Solutions to Assignment 7
PROBLEM 1 – 10 points
The picture above shows the x-axis and a ball of mass m placed at x = 3 (let’s call it the first
ball).
(a) [2 points] Where on the axis should we place a second ball, with mass 2m, so the center of
mass of the system is located at x = 7?
The position of center of mass x is given by the equation,
𝑚1 𝑥1 + 𝑚2 𝑥2
𝑥𝑐𝑚 =
𝑚1 + 𝑚2
Substituting for the position of center of mass to be 7 with mass of first ball as ‘m’ and the
second mass to be ‘2m’, we get the position of the second ball to be 9.
(b) [2 points] Negative mass does not exist in nature, but we can use our imagination to find out
what would happen if it did. Let’s say we have a third ball which has negative mass m3 = -m.
Where should we place it so the center of mass of the system is located at the position of the
second ball?
The position of center of mass x is given by the equation, in this situation,
𝑚1 𝑥1 + 𝑚2 𝑥2 − 𝑚𝑥3
𝑥𝑐𝑚 =
𝑚1 + 𝑚2 − 𝑚
Substituting for the position of center of mass to be 9, we get the position of the third ball with
negative mass to be 3.
(c) [2 points] Now let’s use a regular ball with a positive mass. Where should we place the third
ball, which has the same mass as the first ball, so the center of mass of the system is located at
the position of the second ball?
The position of center of mass x is given by the equation, in this situation,
𝑚1 𝑥1 + 𝑚2 𝑥2 + 𝑚𝑥3
𝑥𝑐𝑚 =
𝑚1 + 𝑚2 + 𝑚
Substituting for the position of center of mass to be 9, we get the position of the third ball with
positive mass to be at x = 15
o be -3.
The picture on the right shows the first ball only, but add the second ball to the picture at the
location you found in part (a). Now we need to add a third ball of unknown mass in such a way
that the center of mass of the system is located at the point (x = 5, y = 7).
We know that the third ball is placed somewhere along the dashed line
(d) [4 points] Find the exact location and the mass (in terms of m) of the third ball.
The position of the second ball is shown in the figure above. The position of the center of mass,
along the x-axis is given by,
𝑚1 𝑥1 + 𝑚2 𝑥2 + 𝑚3 𝑥3
𝑚1 + 𝑚2 + 𝑚3
The problem states that the position of the third mass is along the dotted line in the figure given
above. Therefore the x-co-ordinate of the third mass is 3. We also, know that the x co-ordinate of
the center of mass of the system is 5. Therefore substituting this information in the above
equation, we get the mass of the third ball to be 3m.
𝑥𝑐𝑚 =
The y co-ordinate of the third ball is given by the equation,
𝑚1 𝑦1 + 𝑚2 𝑦2 + 𝑚3 𝑦3
𝑦𝑐𝑚 =
𝑚1 + 𝑚2 + 𝑚3
We know that the y co-ordinates of the first and second ball are zero and the y co-ordinate of the
center of mass to be 7. Substituting this information in the above equation, we get the y coordinate of third ball to be 14. Therefore the exact position of the third ball is (3,14).
PROBLEM 2 – 10 points (1 point for every answer)
A cart that can move along a straight track (aligned
with our x axis) has a mass of 8.00 kg and,
starting from an initial position of x = +20.0 m,
is given an initial velocity of 4.00 m/s in the
positive x-direction. The cart is subjected to a net
force that is either in the positive or negative x
direction, depending on the position of the cart, as
shown in the graph. Note that this graph is very
similar to the one in Assignment 6, but the force
here is plotted as a function of position, rather than
time.
[6 points] (a) Complete the table below, to show the cart's kinetic energy and speed at the
indicated positions.
At x = +20m , we know that the speed of the cart to be 4.00m/s and the mass to be 8.00kg,
we get the kinetic energy, using the definition of Kinetic energy(i.e., K = ½ mv2), as 64J.
The equation that relates the kinetic energy and the work done by force is given by
𝐾𝑓 − 𝐾𝑖 = 𝑊
Where 𝐾𝑓 and 𝐾𝑖 are the final and initial kinetic energies and the W is the work done by the force
when the object moves from the initial position to final position.
The displacement of the cart when it goes from x = 20m to x = 10 m is the negative of the
displacement when it goes from x = 10m to x = 20m. Therefore Work done, when the cart goes
from x =20 m to x = 10m is negative of the work done when the cart goes from x = 10m to x =
20m. Therefore, by knowing the Kinetic energy at x = 20m and knowing the work done
when the cart goes from x = 20 to x =10m(taking the correct sign!) we get the Kinetic
energy at x = 10m as 114J. Using the definition of Kinetic energy(i.e., K = ½ mv2), we get
the speed to be 5.34 m/s.
Similarly, we get the Kinetic energy at the x = 5 m to be 64J and speed to be 4.00 m/s
Position
Kinetic energy
Speed
x = +20.0 m
64 J
4.00 m/s
x = +10.0 m
114 J
5.34 m/s
x = +5.00 m
64 J
4.00 m/s
(b) Where does the cart reverse direction for the first time?
From the data given in the problem we know that the cart is initially moving in the +x direction.
Therefore the cart will reverse the direction at some point greater than x = +20m. At the turning
point, the speed of the particle will be zero. Therefore, we need to find x when the work done by
the force is exactly negative of the initial kinetic energy. Therefore, we know that the force is
10N acting in negative x direction and hence we find that a displacement of 6.4m towards
positive x direction gives the work done to be -64J. Therefore the cart turns around for the
first time at x = 26.4 m.
(c) What is the change in momentum experienced by the cart as it moves from its initial position,
at x = +20.0 m, to the position where it reverses direction for the first time?
At the point where the cart reverses direction, the momentum of the cart is zero because
the speed of the cart is zero. Therefore we get the change in momentum, defined by (final
momentum – initial momentum), as -32.00 kg m/s.
(d) How long does the cart take to move from x = +20.0 m to the position where it reverses
direction for the first time?
We know that the magnitude of the force acting on cart for positions beyond x = 20m is 10N.
Therefore the acceleration of the cart is 10/8 m/s2= 1.25 m/s2 pointing to the negative x-direction.
Therefore, using the equation xf = xi + vi t + ½ a t2, we find that the cart takes 3.2 seconds to
travel a distance of 6.4 m, with a = - 1.25 m/s2.
(e) What is the closest the cart gets to x = 0 as it moves back and forth on the track?
From the graph, we see that the cart will reverse direction at x = 26.4m and will move
towards negative x-direction as the force is acts towards negative x-direction. The cart will
speed up till x =10m as the force acts towards negative x direction. For positions x < 10m,
the force acts in the postive direction and the cart moves towards negative x direction and
therefore the cart will slow down(still moving towards negative x direction). At x = 5m, we
have already calculated that the kinetic energy of the cart to be 64J. Therefore the cart will
travel in negative x-direction approaching x= 0 till the work done by the force is -64J. We
know that the force acting on the cart to be 20N pointing in +x direction. Therefore we find
that a displacement of 3.2 m in negative x direction gives the work done to be -64J.
Therefore, the closest the cart will get to x =0, is x =(5.0 -3.2)m = 1.8m.
6m
PROBLEM 3 – 10 points
6m
C
B
A
30
An object at position A has kinetic energy = 36 J and potential energy  0 J. It slides a distance
of 6 m upward along the 30 frictionless incline, before turning around at B. It then slides to a
halt at C which is 6 m along the slope below A. The surface below A, between A and C, is
rough, rather than frictionless.
(i) What is the potential energy of the object at its highest point (B)? ________
We know that there are no non-conservative forces acting on the block when it goes from
point A to point B. Therefore the total mechanical energy (Kinetic energy + Potential
energy) of the block is conserved. At the highest point, we know that the kinetic energy of
the block is zero. Therefore we get the potential energy of the block at the highest point to
be 36J.
(ii) What is the weight W of the object? _______
We know that the block at point B is 6 sin(30) m = 3m above the point B. From the
definition of potential energy(i.e., U = mgh), we get the weight to be 12N.
(iii) When the object passes through point A on its way down the slope, what is its kinetic
energy? _______
Since, the total mechanical energy of the block is conserved when the block slides from
point B to point A, we find that the kinetic energy to be 36J at point A.
(iv) What is the change in the object’s mechanical energy during the entire process, starting at
the initial point A and continuing until the final point C? ________
We know that the initial mechanical energy to be 36 J. When the block reaches the point C, its
kinetic energy is zero, whereas its potential energy is -36J(since, point C is 3m below point A).
Therefore the change in the mechanical energy when the block slides from point A to B to
C is -36 J-36J= -72J (final mechanical energy – initial mechanical energy).
(v) How much work is done on the object by non-conservative forces during the entire process?
Using the relation, Ki + Ui + Wnc = Kf + Uf, we get the Wnc to be -72J.
PROBLEM 4 – 8 points
(a) A 20 gram pine cone falls 1.25 m to the ground, where it lands with a speed of 4.0 m/s.
Use g = 10 m/s2.
(i) With what speed would the pine cone have landed if there had been no air resistance?
The initial kinetic energy of the pine is zero and the initial potential energy is 0.25J (using
mgh, with h = 1.25 m). When the pine reaches the ground its potential energy is zero and its
kinetic energy would be equal to its initial potential energy, since there would be no work
done by non-conservative forces in the absence of air-resistance. Therefore, we get the
speed with which the pine cone would have hit the ground as 5.0 m/s.
(ii) How much work did air resistance do on the pine cone during its fall? Is your answer
positive, negative, or zero? Explain.
Since the final velocity of the pine is 4.0 m/s, its final kinetic energy is 0.16J. Using the
relation, Ki + Ui + Wnc = Kf + Uf, we get the work done by the air-resistance as -0.09J.
(b) The human brain consumes about 20 W of power under normal conditions, though more
power may be required during exams.
(i) How long can one Snickers bar power the normally functioning brain? One Snickers bar
has about 300 Cal, and 1 Cal is actually equal to 1000 calories. The conversion factor is:
1 Cal = 1000 cal = 1 kcal = 4186 J, although you can use a conversion factor of 1 Cal = 4000
J for this question to practice calculating without a calculator.
The energy of the Snickers bar in terms of joules is 300*4000J = 1200000J. Since, power
is energy divided time, we get the time that a Snickers bar can give energy is
1200000J/20W = 60000s = 1000 minutes.
(ii) At what speed must you lift a 2.0 kg container of water (a 2-liter bottle) if the power
output of your arm is to be 20 W?
When the 2.0 kg container is lifted at constant speed, the net force on it is zero. Therefore,
the force that your arm has to apply on the container is equal to its weight i.e., 20N. Since
P = F * v, we get the speed with which the container has to be lifted to be 1.0m/s.
PROBLEM 5 – 10 points
As shown in the figure, two frictionless ramps are joined by a rough horizontal section that is
4.0 m long. A block is placed at a height of 124 cm up the ramp on the left and released from
rest, reaching a maximum height of 108 cm on the ramp on the right before sliding back down
again.
[2 points] (a) How far up the ramp on the left does the block get in its subsequent motion?
The mass of the block is taken to be ‘m’. The initial kinetic energy of the block, when released is
zero and is also zero when the block reaches the highest point on the right hand side. Using the
relation, Ki + Ui + Wnc = Kf + Uf, with initial height being taken to be 124 cm and final height to be
108cm, we get the Wnc to be -0.16m*g.
Now, taking the initial height to be 108 cm and final height to be the one reached on the left,
and using the above realtion, we get the height on the left as 92cm.
[2 points] (b) What is the coefficient of kinetic friction between the block and the rough surface?
The work done by the non conservative force is found to be -0.16*mg.
Since frictional force and displacement are in opposite directions, we get θ, the angle between the
force and the displacement vectors to be 180o. Also, magnitude of displacement, when frictional
force does work, is 4.0 m. The magnitude of the frictional force is µk*mg. Therefore, using the
relation, W = F * Δx*Cos(θ), we get co-efficient of kinetic friction as 0.040.
[2 points] (c) How many times will the block pass completely across the 4.0 m wide rough section?
(We’re looking for an integer here)
Each time the block passes through the rough surface, 0.16*mg amount of energy is removed from
the block. Therefore, block has to pass ‘n’ times through the rough surface for it to loose all its initial
mechanical energy. Its initial mechanical energy is 1.24*mg. Therefore, we get n =
1.24*mg/(0.16*mg) = 7.75.
Hence the block passes through the rough surface completely for 7 times.
[2 points] (d) At what location does the block eventually come to a permanent stop?
After the block has passed through the rough surface for 7 times, the mechanical energy left in the
block is 1.24*mg – 7*0.16*mg = 0.12*mg. Taking this as the initial mechanical energy of the block
and the final mechanical energy to be zero(because the kinetic and potential energies of the block are
zero when it comes to a stop.) and using the relation Ki + Ui + Wnc = Kf + Uf, we get the Wnc to be
– 0.12*mg. Taking the work done to be W = F * Δx*Cos(θ), with F being the frictional force and θ
being 180o, we get the magnitude of displacement to be 3m. The block enters the rough surface for
the 8th time from the right side of the track. Therefore the block stops at 1.0m from the left end
of the rough surface, or, equivalently, 3 m from the right end.
[2 points] (e) Going back to the very first time the block passes across the rough section, what is the
speed of the block when it passes the middle of that rough section the first time across? Use g = 10
m/s2.
We know that the initial potential energy of the block is 1.24m*g and the magnitude of the frictional
force to be 0.40*mg. The block travels a distance of 2.0m when it reaches the middle part of the
rough surface. Therefore, work done by the non-conservative force to be -0.8*mg. Also, the
potential energy of the block is zero when it reaches the middle of the rough surface. Therefore,
using the relation, Ki + Ui + Wnc = Kf + Uf, we get the kinetic energy to be 1.16mg and hence the
speed to be 𝟐𝟑. 𝟐 = 𝟒. 𝟖𝟐 𝒎/𝒔, when it passes the middle of the rough section for the first
time.