A bullet moving with a velocity of 200 , cm/s | vedclass

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(C) Using the equation of motion $v^2 = u^2 - 2as$,where $v$ is the final velocity,$u$ is the initial velocity,$a$ is the retardation,and $s$ is the d

Vedclass · Accepted Answer

$300$

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(C) Using the equation of motion $v^2 = u^2 - 2as$,where $v$ is the final velocity,$u$ is the initial velocity,$a$ is the retardation,and $s$ is the distance.Since the bullet comes to rest,$v = 0$.Therefore,$0 = u^2 - 2as$,which implies $u^2 = 2as$.Assuming the retardation $a$ is constant for the same block,we get $u^2 \propto s$.This gives the ratio: $\frac{u_2^2}{u_1^2} = \frac{s_2}{s_1}$.Given $u_1 = 200 \, cm/s$,$s_1 = 4 \, cm$,and $s_2 = 9 \, cm$.Substituting the values: $\frac{u_2^2}{200^2} = \frac{9}{4}$.Taking the square root on both sides: $\frac{u_2}{200} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.$u_2 = 200 	imes \frac{3}{2} = 300 \, cm/s$.

$A$ bullet moving with a velocity of $200 \, cm/s$ penetrates a wooden block and comes to rest after traversing $4 \, cm$ inside it. What velocity is needed for travelling a distance of $9 \, cm$ in the same block? (in $cm/s$)

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