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A cube of wood of side a and mass M is resting on a rough horizontal surface. A bullet of mass (m <<M) and moving with velocity $v_{0}$ strikes the block (cube). And gets embedded in it. Assume that wood offers constant resistance force and cube cannot slide due to high friction. The time of penetration of bullet into the cube so that the normal reaction passes through D is $\frac{x v_{0} m}{y M g}$ ( here $x$ and $y$ are co-primes) then find $x + y$ .

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Detailed Solution

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If the time of penetration is $△ t$
Then resistance force
$F = \frac{m v_{0}}{△ t}$
For reaction at the end
$\begin{array}{l} F \times \frac{3 a}{4} = M g \times \frac{a}{2} \\ \frac{m v_{0}}{△ t} \times \frac{3 a}{4} = \frac{M g a}{2} \\ △ t = \frac{3}{2} \frac{m v_{0}}{M g} \end{array}$

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A cube of wood of side a and mass M is resting on a rough horizontal surface. A bullet of mass (m <<M) and moving with velocity v0 strikes the block (cube). And gets embedded in it. Assume that wood offers constant resistance force and cube cannot slide due to high friction. The time of penetration of bullet into the cube so that the normal reaction passes through D is xv0myMg ( here x and y are co-primes) then find x+y.