A body of mass 50 , kg is projected verticall | vedclass

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(A) The initial velocity of the $50 \, kg$ mass is $u = 100 \, m/s$. Using $v = u - gt$ with $g = 10 \, m/s^2$ (standard approximation),the velocity a

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$15 \, m/s$ downwards

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(A) The initial velocity of the $50 \, kg$ mass is $u = 100 \, m/s$. Using $v = u - gt$ with $g = 10 \, m/s^2$ (standard approximation),the velocity after $t = 5 \, s$ is $v = 100 - 10 	imes 5 = 50 \, m/s$.Initial momentum $P_i = m 	imes v = 50 	imes 50 = 2500 \, kg \cdot m/s$ (upwards).Let the velocity of the $30 \, kg$ piece be $V$. According to the law of conservation of momentum,$P_i = P_f$.$2500 = (20 	imes 150) + (30 	imes V)$.$2500 = 3000 + 30V$.$30V = 2500 - 3000 = -500$.$V = -500 / 30 = -16.67 \, m/s$. Note: Using $g = 9.8 \, m/s^2$ as per the original prompt: $v = 100 - 9.8 	imes 5 = 51 \, m/s$.$P_i = 50 	imes 51 = 2550 \, kg \cdot m/s$.$2550 = (20 	imes 150) + 30V$.$2550 = 3000 + 30V$.$30V = -450$.$V = -15 \, m/s$.The negative sign indicates the $30 \, kg$ piece moves downwards at $15 \, m/s$.

$A$ body of mass $50 \, kg$ is projected vertically upwards with a velocity of $100 \, m/s$. After $5 \, s$,the body breaks into two pieces of mass $20 \, kg$ and $30 \, kg$. If the $20 \, kg$ piece travels upwards with a velocity of $150 \, m/s$,what is the velocity of the $30 \, kg$ block?

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