A Box and a Wedge (ITA-IME Brazil)

Classical Mechanics Level 2

A box of mass m moves with speed v toward a wedge of mass M, which moves with speed u in the opposite direction. The box slides on the frictionless wedge and reaches a height h. There is no friction between the ground and the box. Find h.

(M/m+M)(u-v)²/2g (M/m-M)(u+v)²/2g (M/m-M)(u-v)²/2g (M/m+M)(u+v)²/2g

Samuel Nascimento by Samuel Nascimento , 24, Brazil

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1 solution

Samuel Nascimento
Jul 17, 2019

In physics, the reduced mass (m') is the "effective" inertial mass appearing in the two-body problem of Newtonian mechanics. It is a quantity which allows the two-body problem to be solved as if it were a one-body problem. Note, however, that the mass determining the gravitational force is not reduced.

m' = m M m + M \frac{mM}{m+M}

Let v' be the relative velocity between the box and the wedge: (v' = u+v)

From the conservation of mechanical energy, we have:

m ′ ( v ′ ) ² 2 \frac{m'(v')²}{2} = mgh

h = m ′ ( v ′ ) ² 2 m g \frac{m'(v')²}{2mg}

h = m M ( u + v ) ² 2 m g ( m + M ) \frac{mM(u+v)²}{2mg(m+M)}

h = M ( u + v ) ² 2 g ( m + M ) \frac{M(u+v)²}{2g(m+M)}

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