Question

Bomb of mass 3.0 kg explord in air into two pieces of masses 2.0 kg and 1.0 kg the smaller mass goes at the speed of 80 metre per second the total energy impa to the two fragments

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Correct \: Question }}} \mid}}$

Bomb of mass 3.0 kg explord in air into two pieces of masses 2.0 kg and 1.0 kg the smaller mass goes at the speed of 80 metre per second the total energy imparted on the two fragments is?

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Total Mass = 3 kg.
• Speed of Smaller mass (1kg) After Explosion (u) = 80 m/s.
• Let the speed of Heavier Fragment be "v".

$\huge{\bold{\underline{Explanation:-}}}$

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Let The Final Velocity After Explosion of the Two masses be v and u.

From Law of conservation of momentum.

$\large{\boxed{\tt P_i = P_f}}$

But The Initial momentum will be Zero. As the Total mass was rest Before Explosion.

Therefore,

$\large{\tt \leadsto P_f = 0}$

Now,

$\large{\tt \leadsto Mv + mu = 0}$

• "M" = Mass of Heavier Fragment.
• "m" = Mass of Lighter Fragment.

Substituting the values

$\large{\tt \leadsto 2 \times v + 1 \times 80 = 0}$

$\large{\tt \leadsto 2v + 80 = 0}$

$\large{\tt \leadsto 2v = - 80}$

$\large{\tt \leadsto v = \dfrac{- 80}{2}}$

$\large{\tt \leadsto v = \dfrac{\cancel{- 80}}{\cancel{2}}}$

$\large{\tt \leadsto v =- 40 \: m/s}$

Here Negative sign indicates the Motion of the body is opposite to the other body.

Taking Magnitude

$\large{\boxed{\tt v = 40 \: m/s}}$

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Now,

Total Energy of the bodies.

$\large{\boxed{\tt T.E = K.E_1 + K.E_2}}$

• T.E = Total Energy

$\large{\tt \leadsto T.E = \dfrac{1}{2}Mv^2 + \dfrac{1}{2}mu^2}$

$\large{\tt \leadsto T.E = \dfrac{1}{2} \bigg( Mv^2 + mu^2 \bigg)}$

Substituting the values,

$\large{\tt \leadsto T.E = \dfrac{1}{2} \bigg( 2 \times (40)^2+ 1 \times (80)^2 \bigg)}$

$\large{\tt \leadsto T.E = \dfrac{1}{2} \bigg( 2 \times 1600+ 1 \times 6400 \bigg)}$

$\large{\tt \leadsto T.E = \dfrac{1}{2} \bigg( 3200+ 6400 \bigg)}$

$\large{\tt \leadsto T.E = \dfrac{1}{2} (9600)}$

$\large{\tt \leadsto T.E = \dfrac{1}{\cancel{2}}\cancel{(9600)}}$

$\large{\tt \leadsto T.E = 4800 \: J}$

$\huge{\boxed{\boxed{\tt T.E = 4.8 \: KJ }}}$

So, the Total Energy of the Two fragments is 4.8 KiloJoules.

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