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A block of mass m is pushed towards a movable wedge of mass ηm and height h, with a velocity u. All surfaces are smooth. The minimum value of u for which the block will reach the top of the wedge is(a) √(2gh)(b) η √(2gh)(c) √(2gh (1 + 1/η))(d) √(2gh ( 1 - 1/η)) |
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Answer» Correct answer is: (c) √(2gh (1 + 1/η)) The center of mass of the ‘block plus wedge’ must move with speed mu / m + ηm = u/1+η = vCM. ∴ 1/25 mu2 - mgh = 1/2 ( m + ηm) v2CM. |
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