An iron block with mass m B slides down a frictionless hill of height H . At the

Question

An iron block with mass mB slides down a frictionless hill of height H. At the base of the hill, it collides with and sticks to a magnet with mass mM.

a)

What is the speed v of the block and magnet immediately after the collision?

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What is the speed  of the block and magnet immediately after the collision?

v=(mBmM)2gH

b)

Now assume that the two masses continue to move at the speed v from Part A until they encounter a rough surface. The coefficient of friction between the masses and the surface is . If the blocks come to rest after a distance s, which of the following equations would you use to find s?

View Available Hint(s)

Now assume that the two masses continue to move at the speed  from Part A until they encounter a rough surface. The coefficient of friction between the masses and the surface is . If the blocks come to rest after a distance , which of the following equations would you use to find ?

v=(mB+mMmB)2gH v=(mBmB+mM)2gH v=(mBmM)2gH v=(mB+mMmB)2gH

v=(mBmM)2gH

b)

Now assume that the two masses continue to move at the speed v from Part A until they encounter a rough surface. The coefficient of friction between the masses and the surface is . If the blocks come to rest after a distance s, which of the following equations would you use to find s?

View Available Hint(s)

Now assume that the two masses continue to move at the speed  from Part A until they encounter a rough surface. The coefficient of friction between the masses and the surface is . If the blocks come to rest after a distance , which of the following equations would you use to find ?

((mB)2mB+mM)gH=mBgs ((mB)2mB+mM)gH=mMgs ((mB)2mB+mM)gH=(mB+mM)gs ((mB)2mB+mM)gH=(mB+mM)gs ((mB)2mB+mM)gH=(mB+mM)g

Explanation / Answer

(A) speed of block just before collision,

Applying energy conservation,

m g H + 0 = 0 + m v^2 /2

v = sqrt(2 g H)

Now applying momentum conservation for the collision,

mB v = (mB + mM) vf

vf = mB sqrt(2 g H) / (mB + mM)

(B) Work done by friction = - f d

= - u (mB + mM) g d

so - (mB + mM) vf^2 /2 = = Wf

mB^2 (g H) = u (mB + mM) g d

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