Question

A stationary bomb explodes into two fragments of masses 3 kg and 1 kg total k.e is 2400 J find k.e energy of smaller part

Answer 1

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

$\large\red{\boxed{\sf KE=600J}}$

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

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

Mass ($\sf{m_1}$)=1kg

Mass ($\sf{m_2}$)=3kg

Total kinetic energy ($\sf{K_{total})=2400J}$

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

Kinetic Energy of larger part = ?

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According to law of conservation of momentum :

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

$\large\implies{\sf v_1+3v_2=0 }$

$\large\boxed{\sf v_1=-3v_2→(1)}$

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Now ,

$\large{\boxed{\sf {\frac{1}{2}}mv^2={\frac{1}{2}}mv_1^2+{\frac{1}{2}}mv_2^2 }}$

$\large\implies{\sf 2400={\frac{1}{2}}(-3v_2)^2+{\frac{1}{2}}×3v_2^2}$

$\large\implies{\sf 2400={\frac{9}{2}}v_2^2+{\frac{3}{2}}v_2^2}$

$\large\implies{\sf 2400=6v_2^2 }$

$\large\implies{\sf v_2^2={\frac{2400}{6}} }$

$\large\boxed{\sf v_2^2=20m/s }$

On putting value of $\sf{v_2}$ in Equation first .

$\large\implies{\sf v_1=3v_2 }$

$\large\implies{\sf v_1=3×20 }$

$\large\boxed{\sf v_1=60m/s }$

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Kinetic Energy of larger part i.e of 3kg is :

$\large{♡}$$\large{\boxed{\sf KE={\frac{1}{2}}m_2v_2^2 }}$

$\large\implies{\sf KE={\frac{1}{2}}3×20×20 }$

$\large\implies{\sf KE=600J}$

$\huge\red{♡}$$\large\red{\boxed{\sf KE=600J}}$

Kinetic Energy of larger part is 600J