A block of ice and a rubber ball of identical mass are released from rest at the

Question

A block of ice and a rubber ball of identical mass are released from rest at the top of an inclined plane and allowed to descend under the influence of gravity Assuming that the block of ice slides without friction and that the rubber ball rolls without slipping. Compute the translational speed of the ball when it reaches the bottom of the incline in terms of the translational speed of the block of ice at the bottom of the incline. Treat the ball as a solid sphere of uniform mass density b) a 0.35)v c) v (0.40)V d) v- (0.70)v.. As each object reaches the bottom of the incline, the speed of the ball is 59% of the speed of the ice As each object reaches the bottorm of the incline, the speed of the ball is 35% of the speed of the ice As each object reaches the bottom of the incline, the speed of the ball is 40% of the speed of the ice As each object reaches the 70%of the speed of thei bottorm of the incline the speed of the ballis

Explanation / Answer

Hello,

The ball will have translational as well as rotational kinetic energy. For rolling without slipping, v = r*w

for ice block

mgh = 1/2*m*v2

m cancels out,

therefore vice2 = 19.6h (for ice) ------- (1)

Now, for rubber ball (solid sphere)

mgh = 1/2*m*v2 + 1/2*I*w2

for solid sphere, I = 2/5 mr2

mgh =  1/2*m*v2 + 1/2*2/5 mr2*(v/r)2

mgh = 1/2*m*v2 + 1/5*m*v2

m will cancel out

gh = 1/2*v2 + 1/5*v2

v2 = gh / 0.7

vball2 = 14h ( ball) --------------- (2)

Now compare 1 and 2 to get the answer.

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