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A block of mass `m` is pushed towards a movable wedge of mass `nm` 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. Block will reach top of the wedge if `u=sqrt(2gh(1-(1)/(n)))`B. Block will reach top of the wedge if `u=sqrt(2gh(1+(1)/(n)))`C. If the block overshoots `P`, the angle of projectile is `alpha`.D. If the block overshoots `P`, the angle of projectile is less than `alpha`. |
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Answer» When the particle just reaches the top of the wedge `mu=(m+n)v`…..(`1`) `u=sqrt(2gh(1+(1)/(n)))` Angle of projection as observed by the ground will be loss than `alpha` `vecv_(block//ground)=vecv_(block//"wedge")+vecv_(block//ground)` |
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