A small block of mass m 1 = 0.300 kg is released from rest at the top of a curve

Question

A small block of mass m1 = 0.300 kg is released from rest at the top of a curved wedge of mass m2 = 5.00 kg, which sits on  a frictionless horizontal surface as in figure (a) below.  When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right, as in figure (b) below.  

A small block of mass m1 = 0.300 kg is released from rest at the top of a curved wedge of mass m2 = 5.00 kg, which sits on a frictionless horizontal surface as in figure (a) below. When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right, as in figure (b) below. What is the velocity of the wedge after the block reaches the horizontal surface? (Take right to be positive.) m/s What is the height h of the wedge? h = m

Explanation / Answer

a. Use conservation of momentum. So, write out the momentum of the block and set it equal to the momentum of the wedge.

mBvB=mWvW

(.3)(4)=(5)(vW)

vW=0.24m/s

vW= -0.24m/s The wedge is going to go the opposite direction as the block, so make it negative.

b. Use conservation of energy. The gravitational potential energy equals the kinetic energy of the block.

mgh=1/2mvB^2

9.8h=1/2(4)^2

h=.816 m

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