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.
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 μ. 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=μmBgs((mB)2mB+mM)gH=μmMgs((mB)2mB+mM)gH=μ(mB+mM)gs((mB)2mB+mM)gH=−μ(mB+mM)gs((mB)2mB+mM)gH=μ(mB+mM)g

The conditions of the problem require us to apply both principles. When the iron block is moving alone with no friction force, the energy is conserved. But when it sticks with the magnet, mechanical energy is not conserved, but the momentum is.

When the iron block it at a height H (assumed at rest), the total mechanical energy is

[tex]E_m=m_BgH[/tex]

when it reaches the base of the hill, all the potential gravitational energy is transformed into kinetic energy, thus

[tex]\displaystyle m_BgH=\frac{m_Bv_b^2}{2}[/tex]

Solving for vB

[tex]\displaystyle v_b^2=2gH[/tex]

This is the initial speed just before it hits and sticks with the magnet. The total momentum before and after the collision is conserved, thus

[tex]p_o=p_1[/tex]

[tex]m_Bv_B=(m_b+m_M)v[/tex]

Squaring

[tex]m_B^2v_B^2=(m_b+m_M)^2v^2[/tex]

Replacing the expression for vB

[tex]m_B^22gH=(m_b+m_M)^2v^2[/tex]

Solving for v

[tex]\displaystyle v^2=\frac{2m_B^2gH}{(m_b+m_M)^2}[/tex]

Now let's consider the dynamic equations. The final speed is related to the distance s with the formula

[tex]v_f^2=v_o^2+2as[/tex]

Since the objects eventually stop, vf=0. Solving for s

[tex]\displaystyle s=\frac{v_o^2}{-2a} \text{ ........[1]}[/tex]

Where vo is the speed we have already found for the both objects after the collision, and a is the braking acceleration, that can be found by using the fact that the objects are being stopped by the force of friction, thus

[tex]F_r=(m_B+m_M).a[/tex]

Since

[tex]F_r=-\mu (m_B+m_M).g[/tex]

Then

[tex]-\mu (m_B+m_M).g=(m_B+m_M).a[/tex]

Solving for a

[tex]\displaystyle a=-\mu g[/tex]

Replacing all in the equation [1]

[tex]\displaystyle s=\frac{\frac{2m_B^2gH}{(m_b+m_M)^2}}{2\mu g}[/tex]

Operating and simplifying

[tex]\boxed{\displaystyle s=\frac{m_B^2H}{\mu (m_b+m_M)^2}}[/tex]

Note: The options presented are not clearly seen. We give the answer and the student can now safely pick the correct choice.