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E-3.s A bullet of mass m moving vertically upwards instantaneously with a velocity 'u' hits the hanging blockof mass 'm' and gets embedded in it, as shown in the figure. The height through which the block nisesafter the collision. (assume sufficient space above block) isul(A) u/2g(B) u'/g(c) u'iBg(D) u14g |
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Answer» From conservation of momentum , the final velocity after the bullet gets embedded is V' = mu/(m+m) = u/2 now we know tha velocity is = u/2 so, using Newton's 3rd formula of motion, at maximum height , velocity will be 0 => 0² = (u/2)² -2gH=> H = u²/8g option C |
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