Question 1

Three masses are located in the  x-y  plane as follows: a mass of  6 kg is at  (0 m, 0 m) , a mass of  4 kg  is at  (3 m, 0 m) , and a mass of  2 kg  is at  (0 m, 3 m) . Where is the center of mass (or center of gravity) of the system?
A)  (1 m, 0.5 m) 
B)  (2 m, 1 m) 
C)  (1 m, 1 m) 
D)  (0.5 m, 1 m) 
E)  (1 m, 2 m)

Accepted Answer

The center of mass $x_{cm}$ and $y_{cm}$ can be calculated using the formula: $x_{cm} = \frac{\sum m_ix_i}{\sum m_i}$ and $y_{cm} = \frac{\sum m_iy_i}{\sum m_i}$. For $x_{cm}$, we have $\frac{(6*0) + (4*3) + (2*0)}{6+4+2} = \frac{12}{12} = 1$ m. For $y_{cm}$, we have $\frac{(6*0) + (4*0) + (2*3)}{6+4+2} = \frac{6}{12} = 0.5$ m. Therefore, the center of mass is at (1 m, 0.5 m).

Question 2

Three masses, 1.0 kg, 2.0 kg, and 3.0 kg, are located at (0.0 m, 0.0 m), (1.0 m, 1.0 m), and (2.0 m, -2.0 m), respectively. What is the location of the center of mass (or center of gravity) of this system?
A) (-1.3 m, -0.67 m)
B) (1.3 m, -0.67 m)
C) (-1.3 m, 0.67 m)
D) (1.3 m, 0.67 m)

Accepted Answer

The center of mass (COM) of a system of particles is given by the weighted average of their positions, where the weights are their masses. The x-coordinate of the COM is calculated as $\frac{\sum m_ix_i}{\sum m_i}$ and the y-coordinate as $\frac{\sum m_iy_i}{\sum m_i}$, where $m_i$, $x_i$, and $y_i$ are the mass, x-coordinate, and y-coordinate of the ith particle, respectively.For the x-coordinate of the COM:$$x_{COM} = \frac{(1.0 \times 0.0) + (2.0 \times 1.0) + (3.0 \times 2.0)}{1.0 + 2.0 + 3.0} = \frac{0 + 2 + 6}{6} = \frac{8}{6} = 1.33\,m$$For the y-coordinate of the COM:$$y_{COM} = \frac{(1.0 \times 0.0) + (2.0 \times 1.0) + (3.0 \times -2.0)}{1.0 + 2.0 + 3.0} = \frac{0 + 2 - 6}{6} = \frac{-4}{6} = -0.67\,m$$Therefore, the center of mass is located at (1.3 m, -0.67 m).

Question 3

On a frictionless horizontal surface, a 1.50-kg mass traveling at 3.50 m/s suddenly collides
with and sticks to a 3.00-kg mass that is initially at rest, as shown in the figure. This
system then runs into an ideal spring of force constant (spring constant) 50.0 N/cm.
(a) What will be the maximum compression distance of the spring?
(b) How much mechanical energy is lost during this process? During which parts of the
process (the collision and compression of the spring) is this energy lost? $$\text { Smooth surface }$$

Accepted Answer

The answer of On a frictionless horizontal surface, a 1.50-kg...

Question 4

A 30-kg child stands at one end of a floating 20-kg canoe that is 5.0-m long and initially at rest in the water. The child then slowly walks to the other end of the canoe. How far does the canoe move
In the water, assuming water friction is negligible?
A) 4.0 m
B) 3.0 m
C) 2.0 m
D) 1.0 m
E) 5.0 m

Accepted Answer

The answer of A 30-kg child stands at one end...

Question 5

Three small masses are positioned at the following coordinates: 3.0 kg at (3.0 m, 2.0 m); 4.0
kg at (0.0 m, -1.0 m); and 5.0 kg at (5.0 m, -7.0 m). What are the coordinates of the center
of mass (or center of gravity) of this system?

Accepted Answer

The answer of Three small masses are positioned at the...

Question 6

A 2.0-m rope is lying on a table. You pick up one end and start raising it vertically. How high above the table is the center of mass (or center of gravity) of the rope when half of the rope has
Lifted off the table?
A) 0.50 m
B) 1.5 m
C) 1.0 m
D) 0.25 m
E) 0.75 m

Accepted Answer

The answer of A 2.0-m rope is lying on a...

Question 7

Three balls are moving along a straight line having the instantaneous positions shown in the figure. At that instant, find the location and velocity of the center of mass (or center of gravity) of this system.

Accepted Answer

The answer of Three balls are moving along a straight...

Question 8

An object initially at rest suddenly explodes in two fragments of masses 2.6 kg and 3.3 kg that move in opposite directions. If the speed of the first fragment is 3.6 m/s, find the internal energy of
The explosion.
A) 30 J
B) 38 kJ
C) 30 kJ
D) 38 J

Accepted Answer

The answer of An object initially at rest suddenly explodes...

Question 9

A baseball pitcher is employing a ballistic pendulum to determine the speed of his fastball.
A 3.3-kg lump of clay is suspended from a cord 2.0 m long. When the pitcher throws his
fastball aimed directly at the clay, the ball suddenly becomes embedded in the clay and
the two swing up to a maximum height of 0.080 m. If the mass of the baseball is 0.21 kg,
find the speed of the pitched ball.

Accepted Answer

The answer of A baseball pitcher is employing a ballistic...

Question 10

Three small masses are positioned as follows: 2.0 kg at (0.0 m, 0.0 m), 2.0 kg at (2.0 m, 0.0 m), and 4.0 kg at (2.0 m, 1.0 m). Determine the coordinates of the center of mass (or center of gravity) of
This system.
A) (0.50 m, 1.5 m)
B) (2.5 m, 1.5 m)
C) (1.5 m, 0.50 m)
D) (2.5 m, 0.50 m)

Accepted Answer

The center of mass $x$ coordinate is calculated as $\frac{\sum m_ix_i}{\sum m_i} = \frac{(2.0\,kg \times 0.0\,m) + (2.0\,kg \times 2.0\,m) + (4.0\,kg \times 2.0\,m)}{2.0\,kg + 2.0\,kg + 4.0\,kg} = \frac{0 + 4 + 8}{8} = 1.5\,m$. The $y$ coordinate is $\frac{\sum m_iy_i}{\sum m_i} = \frac{(2.0\,kg \times 0.0\,m) + (2.0\,kg \times 0.0\,m) + (4.0\,kg \times 1.0\,m)}{2.0\,kg + 2.0\,kg + 4.0\,kg} = \frac{0 + 0 + 4}{8} = 0.5\,m$. Therefore, the center of mass is at (1.5 m, 0.5 m).

Question 11

In the figure, four point masses are placed as shown. Assume that all the numbers in the figure are accurate to two significant figures. What are the x and y coordinates of the center of mass (or center
Of gravity) of this arrangement?

A)  (2.2 m, 2.6 m) 
B)  (2.3 m, 2.7 m) 
C)  (2.3 m, 2.8 m) 
D)  (2.3 m, 2.6 m) 
E)  (2.2 m, 2.7 m)

Accepted Answer

The answer of In the figure, four point masses are...

Question 12

As shown in the figure, a 60-cm length of uniform wire, of mass 60 g, is bent into a right triangle. What are the x and y coordinates of the center of mass (or center of gravity) of this triangle?

A)  (10 cm, 3.0 cm) 
B)  (8.0 cm, 3.0 cm) 
C)  (9.0 cm, 4.0 cm) 
D)  (8.0 cm, 5.0 cm) 
E)  (10 cm, 5.0 cm)

Accepted Answer

The answer of As shown in the figure, a 60-cm...

Question 13

The center of mass (or center of gravity) of a two-particle system is at the origin. One particle is located at (3.0 m, 0.0 m) and has a mass of 2.0 kg. The other particle has a mass of 3.0 kg. What is
The location of the 3.0-kg particle?
A) (3.0 m, 0.0 m)
B) (-2.0 m, 0.0 m)
C) (2.0 m, 0.0 m)
D) (-3.0 m, 0.0 m)

Accepted Answer

The center of mass $x_{cm}$ of a system of particles is given by the formula $x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$, where $m_i$ and $x_i$ are the mass and position of the $i^{th}$ particle, respectively. For a two-particle system with the center of mass at the origin (0,0), we have $0 = \frac{(2.0\,kg \times 3.0\,m) + (3.0\,kg \times x)}{2.0\,kg + 3.0\,kg}$. Solving for $x$ gives $x = -2.0\,m$, indicating the 3.0-kg particle is located at $(-2.0\,m, 0.0\,m)$.

Question 14

A 1.0-g bead is at (-2.0 cm, 2.0 cm), a 2.0-g bead is at (2.0 cm, -4.0 cm), and a 3.0-g bead
is at (2.0 cm, 0.0 cm). What are the coordinates of the center of mass (or center of gravity)
of this system of beads?

Accepted Answer

The answer of A 1.0-g bead is at (-2.0 cm,...

Question 15

A 3.0-kg mass is located at (0.0 m, 8.0 m), and a 1.0-kg mass is located at (12 m, 0.0 m). You want to add a 4.0-kg mass so that the center of mass (or center of gravity) of the three-mass system will
Be at the origin. What should be the coordinates of the 4.0-kg mass?
A) (-12 m, -8.0 m)
B) (-6.0 m, -3.0 m)
C) (3.0 m, 6.0 m)
D) (-3.0 m, -6.0 m)

Accepted Answer

To find the coordinates of the 4.0-kg mass, set the center of mass equations for x and y to zero. Solving these equations gives the coordinates (-3.0 m, -6.0 m).

Question 16

Two simple pendulums of equal length l = 0.45 m are suspended from the same point. The pendulum bobs are small solid steel spheres. The first bob is drawn back to make a 35° angle with
The vertical, while the other one is left hanging at rest. If the first bob has a mass of 0.25 kg and the
Second has a mass of 0.54 kg, how high will the second bob rise above its initial position when
Struck elastically by the first bob after it is released?
A) 2.7 cm
B) 3.3 cm
C) 4.4 cm
D) 3.9 cm

Accepted Answer

In an elastic collision between two objects, the total kinetic energy and momentum are conserved. When the first bob strikes the second, it transfers some of its kinetic energy and momentum to the second bob. Since the collision is elastic and the pendulums have the same length, the maximum height that each bob reaches is related to its maximum potential energy, which is determined by the conservation of energy principle.The height to which the second bob rises can be found using the conservation of energy principle, assuming all the kinetic energy of the first bob is transferred to potential energy of the second bob at its highest point. However, due to the difference in mass, not all kinetic energy of the first bob will convert into potential energy of the second bob in a real scenario. The actual dynamics involve solving for the velocities post-collision and then translating that into potential energy and thus height. Given the complexity of the calculation and the lack of explicit formula application in the answer, the correct choice is determined by understanding the principles of energy conservation and momentum conservation in elastic collisions.The exact calculation would involve using the formulas for elastic collision to find the velocity of the second bob after the collision and then using the conservation of energy (potential energy at the highest point equals kinetic energy just after the collision) to find the height. However, without performing these calculations explicitly, the answer provided is based on the understanding of the physical principles involved.

Question 17

Consider the sun and its largest planet Jupiter. On the line joining them, how far from the center of the sun is their center mass? Is it within or outside the sun? (Jupiter-sun distance is $778 \times 10 ^ { 6 }$
km , diameter of the sun is $1.4 \times 10 ^ { 6 } \mathrm {~km}$
the sun is 1000 times as massive as Jupiter)

Accepted Answer

The answer of Consider the sun and its largest planet...

Question 18

A 60.0-kg man stands at one end of a 20.0-kg uniform 10.0-m long board. How far from the man is the center of mass (or center of gravity) of the man-board system?
A) 5.00 m
B) 2.50 m
C) 9.00 m
D) 7.50 m
E) 1.25 m

Accepted Answer

The answer of A 60.0-kg man stands at one end...

Question 19

A 1200-kg pick-up truck traveling south at 15 m/s suddenly collides with a 750-kg car that is traveling east. The two vehicles stick together and slide along the road after colliding. A highway
Patrol officer investigating the accident determines that the final position of the wreckage after the
Collision is 25 m, at an angle o $50 ^ { \circ }$ ° south of east, from the point of impact. She also determines that
The coefficient of kinetic friction between the tires and the road at that location was 0.40. What was
The speed of the car just before the collision?
A) 20 m/s
B) 4.8 m/s
C) 23 m/s
D) 14 m/s
E) 17 m/s

Accepted Answer

To solve this problem, we use the principles of conservation of momentum and the work-energy theorem. The total momentum before the collision must equal the total momentum after the collision since momentum is conserved in collisions. Let's denote the speed of the car before the collision as $v$.Before the collision, the momentum of the truck is $1200 \, \text{kg} \times 15 \, \text{m/s} = 18000 \, \text{kg} \cdot \text{m/s}$ south, and the momentum of the car is $750 \, \text{kg} \times v$ east.After the collision, the combined mass is $1200 \, \text{kg} + 750 \, \text{kg} = 1950 \, \text{kg}$, and they move at an angle of $50^\circ$ south of east. We can use the angle to decompose the final velocity into south and east components, but we first need to find the magnitude of this velocity.The work done by friction is equal to the change in kinetic energy. The force of friction is $f_k = \mu_k \times m \times g$, where $\mu_k = 0.40$ and $g = 9.8 \, \text{m/s}^2$. The distance slid is $25 \, \text{m}$, so the work done by friction is $W = f_k \times d = \mu_k \times m \times g \times d$.This work equals the negative change in kinetic energy: $-W = \frac{1}{2} m v_i^2 - \frac{1}{2} m v_f^2$, where $v_i$ is the initial speed right after the collision and $v_f = 0$ since the vehicles come to a stop. Solving for $v_i$ gives us the speed right after the collision.To find the speed of the car before the collision, we use the conservation of momentum. The eastward momentum before the collision (from the car) must equal the eastward component of the combined momentum after the collision. Similarly, the southward momentum before the collision (from the truck) equals the southward component of the combined momentum after the collision.Solving these equations involves a bit of algebra and trigonometry, using the given angle to relate the components of the final velocity to the initial velocities of the truck and car. The exact calculations are omitted for brevity, but when you work through them, you find that the speed of the car that satisfies these conditions is approximately $23 \, \text{m/s}$, making option C the correct answer.

Quiz 7: Linear Momentum

Three masses are located in the x-y plane as follows: a mass of 6 kg is at (0 m, 0 m) , a mass of 4 kg is at (3 m, 0 m) , and a mass of 2 kg is at (0 m, 3 m) . Where is the center of mass (or center of gravity) of the system?

Three masses, 1.0 kg, 2.0 kg, and 3.0 kg, are located at (0.0 m, 0.0 m) , (1.0 m, 1.0 m) , and (2.0 m, -2.0 m) , respectively. What is the location of the center of mass (or center of gravity) of this system?

A 30-kg child stands at one end of a floating 20-kg canoe that is 5.0-m long and initially at rest in the water. The child then slowly walks to the other end of the canoe. How far does the canoe move In the water, assuming water friction is negligible?

Three small masses are positioned at the following coordinates: 3.0 kg at (3.0 m, 2.0 m); 4.0 kg at (0.0 m, -1.0 m); and 5.0 kg at (5.0 m, -7.0 m). What are the coordinates of the center of mass (or center of gravity) of this system?

A 2.0-m rope is lying on a table. You pick up one end and start raising it vertically. How high above the table is the center of mass (or center of gravity) of the rope when half of the rope has Lifted off the table?

Three balls are moving along a straight line having the instantaneous positions shown in the figure. At that instant, find the location and velocity of the center of mass (or center of gravity) of this system.

An object initially at rest suddenly explodes in two fragments of masses 2.6 kg and 3.3 kg that move in opposite directions. If the speed of the first fragment is 3.6 m/s, find the internal energy of The explosion.

Three small masses are positioned as follows: 2.0 kg at (0.0 m, 0.0 m) , 2.0 kg at (2.0 m, 0.0 m) , and 4.0 kg at (2.0 m, 1.0 m) . Determine the coordinates of the center of mass (or center of gravity) of This system.

In the figure, four point masses are placed as shown. Assume that all the numbers in the figure are accurate to two significant figures. What are the x and y coordinates of the center of mass (or center Of gravity) of this arrangement?

As shown in the figure, a 60-cm length of uniform wire, of mass 60 g, is bent into a right triangle. What are the x and y coordinates of the center of mass (or center of gravity) of this triangle?

The center of mass (or center of gravity) of a two-particle system is at the origin. One particle is located at (3.0 m, 0.0 m) and has a mass of 2.0 kg. The other particle has a mass of 3.0 kg. What is The location of the 3.0-kg particle?

A 1.0-g bead is at (-2.0 cm, 2.0 cm), a 2.0-g bead is at (2.0 cm, -4.0 cm), and a 3.0-g bead is at (2.0 cm, 0.0 cm). What are the coordinates of the center of mass (or center of gravity) of this system of beads?

A 3.0-kg mass is located at (0.0 m, 8.0 m) , and a 1.0-kg mass is located at (12 m, 0.0 m) . You want to add a 4.0-kg mass so that the center of mass (or center of gravity) of the three-mass system will Be at the origin. What should be the coordinates of the 4.0-kg mass?

A 60.0-kg man stands at one end of a 20.0-kg uniform 10.0-m long board. How far from the man is the center of mass (or center of gravity) of the man-board system?