because they are opposite to each other, R’ will have an angle of $= 180 - 45 = {135^o}$ with each of the other two bodies.
Hence, the correct answer is option D.
Note: It should be remembered that, always make sure to take negative sign velocity as in opposite direction to that of positive one as velocity is a vector quantity therefore momentum is also a vector quantity and always draw rough diagram of such kind of problems to solve easily.
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