Two block of mass M = 30 kg and m = 10 kg slide without friction down the sides

Question

Two block of mass M = 30 kg and m = 10 kg slide without friction down the sides of a hemispherical bowl. They each start simultaneously from rest at the lip of the bowl, a height h = 6.0 m above the bottom of the bowl. (You may assume that the size of the blocks is small compared to h.)(There is gravity in this problem. You may take g = 10 m/s^2.) a) If the collision is perfectly inelastic (they stick together), what is the maximum height that the stuck-together blocks will reach? b) If the collision between the blocks is elastic, what is the maximum height that the 10 kg block will reach

Explanation / Answer

Here Velocity attained by the both te blocks at the bottom = sqrt(2gh) = sqrt(2*9.8*6) = 10.844 m/sec Now Apply LAw of Conservation of Momentum 30*10.844 - 10*10.844 = (30+10)*V V = 5.422 m/sec Now H = V^2/2g = (5.422)^2/19.6 = 1.5 m

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