Download 2012 F=ma Solutions - Art of Problem Solving

2012 F=ma Solutions
1. When the first drop hits the sink, a time
√
√
passes. When this drop hits the sink
the second drop is at a height above the sink after falling a distance . The time it takes the
second drop to hit the sink is the difference between the time of the first drop and the time it
takes the second drop to fall a distance
(√
√
(
)
)
( )(
√
or
)
√
( )
2. The equation for the maximum height of a projectile is
( )
( )
so
( )
(
)
Therefore,
√
√
Now use the quadratic equation to yield the only positive solution,
is between
.
of a projectile is
√
so the answer
and the maximum range
for
3. The triangle will topple when the line connecting its center of mass to the lowermost vertex is
perpendicular to the incline. As the angle between the CoM and the vertex is
the incline
needs to be
4. Momentum is conserved, and the initial momentum of the system is zero as it is at rest. When
two pieces emerge perpendicular to each other, their resultant speed is √
√ . The
third piece must travel in an equal and opposite direction to the resultant, so the speed of the
third particle is √ .
5. First we find the initial kinetic energy of the system, which is the sum of the individual kinetic
(
energy of each block or
)( )
( )( )
The blocks stick together so
they have a combined mass of
This is an inelastic collision, so we can find their
( )
speed by using conservation of momentum or ( )
The kinetic
energy of the combined blocks is then
(
)( )
The kinetic energy lost is the
difference of the initial and final kinetic energy or
6. For the upper cannon,
and for the lower cannon,
. We
know the sum of the distances each cannon travels must be
so
( ) (
(
))
7. This is the distance the upper cannonball travels or
8. The initial kinetic energy before the spring is compressed is equal to the potential energy of the
spring plus the work due to friction as the spring is compressed. We have
( )( )
to
(
)
Now apply the quadratic equation
to yield the only positive solution,
9. When an object escapes from a planet, we can consider its distance to be infinite and its
velocity to be negligibly small. So equating the initial and final energy, we have
√
. However, we know that the acceleration due to gravity, g, is equal to
√
so
10.
. The moment of inertia takes the form
and
for some constant
( )
so
We see that energy doesn’t depend on mass or radius, so times A,B and D should
have the same time as they are all solid balls and C should have the greatest time because the
hollow ball has the greatest moment of inertia as is greater than for a solid ball.
11. The horizontal forces on the box gives
( )
( )
The cosine of an angle
decreases as the angle increases, so as the tension in the rope has a constant magnitude, the
frictional force must be decreasing.
12. At the end of rotation, is vertical, the weight of the box, is directed down, and is at an
angle below the horizontal. So as is the only force with a horizontal component, the net
force of this system is zero, so must be zero.
13. The total work of this system is the area under the curve, so using basic geometry,
( )
( )( )
( )
This is equal to the change in kinetic energy or
( )( )
14. The torque at the bottom of the cylinder must be zero and balanced by the torque from the
removed mass and the applied force. The torque of the removed mass is
and the torque from the applied force is
to
(
) So
(
)
which is closest
15. There is no friction, so the component of force along the direction of motion is only
Power is constant, so
( )
( )
16. The springs are considered to be connected in parallel, so we can reduce this to a system with
one spring in which
√
. Therefore, the frequency is
17. The slope of the line, from any two points, is . The equation of a line is
. From the
( )
( ) so
( )
( )
graph,
. Manipulate this equation to
( )
solve for
(
)
( )
(
)
( )
18. The period of oscillation of a spring depends only on the mass of the object and the spring
constant, neither of which change. So
√ does not change. The other choices cannot
change without the other in order to keep energy constant.
19. The water falls due to gravity, so
. The mass of the water is equal to the density
times the cross-sectional area and height of the pipe or
Putting everything together,
√
20. The weight of the immersed block balances the buoyant force or
the volume of the immersed wood. The mass of the immersed wood is half or
density is
so the volume is
where is
, and the
. Therefore, the mass of the displaced water is
so the scale of the tank shows
and the scale on the
block shows
21. The two springs on the bottom are in parallel and have an equivalent spring of
This equivalent spring is in series with the spring above and the equivalent spring of these two
is
( ) (
)
22. We want to convert
. By hookes law,
(
(
)
)( )
to the fundamental SI units. Watts is the unit of power, which equals
. Finally,
23. A: Attach the mass to the spring scale, which gives you the weight of the object. Divide the
weight by the known mass to get the acceleration due to gravity,
.
B: Consider this to be a bar pendulum in which the period of motion is
√ The
stopwatch can be used to calculate the period of motion as the rod swings and then solve for g.
D:
( ) Use the meterstick to find the maximum range and maximum height to solve
for the angle. Next, the range is equal to
(
)
and knowing the velocity, angle, and
range, you can solve for g.
E: One possible experiment could be to propel the mass to free fall.
and knowing power, time, and the mass, you can calculate g.
(
)
C: This is not possible because you do not know the angle of inclination and also the coefficient
of friction of the plane (if there is friction, which isn’t specified).
24. Each point mass is traveling in a circle which has a radius that is the circumradius of the
equilateral triangle. Thus, there is centripetal force on each mass which is provided by each
spring. First we need to calculate the circumradius of the equilateral triangle of side length
Draw the circumcircle, in which the circle passes through each vertex of the triangle. Suppose
the radius of the circumcircle is . The area of the circle is
and the area of the equilateral
triangle is ( )( )
(
√
)
. Draw a line from each vertex to the center of the triangle.
This separates the equilateral triangle into 3 triangles of equal area. The area of each triangle,
in terms of
√
is ( )( )
(
)
√
. Thus the total area of the triangle in terms of
. This is equal to the area in terms of
once rotating, has a length of
so
so the circumradius is
(
point mass is then
)
√
√
√
√
is
The new triangle,
The centripetal force on a
√
. The spring force provided
by the two springs is directed towards the center (same as the centripetal force), or
(
25. Let’s first compare
)
(
and
where
)√
√ . Finally,
√
√
. The total mechanical energy of the object at point A is
and the energy at point C is
. Since k
is less than one, the PE at point A is less than the PE at point C so the KE at point A is higher and
as an elliptical orbit contains greater mechanical energy than in a circular orbit, the KE at A is
higher than the KE at point C so
. Now angular momentum is conserved in elliptical
orbit, so
. Choose the boundary conditions for the radius so
. Since is the minimum distance at point A, it follows that
so
,