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. v_tou = (0.59)v_ico v_tou = (0.35)v_ico v_tou = (0.40)v_ico v_tou = (0.70)v_ico 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 bottom 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 bottom of the incline, the speed of the ball is 70% of the speed of the ice

Explanation / Answer

let h is the vertical height of the incline.

for Ice

Apply conservation of energy

final kinetic energy = initial potentila energy

(1/2)*m*v^2 = m*g*h

v_Ice = sqrt(2*g*h)

for sphere

let r is the radius of the sphere.

let w is the angular speed of the bal when it reaches the bottom.

Apply conservation of energy

final kinetic energy(trnslational + rotational) = initial potentila energy

(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h

(1/2)*m*v^2 + (1/2)*(2/5)*m*r^2*w^2 = m*g*h

(1/2)*m*v^2 + (1/5)*m*(r*w)^2 = m*g*h

(1/2)*m*v^2 + (1/5)*m*v^2 = m*g*h (since r*w = v)

(7/10)*m*v^2 = m*g*h

v_bal = sqrt(10*g*h/7)

= sqrt(5/7)*sqrt(2*g*h)

= 0.845*V_Ice

The above answer is not given in the options.

I checked many times I am getting the same answer.

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