2010 F=ma Solutions
1. The slope of a position vs time graph gives the velocity of the object
So you can see that the
position from B to D gives the steepest slope, so the speed is the greatest in that area.
2. Speed is the magnitude of velocity and as we are given a velocity v. time graph, we are looking
for the maximum and minimum points and on this graph we see they occur at points B and D
and as the magnitude of the velocity are equivalent at those two points, the greatest speed is at
points B and D.
3. The area under an acceleration vs time gives the velocity, so from the choices listed we see that
the area under the curve at B gives the greatest area. After time B, the speed decreases because
the object reverses direction so the overall area will decrease.
√
4. The time for the piano to hit the ground is
movers notice the piano after it falls
that is √
√
so the piano has to fall another
so we are looking at the time it takes for the projectile
( )
to cross this point again as that is the difference in time or
6. At the point of maximum height,
because the object reverses vertical direction, so using
(
in the vertical direction,
( )
is therefore twice this or
( )
( )
(
( )
( )
)
)
( )
The total time of motion
Horizontal velocity is constant so
. For maximum height, the projectile is in free fall, so
Therefore
7. Angular momentum is defined as
withstands the force from the carousel of
(
and the time for
so
5. Define the initial launch point at
( )
. The
( )
so we need to solve for
so
(
Centripetal force
)
so
)( )
8. The only forces acting in the x direction are friction and the horizontal component of gravity.
Friction force prevents the car from accelerating up the hill, so it is greater than the horizontal
( )
component of gravity. Therefore, from the free body diagram,
( )
( )
( )
( ))
(
9. The only acceleration is in the x-direction and the forces acting on the ball are at an angle and
( )
gravity down, so writing the vertical and horizontal net forces,
( )
( )
( )
10. Because block 1 slips on block 2, there is friction on block 1 acting right (block 1 slips left) so by
Newton’s third law, the same friction acts on block 2 left. We also have the friction from block 2
being pulled right, so summing the forces,
(
(
)
)
11. Tension is uniform through this one string, so from the first point mass, the forces are T
up and gravity down, so
the center mass to the left string be
center mass to the right string is
. Now let the horizontal distance from
Therefore, the horizontal distance from the
. The net torque of this system is zero, so we can
(
)
find by taking the torque about the center mass, so
Now we find the angle formed from the string about the center mass and the horizontal
( )
by taking the net vertical force of the center mass, so (
)
and
from earlier, so
( )
( )
. Therefore, we have a
right triangle on both sides of the center mass with legs
√
and , so
(
)
√
12. Apply the work kinetic energy theorem. The work due to gravity is the only work over
this time so
ball must lose
Therefore for the block to gain
in potential energy. The initial energy of the ball is
and
the
√
so
13. The initial PE is equal to the total kinetic energy including rotational energy, so
(
projected up, it has a velocity of
)( )
When the ball is
and rises to
with no rotational energy so,
(
maximum PE = maximum KE so
)
14. We find the velocity of the 5kg block, call that block a, by using the work kinetic energy
theorem, from the onset to right before block a hits the second block, b. The work is due to
(
friction, so
)
(
. This is the speed right before collision so call it
)( )(
)( )
( )(
)
Now this is an elastic collision, so
momentum and kinetic energy are conserved, so call the velocity of block b right after the
collision to be and the velocity of block a after the collision to be
So we have
and
(
)
Solve the first equation for
plug it into the energy equation to get
again except
( (
))
and
so
and
Now apply the work kinetic energy theorem
(
)
(
)(
)
15. There are no external forces, so momentum and kinetic energy are conserved. First
thing is to solve for the speed of the combined masses by using conservation of
(
)
momentum, so
. Energy is conserved and we have the
initial kinetic energy of
and the kinetic energy of the combined masses and the
(
potential energy of the small mass on the larger mass, or
(
)(
)
(
)
)
16. This is an elastic collision, so momentum and kinetic energy are conserved before the
( )
collision and while the blocks separate again, so we have
. The
negative sign indicates the smaller block is moving left and
is the mass of the larger
block moving right. Energy is conserved, so
(
equation,
(
(
)
(
)
and
)
(
)
(
(
)
and substitute that in the second equation to get
)
(
(
)
. Now complete the square or use the quadratic equation to get
√
(
)
and from the first
)(
)
)
(
( (
(
(
√
)) (
)
)
)
√
(
)
) from which the two solutions are
however we already defined
negative sign because it’s moving left, so
(
)
as positive by the inclusion of the
indicates the block
is moving right,
therefore the answer is
17. Each mass is connected to the other masses, so the combined gravitational energy is
the sum of the individual energies without overlap, so since there are 4 pairs, there are
6 possible pairs, so the total energy is
18. Force is defined as the derivative of the potential energy or
so the
corresponding force graph is found by calculating the individual slopes of the line
segments on the potential energy graph, and doing so, in order, gives
and 0
constant over the appropriate distances, so the correct graph is
19. For this question, compare the positioning in each graph with the energy diagram to
determine which ones are possible. For graph 1, x is constant at 15 m so v=0 and on the
energy diagram,
at x=15 m, so that means the kinetic energy is either at a
maximum or 0 so, graph 1 is possible. For graph 2, x is constant at -5m, so v=0 and K=0,
but on the energy diagram, U is decreasing so by conservation of energy, there is KE
increasing so graph 2 is not possible. For graph 3, x increases linearly from 10 to 15m, so
there is constant velocity. From graph 1 we deduced that since U is 0 at 10-15 m, KE is at
a maximum or 0, so graph 3 is reasonable because the KE must be at a maximum.
Therefore I and III are possible.
20. Recall conservation of energy or
. For this question, we need to find where
the potential energy is at a maximum so that K is 0. The kinetic energy is zero when the
velocity is zero, so the position when the velocity is zero can be deduced by the
maximum and minimum of the position graph, since velocity is the derivative or slope of
the position graph, which is zero at a maximum and minimum. Therefore, the maximum
and minimum of the position graph is 5 and -5 m, and the corresponding potential
energy is -5 J, which is the same for both positions as expected.
21. This problem implies that the gravitational self potential energy depends on the
universal gravitational constant, the radius of the object, the density, and some
numerical constant. To solve this problem, we need to figure out how energy scales with
the radius by determining the dimension quantities of this equation. Energy is in terms
of
, density is in terms of
constant is in terms of
radius is in terms of
Now let
Therefore our
(
equation is
and the gravitational
) (
) ( ) where k is the numerical
constant. Each quantity increases exponentially by some constants, a, b and c.
Therefore, by matching the exponents, our 3 equations are
. Therefore,
2
and
, so the radius increases exponentially as
so increasing the radius by a
factor of while keeping everything else constant increases the energy by a factor of
so the new energy is
22. Helium is less dense than air (air is mostly made up of nitrogen, which is heavier than
helium), so as the car is traveling around the circle, air will move outwards, towards B,
so as helium is less dense, the balloon will be pushed inwards, towards D.
23. Because the net force on each tube is the same, the applied forces to each tube are
equal, so we know that
momentum is equal
or
where
so
(
For the left tube, the change in
)
where is the mass of the flowing water
is the mass flow rate (mass of water moving per unit time)
so equating the forces gives
√
24. The moment of inertia of a uniform solid disk is
A circular section with radius
is removed, so the cross sectional area of the disk is removed, so the mass of the
removed section is
The axis of rotation is units from the center of the removed
section, so by the parallel axis theorem, the inertia of the removed section is
( )( )
( )
. Therefore, the moment of inertia of the remaining disk
is
25. First find the speed of the ship in the circular orbit of radius
centripetal and gravitational force or
(
by equating the
. Now in elliptical orbit,
)
energy is conserved so at the minimum and maximum radius, the respective velocities
are
and
. By conservation of energy, we have
Angular momentum is conserved in orbit so
Plugging in this speed gives
√
(
)
Finally,