# PHYSICS

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A merry-go-round (pictured) is sitting in a playground. It is free to rotate, but is currently stationary. You can model it as a uniform disk of mass 190 kg and radius 100 cm (consider the metal poles to have a negligible mass compared to the merry-go-round). The poles near the edge are 90 cm from the center.

Someone hits one of the poles with a 8 kg sledgehammer moving at 18 m/s in a direction tangent to the edge of the merry-go-round. The hammer is not moving after it hits the merry-go-round.

How much energy (modulus of (deltaE)) is lost in this collision? (enter a positive number for the absolute value in Joules)

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example of inelastic collision, where momentum before = momentum after, and simply calculate the difference in energy then!

• PHYSICS -

hi enjoy, what is the use of radius here?

• PHYSICS -

1/2 *[(m_1*m_2)/m_1+m_2)]*v_1^2

• PHYSICS -

So Anonymous you got 964.29 N??

• PHYSICS -

First, get all the variables:

m_1: The mass of the merry-go-round
m_2: The mass of the sledgehammer
v_1: The velocity of the merry-go-round before the collision
v_2: The velocity of the sledgehammer before the collision
v_1': The velocity of the merry-go-round after the collision
v_2': The velocity of the sledgehammer after the collision

Apply the conservation of momentum:

(m_1*v_1)+(m_2*v_2)=(m_1*v_1')+(m_2*v_2')

You know everything except for v_1', so you can solve the above equation for v_1'. Then look at the difference in kinetic energy before and after the collision:

((0.5*m_1*v_1^2)+(0.5*m_2*v_2^2))-((0.5*m_1*v_1'^2)+(0.5*m_2*v_2'^2))

• PHYSICS -

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so the radius is not important???
and we'll have to consider v_1=0 and v_2'=0

• PHYSICS -

First Calculate the KE of the sledge hammer.
2: Calculate the speed of the Merry-go-round (MGR) from m1.v1+m2.v2=m1.v1'+m2.v2'. Note: v2 and v1' are zero
3: Calculate angular KE of the MGR
4: Lost energy =1-3

So if you can't solve this then you should go to elementary school

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