A bullet weighing 10 ,g and moving with a vel | vedclass

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

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) According to the law of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision.Let $m_1 = 10 \,g = 0.01 \,kg$ be the mass of the bullet and $v_1 = 300 \,m/s$ be its initial velocity.Let $m_2 = 5 \,kg$ be the mass of the ice block and $v_2 = 0 \,m/s$ be its initial velocity.After the collision,the bullet gets embedded in the block,so the total mass is $M = m_1 + m_2 = 0.01 \,kg + 5 \,kg = 5.01 \,kg$.Let $V$ be the final velocity of the system.$m_1 v_1 + m_2 v_2 = (m_1 + m_2) V$$(0.01 \,kg)(300 \,m/s) + (5 \,kg)(0 \,m/s) = (5.01 \,kg) V$$3 = 5.01 V$$V = \frac{3}{5.01} \approx 0.5988 \,m/s \approx 0.6 \,m/s$.To convert this into $cm/s$,we multiply by $100$:$V = 0.6 	imes 100 = 60 \,cm/s$.

$A$ bullet weighing $10 \,g$ and moving with a velocity $300 \,m/s$ strikes a $5 \,kg$ block of ice and gets embedded in it. The ice block is kept on a smooth surface. The speed of the block after the collision is ........ $cm/s$.

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