2010: Problems 15-16

Topic: System of Masses
Concepts: Conservation of energy, Conservation of linear momentum


Solution:

When the block reaches its maximum height, it is at rest with respect to the incline. Thus, they both have the same velocity vv. Conserving linear momentum,

mv0=(m+M)vmv_0=(m+M)v
v=mv0m+Mv=\frac{mv_0}{m+M}

To find the height hh, we conserve energy,

12mv02=mgh+12(m+M)v2=mgh+m2v022(m+M)\frac{1}{2}mv_0^2=mgh+\frac{1}{2}(m+M)v^2=mgh+\frac{m^2v_0^2}{2(m+M)}
h=v022g(1mm+M)=Mv022g(m+M)h=\frac{v_0^2}{2g}\left(1-\frac{m}{m+M}\right)=\frac{Mv_0^2}{2g(m+M)}

so the answer is C.

Topic: System of Masses
Concepts: 1D elastic collision, Speed vs. velocity


Solution:

After the block leaves the inclined plane, the system is equivalent to an elastic collision (since energy and momentum are both conserved). The velocity of the block is given by the elastic collision equation,

v1=mMm+Mv0v_1=\frac{m-M}{m+M}v_0

The speed is the magnitude of the velocity,

v=|v1|=Mmm+Mv0v=|v_1|=\frac{M-m}{m+M}v_0

so the answer is E.

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