A bomb of mass 9 ,kg explodes into two pieces | vedclass

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(C) According to the law of conservation of linear momentum,the initial momentum of the bomb is zero because it is at rest.Let $m_1 = 3 \,kg$ and $m_2

Vedclass · Accepted Answer

$192$

Vedclass · Answer

(C) According to the law of conservation of linear momentum,the initial momentum of the bomb is zero because it is at rest.Let $m_1 = 3 \,kg$ and $m_2 = 6 \,kg$. The velocity of $m_1$ is $v_1 = 16 \,m/s$.Since $m_1 v_1 + m_2 v_2 = 0$,we have $3 	imes 16 + 6 	imes v_2 = 0$.$48 + 6 v_2 = 0 \implies v_2 = -8 \,m/s$.The kinetic energy of the $6 \,kg$ mass is $K.E. = \frac{1}{2} m_2 v_2^2$.$K.E. = \frac{1}{2} 	imes 6 	imes (-8)^2 = 3 	imes 64 = 192 \,J$.

$A$ bomb of mass $9 \,kg$ explodes into two pieces of masses $3 \,kg$ and $6 \,kg$. The velocity of mass $3 \,kg$ is $16 \,m/s$. The kinetic energy of mass $6 \,kg$ in joule is

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