2009 F=ma Solutions
√
1. Find the speed through conservation of energy
Pressure equals force over area and the force on impact is equal to the change
in momentum over time or
2.
3.
4.
5.
so the applied pressure is
.
Each collision is perfectly elastic, so the first collision will result in the left block coming
to rest where the center block initially was, which is now moving right at speed v. After
the center block collides with the right block, the center block moves left again with
speed v while the right block moves right with speed v. Thus, when the center block
collides with the left block again, it will come to rest on its initial position as the left
block will move left with speed v. D
The initial momentum of the system is zero, and the blocks will stick in a perfectly
inelastic collision, so the center block will be at rest a long time after. Since the left block
collides first with the center block, the system will come to rest somewhere to the right
as the right block collides and brings the left and center block to rest. E
( )
Summing the forces,
For an elliptical orbit, B, angular momentum is conserved. Therefore, at the point
farthest from the planet on orbit B,
so
. Similarly, at the point closest,
so
and
( ) so √
6. The speed is independent of the angle.
( ) and
eliminates the angle (trig identity) so the speed is independent and is constant.
7.
so
8. Angular acceleration is the slope of the angular velocity v. time graph, so choose two
points to calculate the slope
( )( )
9. The net angle is the sum of the two triangle areas or
10.
and so for the apple going up,
( )
apple going down
11.
∫
∫
(
)(
( )
(
( )(
)(
)
)
and the
) and then
(
12. The total work is the work to get on the roof then to pull robin up or
)
13. Lucy weighs more than Mary, so in order for Mary’s torque to balance Lucy’s, Henry
must sit closer to Mary than Lucy so that
so Lucy exerts the most torque.
14. No momentum is conserved because we are looking at the system from -10 to 10s and
the collision is instantaneous and forces like tension act on the system after the
collision. And since this is an inelastic collision (bullet is embedded in the block),
mechanical energy is lost. So E
15. Constant speed means acceleration is zero so the net force is zero. The forces acting on
the suitcase are gravity down, the normal force up, friction to the left, and the pull force
( )
at a 30 degree angle. So we sum the forces in the y direction to get
( )
. Now we some the x direction forces to get
( )
16. Consider this a one body system by using reduced mass
. You can
easily derive this using newton’s second and third law. So for this system, both masses
are the same so
√
√
so
√ you
17. Attach the mass to the spring and as the frequency of a spring is given by
know and the tuning fork allows you to find the frequency, so you have the necessary
information to find the spring constant.
18. From the initial peg point, the period to return to the initial starting point is the total
√
period (if the small peg wasn’t in place, try and visualize it) or
the small peg, the length of the swinging string is now
√
for this movement is
√
√ (
√
√
so likewise, the period
so the total period is the sum or
√
)
(
19. The maximum range of a projectile is given by
simple projectile equations). Range is maximized when
so from the given information,
projectile is
√ From
so
)
(
(you can derive this using
)
or when
. The maximum height of a
and
20. Vertical and horizontal momentum is conserved. The initial horizontal momentum is
(
from lump 2 and is
)
For the vertical, the
momentum comes from lump 1, or
√(
the speed is vector sum or
Therefore
)
√
( )
(
21. The gravitational potential energy of such a system is
22. Let the midpoint,
( )
(
)( )
be the origin, so find the center of mass or
(
))
Therefore the center of mass is a distance from the
larger mass so the radius of orbit for the
mass is
. Now we find the speed by
equating the gravitational force provided by the centripetal force of the 3M mass or
(
)( )
√
. And the velocity is equal to the circumference divided by
( )
√
the period or
√
23. The position, velocity, and acceleration equations of motion of a spring are, respectively,
( )
( ) ( )
( )
( )
( ) Power can be
defined as
( ( ))( ( )) (remember hookes law) so
(
(
)
(
(
)
(
))
(
)
(
)
) This is maximized, keeping the negative sign in mind, when
or
(
)
24. A box will tip if the normal force,
( ) doesn’t go through the area of contact and
( ) So to find the
the box will not if the static friction (lower left corner) equals
critical condition, we set the torque sum on the top right corner equal to zero and the
( )
( )
normal force on the left most edge or
( )
have
( ) so for the box to tip, we must
The object will slide when
( )
or
25. Let the left disc be 1 and the right disc be 2. The forces applied to each disc are the same
so
We know
and torque is the change in angular
momentum over time and as the final momentum is zero and the change in time is
equal for both discs, we get
Since the
materials are the same, the density is equal or
is the thickness (which is equal). Finally, (
(
)
(
)
(
)
)where h