A block of mass 10 kg moving with a velocity | vedclass

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Question

(B) According to the law of conservation of linear momentum,the total momentum before the explosion is equal to the total momentum after the explosion

Vedclass · Accepted Answer

$1.67 \ m \ s^{-1} \hat{i}$

Vedclass · Answer

(B) According to the law of conservation of linear momentum,the total momentum before the explosion is equal to the total momentum after the explosion.Initial momentum $P_i = M \vec{v} = 10 \ kg 	imes 5 \hat{i} \ m \ s^{-1} = 50 \hat{i} \ kg \ m \ s^{-1}$.Let the mass of the first piece be $m_1 = 4 \ kg$ and its velocity be $\vec{v}_1 = 10 \hat{i} \ m \ s^{-1}$.The mass of the second piece is $m_2 = M - m_1 = 10 \ kg - 4 \ kg = 6 \ kg$.Let the velocity of the second piece be $\vec{v}_2$.Final momentum $P_f = m_1 \vec{v}_1 + m_2 \vec{v}_2 = 4 	imes 10 \hat{i} + 6 	imes \vec{v}_2 = 40 \hat{i} + 6 \vec{v}_2$.Equating $P_i = P_f$:$50 \hat{i} = 40 \hat{i} + 6 \vec{v}_2$.$6 \vec{v}_2 = 10 \hat{i}$.$\vec{v}_2 = \frac{10}{6} \hat{i} = 1.67 \hat{i} \ m \ s^{-1}$.

$A$ block of mass $10 \ kg$ moving with a velocity of $5 \hat{i} \ m \ s^{-1}$ on a frictionless horizontal surface suddenly explodes into two pieces. If one piece with mass $4 \ kg$ moves with a velocity of $10 \hat{i} \ m \ s^{-1}$,then the velocity of the second piece is:

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