A mass of 1 ; kg is thrown up with a velocity | vedclass

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(C) Initial velocity $u = 100 \; m/s$. Acceleration $g = -10 \; m/s^2$.Velocity of the mass at $t = 5 \; s$ is $v = u + at = 100 - 10 	imes 5 = 50 \;

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$100 \; m/s$ upward

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(C) Initial velocity $u = 100 \; m/s$. Acceleration $g = -10 \; m/s^2$.Velocity of the mass at $t = 5 \; s$ is $v = u + at = 100 - 10 	imes 5 = 50 \; m/s$ (upward).Let the total mass be $M = 1 \; kg$. The mass splits into $m_1 = 0.4 \; kg$ and $m_2 = 0.6 \; kg$.According to the law of conservation of momentum,the momentum before the explosion equals the momentum after the explosion.$Mv = m_1 v_1 + m_2 v_2$Taking upward direction as positive:$1 	imes 50 = 0.4 	imes (-25) + 0.6 	imes v_2$$50 = -10 + 0.6 	imes v_2$$60 = 0.6 	imes v_2$$v_2 = \frac{60}{0.6} = 100 \; m/s$.Since the result is positive,the velocity is $100 \; m/s$ upward.

$A$ mass of $1 \; kg$ is thrown up with a velocity of $100 \; m/s$. After $5 \; s$,it explodes into two parts. One part of mass $400 \; g$ comes down with a velocity of $25 \; m/s$. Calculate the velocity of the other part.

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