Block A, of mass 0.08 kg, is released from rest at the rim of a frictionless hem

Question

Block A, of mass 0.08 kg, is released from rest at the rim of a frictionless hemispherical bowl of radius R = 0.5m. It slides down the inside of the bowl and strikes block B, of mass 0.04 kg and initially at rest at the bottom of the bowl. The two blocks stick together at this point.

Initial placement.

a) What is the speed of block A just before it reaches block B?

b) What is the magnitude of the normal force exerted on block A just before it reaches block B?

c) What is the speed of the combined object (blocks A and B stuck together) immediately after the collision?

d) What is the maximum vertical distance above the bottom reached by the combined A-B?

Explanation / Answer

a) The speed of block A just before it reaches block B can be obatined by conserving mechanical energy

Initial potential energy = Final kinetic energy

or, m x g xh = 0.5 x m x v2

or, v = (2 x g x h)0.5 = (2 x 10 x 0.5)0.5 = 3.162 m/s

b) Magnitude of the normal force = Centripetal force = mv2/r = 0.08 x 3.1622 / 0.5 = 1.6 N

c) The speed of the combined mass may be obtained by conserving mechanical energy

0.5 x m1 x v12 = 0.5 x (m1 + m2) x v22

or, 0.08 x 3.1622 = (0.08 + 0.04) x v22

or, v22 = 6.67

or, v2 = 2.58 m/s

d) Maximum verticle height can again be computed by conserving mechanical energy

0.5 x (m1 + m2) x v22 =  (m1 + m2) x g xh

or, h =  v22/2g = 0.3335 m

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