Question

A body of mass 4 m at rest explodes into three
pieces. Two of the pieces each of mass m move with
a speed v each in mutually perpendicular directions.
The total kinetic energy released is :-
​

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\red{\boxed{\sf Total\:Kinetic\:Energy={\frac{3mv^2}{2}}}}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Velocity of mass (m) $\bf{v_m}$=v

• Two masses are of m and one is of 2m

$\large\underline\pink{\sf To\:Find: }$

• Total kinetic energy .

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Mass (m) move with speed v in perpendicular direction (Given )

Therefore ,

Resultant Velocity of 2m =$\sf{\sqrt{v^2+v^2}}$

$\implies\sf{{v_{2m}}=\sqrt{2}v}$

According to law of conservation of momentum :

$\large{\boxed{\sf m_1u_1+m_2u_2+m_3u_3=m_1v_1+m_2v_2+m_3v_3 }}$

Body is initially at rest .

Therefore

$\large\implies{\sf 0=mv+mv+2m\sqrt{2}v_m}$

$\large\implies{\sf 0=2m+2m\sqrt{2}v_m}$

$\large\implies{\sf v_m={\frac{-2mv}{2m\sqrt{2}}}}$

$\large{\boxed{\sf v_m={\frac{-v}{\sqrt{2}}}}}$

Total kinetic Energy :

$\implies{\sf K={\frac{1}{2}}mv^2+{\frac{1}{2}}mv^2+{\frac{1}{2}}2m×{\frac{-v^2}{\sqrt{2}}}}$

$\large\implies{\sf K={\frac{2mv^2}{2}}+{\frac{mv^2}{2}} }$

$\large\implies{\sf K={\frac{3mv^2}{2}} }$

$\huge\red{♡}$$\red{\boxed{\sf Total\:Kinetic\:Energy={\frac{3mv^2}{2}}}}$

Answer 2

see the attachement....

#Khushi here