Block A with mass 0.0400 kg is released from rest at the rim of a frictionless h

Question

Block A with mass 0.0400 kg is released from rest at the rim of a frictionless hemispherical bowl that has radius R=0.600 m. Block A slides down the side ofthe bowl and strikes block B which has mass 0.0800 kg and that is initially sitting at rest at the bottom of the bowl, and the two blocks stick together. 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(Block A and B stuck together) immediately after the collision? d) What maximum vertical distance above the bottom of the bowl will the combined object reach after the collision?

Explanation / Answer

a)
mgh = 1/2mv^2

v= 2gh = 2*9.8*0.6 = 3.43 m/s

b)

N = mv^2/r = 0.04*3.432/0.6 = 0.78 N

c)

momentum conservation:

0.04*3.43 = (0.04+0.08)v

v = 1.143 m/s

d)

v = 2gh or

h = v^2/2g = 1.1432/2*9.8 = 0.067 m

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---