Selection of Even–Numbered Problem Solutions

Chapter 7

P–7.12

We use the conservation of energy concept, written as:

(1)

for which we first identify the initial and final instants. Choose them such that the information available at

one instant is complete, and that the unknown parameter occurs at the other instant. It doesn't matter,

however, on which side of Eq. [1] the unknown parameter occurs.

We choose as the initial instant the moment of release when both objects are at rest. For the final

instant we choose the instant at which the object of mass M has fallen 30 cm since we are asked about the

total kinetic energy (unknown variable) at that point. Now we evaluate each term in Eq. [1]:

! The initial kinetic energy is zero since both objects start from rest.

! The initial potential energy can be chosen arbitrarily since there is no predetermined origin along the

vertical axis which defines y = 0 in Epot = m g y for either object. For convenience, we choose y = 0 at the

initial height for each of the two objects and get Epot, init = 0 J. We also define the y–axis as directed upward.

! The final kinetic energy of the two objects is sought. We write this energy as Ekin, final = ½(m + M) v²final.

(2)

! The final potential energy has two contributions. The object of mass M is at this time at y = – 0.3 m, i.e.,

the contribution of this object to the final potential energy is:

(3)

This value is negative since the object is closer to Earth, which is associated with a decrease in the potential

energy. At the same time, the object of mass m has moved 30 cm up the inclined plane. We know this because

the massless string is taut, and therefore, does not change its length during the process. Since the object of

mass m moves along the inclined plane and not vertically up, its position relative to the surface of Earth has

not changed by a length of 0.3 m but by a distance d:

(4)

which is the y–component of the total displacement. Remember, the potential energy is a function of the

position relative to the surface of Earth; thus, only the change in the vertical position matters. From this we

now calculate the final potential energy of the object of mass m:

(5)

Next we enter all four terms we found into Eq. [1]:

(6)

which yields:

(7)

The final kinetic energy is larger than zero since both objects are in motion when the object of mass M passes

the 30 cm marker. The result in Eq. [7] allows us to calculate the final speed:

(8)

The sum (m + M) occurs in the denominator because the objects are linked by the string, and therefore move

with the same speed at every instant.

## Document Summary

We use the conservation of energy concept, written as: (1) for which we first identify the initial and final instants. Choose them such that the information available at one instant is complete, and that the unknown parameter occurs at the other instant. It doesn"t matter, however, on which side of eq. We choose as the initial instant the moment of release when both objects are at rest. For the final instant we choose the instant at which the object of mass m has fallen 30 cm since we are asked about the total kinetic energy (unknown variable) at that point. The initial kinetic energy is zero since both objects start from rest. The initial potential energy can be chosen arbitrarily since there is no predetermined origin along the vertical axis which defines y = 0 in epot = m g y for either object.