Question 1

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 (x, y) can be calculated using the formula: $x_{cm} = \frac{\sum m_ix_i}{\sum m_i}$ and $y_{cm} = \frac{\sum m_iy_i}{\sum m_i}$, where $m_i$, $x_i$, and $y_i$ are the masses and their respective x and y coordinates. For the given masses and coordinates, $x_{cm} = \frac{(1.0*0.0) + (2.0*1.0) + (3.0*2.0)}{1.0 + 2.0 + 3.0} = \frac{0 + 2 + 6}{6} = \frac{8}{6} = 1.33$ m, and $y_{cm} = \frac{(1.0*0.0) + (2.0*1.0) + (3.0*(-2.0))}{6} = \frac{0 + 2 - 6}{6} = \frac{-4}{6} = -0.67$ m. Therefore, the center of mass is at (1.3 m, -0.67 m).

Question 2

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

Accepted Answer

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

Question 3

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 $x_{cm} = \frac{\sum m_ix_i}{\sum m_i}$, where $m_i$ and $x_i$ are the masses and positions of the particles, respectively. Given that the center of mass is at the origin (0,0) and one particle is at (3.0 m, 0.0 m) with a mass of 2.0 kg, we can set up the equation $0 = \frac{(2.0\,kg \cdot 3.0\,m) + (3.0\,kg \cdot x)}{2.0\,kg + 3.0\,kg}$. Solving for $x$ gives $x = -2.0\,m$, indicating the 3.0-kg particle is at (-2.0 m, 0.0 m).

Question 4

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 
How high will the second bob rise above its initial position when struck elastically by the first bob after it is released?
A)3.3 cm
B)2.7 cm
C)3.9 cm
D)4.4 cm

Accepted Answer

The answer of Two simple pendulums of equal length l...

Question 5

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,2 m)
B)(2 m,1 m)
C)(1 m,1 m)
D)(1 m,0.5 m)
E)(0.5 m,1 m)

Accepted Answer

The center of mass $x_{cm}$ and $y_{cm}$ can be calculated using the formulas $x_{cm} = \frac{\sum m_ix_i}{\sum m_i}$ and $y_{cm} = \frac{\sum m_iy_i}{\sum m_i}$, where $m_i$, $x_i$, and $y_i$ are the mass, x-coordinate, and y-coordinate of the ith mass, respectively. For the x-coordinate: $x_{cm} = \frac{(6*0) + (4*3) + (2*0)}{6+4+2} = \frac{12}{12} = 1$ m. For the y-coordinate: $y_{cm} = \frac{(6*0) + (4*0) + (2*3)}{6+4+2} = \frac{6}{12} = 0.5$ m. Therefore, the center of mass is at (1 m, 0.5 m).

Question 6

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)(1.5 m,0.50 m)
C)(2.5 m,1.5 m)
D)(2.5 m,0.50 m)

Accepted Answer

The center of mass $x_{cm}$ and $y_{cm}$ can be calculated using the formulas:$$x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$$$$y_{cm} = \frac{\sum m_i y_i}{\sum m_i}$$Given the masses and their coordinates:- Mass 1: $2.0 \, \text{kg}$ at $(0.0 \, \text{m}, 0.0 \, \text{m})$- Mass 2: $2.0 \, \text{kg}$ at $(2.0 \, \text{m}, 0.0 \, \text{m})$- Mass 3: $4.0 \, \text{kg}$ at $(2.0 \, \text{m}, 1.0 \, \text{m})$Calculating $x_{cm}$:$$x_{cm} = \frac{(2.0 \times 0.0) + (2.0 \times 2.0) + (4.0 \times 2.0)}{2.0 + 2.0 + 4.0} = \frac{0 + 4 + 8}{8} = \frac{12}{8} = 1.5 \, \text{m}$$Calculating $y_{cm}$:$$y_{cm} = \frac{(2.0 \times 0.0) + (2.0 \times 0.0) + (4.0 \times 1.0)}{2.0 + 2.0 + 4.0} = \frac{0 + 0 + 4}{8} = \frac{4}{8} = 0.5 \, \text{m}$$Therefore, the coordinates of the center of mass are $(1.5 \, \text{m}, 0.5 \, \text{m})$.

Question 7

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)1.0 m
B)2.0 m
C)3.0 m
D)4.0 m
E)5.0 m

Accepted Answer

According to the principle of conservation of momentum, when the child starts moving towards the other end, the total momentum of the system (child and canoe) must remain constant. Since the initial momentum is zero, the final momentum must also be zero. This means that the canoe must move in the opposite direction to compensate for the child's motion.

The child moves from one end to the other end, covering a distance of 5.0 m. Since the canoe is initially at rest, the center of mass of the system (child and canoe) is at rest. When the child moves to the other end, the center of mass of the system is displaced by a distance of 2.5 m (half the length of the canoe), towards the end where the child started. In order to maintain the center of mass of the system at rest, the canoe must move 2.5 m in the opposite direction.

Therefore, the canoe moves a total distance of 2.5 m + 0.5 m (the initial distance between the child and the end of the canoe) = 3.0 m in the water.

Hence, the correct choice is C) 3.0 m.

Question 8

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)1.25 m
B)2.50 m
C)5.00 m
D)7.50 m
E)9.00 m

Accepted Answer

The center of mass of the system can be calculated using the formula: (m1*x1 + m2*x2) / (m1 + m2). Here, m1 = 60 kg, x1 = 0 m (man's position), m2 = 20 kg, x2 = 5 m (center of the board). The center of mass is (60*0 + 20*5) / (60 + 20) = 1.25 m from the man.

Question 9

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)0.25 m
B)0.50 m
C)0.75 m
D)1.0 m
E)1.5 m

Accepted Answer

When half of the rope is lifted off the table, the center of mass of the lifted portion is at the midpoint of that half, which is 0.5 meters from the end being lifted. Since this is half of the rope, the midpoint is 1 meter from the end, making the center of mass 0.25 meters above the table.

Question 10

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)(-3.0 m,-6.0 m)
B)(-12 m,-8.0 m)
C)(3.0 m,6.0 m)
D)(-6.0 m,-3.0 m)

Accepted Answer

The center of mass (CM) of a system can be found 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 the x and y coordinates, respectively. For the CM to be at the origin, both $x_{CM}$ and $y_{CM}$ must be 0. Given masses and their coordinates, and setting the desired CM at (0,0), we solve for the coordinates of the 4.0-kg mass. Plugging in the given values and solving the equations $0 = \frac{(3.0)(0) + (1.0)(12) + (4.0)x}{3.0 + 1.0 + 4.0}$ and $0 = \frac{(3.0)(8) + (1.0)(0) + (4.0)y}{3.0 + 1.0 + 4.0}$ leads to the solution for the coordinates of the 4.0-kg mass, which are (-3.0 m, -6.0 m).

Question 11

In a certain nuclear reactor,neutrons suddenly collide with carbon nuclei,which have 12 times the mass of neutrons.In a head-on elastic collision with a stationary carbon nucleus,what fraction of its initial speed does the neutron have after the collision?
A)11/12
B)1/12
C)11/13
D)5/6
E)10/13

Accepted Answer

In an elastic collision, the velocity of a particle after collision can be determined using the formula: $v' = \frac{(m_1 - m_2)}{(m_1 + m_2)}v$, where $v'$ is the velocity of the particle after collision, $m_1$ is the mass of the moving particle (neutron), $m_2$ is the mass of the stationary particle (carbon nucleus), and $v$ is the initial velocity of the moving particle. Given that the carbon nucleus has 12 times the mass of the neutron, we substitute $m_1 = 1m$ and $m_2 = 12m$ into the formula, resulting in $v' = \frac{(1m - 12m)}{(1m + 12m)}v = \frac{-11m}{13m}v$. The negative sign indicates direction change, but the question asks for the fraction of the initial speed, which is the absolute value, hence $|-\frac{11}{13}|$ or $\frac{11}{13}$.

Question 12

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 of 50° 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)14 m/s
D)23 m/s
E)17 m/s

Accepted Answer

The answer of A 1200-kg pick-up truck traveling south at...

Question 13

An object initially at rest suddenly explodes in two fragments of masses 
And 
That move in opposite directions.If the speed of the first fragment is 
Find the internal energy of the explosion.
A)30 J
B)30 kJ
C)38 J
D)38 kJ

Accepted Answer

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

Question 14

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

Accepted Answer

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

Quiz 9: Linear Momentum and Collisions

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?

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?

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 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 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 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 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 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?

In a certain nuclear reactor,neutrons suddenly collide with carbon nuclei,which have 12 times the mass of neutrons.In a head-on elastic collision with a stationary carbon nucleus,what fraction of its initial speed does the neutron have after the collision?

An object initially at rest suddenly explodes in two fragments of masses And That move in opposite directions.If the speed of the first fragment is Find the internal energy of the explosion.

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?