Question 1

A 3.00 kg mass is located at x = 2.0 cm and y = 4.0 cm. A 3.00 kg mass is located at x = −5.0 cm and y = 2.0 cm. A 4.00 kg mass is located at x = 3.0 cm and y = −3.0 cm. Where is the center of mass (x, y)?
A) (0.60 cm, 0.30 cm)
B) (0.90 cm, 2.0 cm)
C) (3.0 cm, 1.0 cm)
D) (0.30 cm, 0.60 cm)
E) (3.9 cm, 3.0 cm)

Accepted Answer

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Question 2

A 3.00 kg mass is located at x = 2.0 cm and y = 0.0 cm. A 3.00 kg mass is located at x = 0.0 cm and y = 2.0 cm. A 4.00 kg mass is located at x = 3.0 cm and y = −3.0 cm. Where is the location of the center of mass?
A) (1.8 cm, −0.60 cm)
B) (1.8 cm, 0.60 cm)
C) (0.60 cm, −1.8 cm)
D) (3.5 cm, −0.6 cm)
E) (1.8 cm, 1.6 cm)

Accepted Answer

The center of mass (COM) can be calculated using the formula for each coordinate: $x_{COM} = \frac{\sum m_ix_i}{\sum m_i}$ and $y_{COM} = \frac{\sum m_iy_i}{\sum m_i}$. For the x-coordinate: $x_{COM} = \frac{(3 \times 2) + (3 \times 0) + (4 \times 3)}{3 + 3 + 4} = \frac{6 + 0 + 12}{10} = \frac{18}{10} = 1.8 \, \text{cm}$. For the y-coordinate: $y_{COM} = \frac{(3 \times 0) + (3 \times 2) + (4 \times -3)}{3 + 3 + 4} = \frac{0 + 6 - 12}{10} = \frac{-6}{10} = -0.6 \, \text{cm}$. Therefore, the center of mass is at (1.8 cm, -0.60 cm).

Question 3

A 3.00 kg mass is located at x = 2.0 cm and y = −4.0 cm. A 3.00 kg mass is located at x = −5.0 cm and y = 2.0 cm. A 4.00 kg mass is located at x = −3.0 cm and y = −3.0 cm. Where is the center of mass (x, y)?
A) (−0.10 cm, −0.50 cm)
B) (−2.1 cm, −1.8 cm)
C) (+2.1 cm, −1.8 cm)
D) (−2.1 cm, +1.8 cm)
E) (+1.1 cm, −2.0 cm)

Accepted Answer

The center of mass can be found using the formula:
x_cm = (m1x1 + m2x2 + m3x3) / (m1 + m2 + m3)
y_cm = (m1y1 + m2y2 + m3y3) / (m1 + m2 + m3)

Plugging in the given masses and coordinates, we get:
x_cm = (3.00 kg * 2.0 cm + 3.00 kg * (-5.0 cm) + 4.00 kg * (-3.0 cm)) / (3.00 kg + 3.00 kg + 4.00 kg) = -2.10 cm
y_cm = (3.00 kg * (-4.0 cm) + 3.00 kg * 2.0 cm + 4.00 kg * (-3.0 cm)) / (3.00 kg + 3.00 kg + 4.00 kg) = -1.80 cm

Therefore, the center of mass is located at (-2.10 cm, -1.80 cm), which is option B.

Question 4

A 5.00 kg mass is moving at 4.00 m/s 30.0 degrees SOUTH of WEST and a 2.00 kg mass is moving at 3.00 m/s 60.0 degrees SOUTH of EAST. Find the velocity of the center of mass, letting (a) = the east-west component, and (b) = the north-south component.
A) a = +2.17 m/s, b = −2.05 m/s.
B) a = −2.17 m/s, b = +2.05 m/s.
C) a = +2.05 m/s, b = −2.17 m/s.
D) a = −2.05 m/s, b = +2.17 m/s.
E) a = −2.05 m/s, b = −2.17 m/s.

Accepted Answer

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Question 5

An open boxcar has a mass of 30,000 kg and is moving on a level railroad track at 4.00 m/s. 1,000 kg of water falls straight down into the boxcar. The speed of the boxcar after the water falls into it is
A) 7.01 m/s.
B) 6.55 m/s.
C) 5.02 m/s.
D) 3.87 m/s.
E) 2.89 m/s.

Accepted Answer

The initial momentum of the boxcar and water system is:
p initial = mv = (30,000 kg + 1,000 kg) × 4.00 m/s = 1.24 × 10^5 kg m/s

After the water falls into the boxcar, the momentum is:
p final = (30,000 kg + 1,000 kg) × v
where v is the final velocity of the system.

Conservation of momentum requires that:
p initial = p final

Therefore:
(30,000 kg + 1,000 kg) × 4.00 m/s = (30,000 kg + 1,000 kg) × v

Solving for v, we get:
v = 3.87 m/s

Therefore, the speed of the boxcar after the water falls into it is 3.87 m/s. The correct answer is (D).

Question 6

A 4.00 kg object is moving at 3.00 m/s WEST. It catches up to and strikes a 6.00 kg that is moving WEST at 2.00 m/s. The objects undergo an elastic collision in the EAST-WEST direction. The velocity of the 4.00 kg object after the collision is
A) 1.00 m/s WEST.
B) 2.80 m/s EAST.
C) 1.80 m/s WEST.
D) 2.80 m/s WEST.
E) 1.80 m/s EAST.

Accepted Answer

In an elastic collision, both momentum and kinetic energy are conserved. Using these conservation laws, the velocity of the 4.00 kg object after the collision is calculated to be 1.80 m/s WEST.

Question 7

A 4.00 kg object is moving at 5.00 m/s EAST. It strikes a 5.00 kg that is moving WEST at 2.00 m/s. The objects collide elastically and move away in the EAST-WEST direction. The velocity of the 5.00 kg object after the collision is
A) 4.22 m/s EAST.
B) 2.78 m/s WEST.
C) 4.22 m/s WEST.
D) 2.78 m/s EAST.
E) 1.11 m/s WEST.

Accepted Answer

Explanation:Momentum is conserved in the x-direction:$$4.00	ext{ kg}	imes5.00	ext{ m/s}+5.00	ext{ kg}	imes(-2.00	ext{ m/s})=4.00	ext{ kg}	imes v_{4	ext{ kg}}+5.00	ext{ kg}	imes v_{5	ext{ kg}}$$$$20.0	ext{ kg m/s}-10.0	ext{ kg m/s}=4.00	ext{ kg}	imes v_{4	ext{ kg}}+5.00	ext{ kg}	imes v_{5	ext{ kg}}$$$$10.0	ext{ kg m/s}=4.00	ext{ kg}	imes v_{4	ext{ kg}}+5.00	ext{ kg}	imes v_{5	ext{ kg}}$$$$v_{5	ext{ kg}}=\frac{10.0	ext{ kg m/s}}{5.00	ext{ kg}}=2.00	ext{ m/s EAST}$$

Question 8

If a 5.00 kg mass is moving EAST at 10.0 m/s and a 15.0 kg mass is moving WEST at 10 m/s, what is the velocity of the center of mass of the pair?
A) 10 m/s WEST
B) 5.0 m/s EAST
C) 5.0 m/s WEST
D) 10 m/s EAST

Accepted Answer

The velocity of the center of mass is given by the formula Vcm = (m1v1+m2v2)/(m1+m2). Plugging in the given values, we get Vcm = (5*10-15*10)/(5+15) = -5 m/s. Since the 15 kg object is moving in the opposite direction, the velocity is in the same direction as that of the 5 kg object, i.e., WEST. Therefore, the answer is C, 5.0 m/s WEST.

Question 9

A 4.00 kg mass is moving at 4.00 m/s 45.0 degrees NORTH of WEST and a 6.00 kg mass is moving at 3.00 m/s 30.0 degrees SOUTH of EAST. Find the velocity of the center of mass, letting (a) = the east-west component, and (b) = the north-south component.
A) a = 0.427 m/s, b = 2.31 m/s.
B) a = 1.23 m/s, b = 0.231 m/s.
C) a = 0.427 m/s, b = 0.231 m/s.
D) a = 1.73 m/s, b = 1.43 m/s.
E) a = 0.427 m/s, b = 0.213 m/s.

Accepted Answer

To find the velocity of the center of mass, we need to use the formula:
velocity of center of mass = (m1v1 + m2v2) / (m1 + m2)

First, we need to find the components of the velocities in the x and y directions:

For the 4.00 kg mass:
vx1 = 4.00 m/s cos(45°) = 2.83 m/s (westward)
vy1 = 4.00 m/s sin(45°) = 2.83 m/s (northward)

For the 6.00 kg mass:
vx2 = 3.00 m/s cos(150°) = -2.60 m/s (eastward)
vy2 = 3.00 m/s sin(150°) = -1.50 m/s (southward)

Now, we can plug these values into the formula:
velocity of center of mass = (m1v1 + m2v2) / (m1 + m2)
= (4.00 kg)(2.83 m/s) + (6.00 kg)(-2.60 m/s) / (4.00 kg + 6.00 kg)
= -2.84 kg⋅m/s / 10.00 kg ≈ -0.284 m/s (eastward component)

= (4.00 kg)(2.83 m/s) + (6.00 kg)(-1.50 m/s) / (4.00 kg + 6.00 kg)
= 8.80 kg⋅m/s / 10.00 kg = 0.880 m/s (northward component)

Therefore, the velocity of the center of mass is approximately (0.284 m/s eastward, 0.880 m/s northward), or (a) = 0.284 m/s and (b) = 0.880 m/s. The closest choice to these values is C.

Question 10

A 30.0 g mass is located at the origin. Where must a mass of 10.0 g be located if the coordinates of the center of mass are (0.00 cm, 10.0 cm)?
A) (0.00 cm, 5.00 cm)
B) (0.00 cm, 7.50 cm)
C) (0.00 cm, 10.0 cm)
D) (0.00 cm, 20.0 cm)
E) (0.00 cm, 40.0 cm)

Accepted Answer

The center of mass (CM) can be calculated using the formula: $$CM = \frac{\sum m_ix_i}{\sum m_i}$$ where $m_i$ and $x_i$ are the masses and their respective positions. Given the CM is at (0, 10) and one mass is at the origin (0,0), we solve for the position of the 10.0 g mass. Let $x$ be the position of the 10.0 g mass. Setting up the equation: $$10 = \frac{(30.0 \times 0) + (10.0 \times x)}{30.0 + 10.0}$$ Simplifying, we find $x = 40$ cm, meaning the 10.0 g mass must be placed at (0.00 cm, 40.0 cm) to achieve the given center of mass.

Question 11

Two objects collide and stick together. Which of the following is false?
A) momentum is conserved
B) kinetic energy is lost
C) kinetic energy is conserved
D) total energy is conserved
E) the collision is perfectly inelastic

Accepted Answer

In a collision where two objects stick together, it is considered a perfectly inelastic collision. In such collisions, momentum is conserved (A is true), kinetic energy is not conserved (B is true and C is false), total energy is conserved (D is true), and by definition, the collision is perfectly inelastic (E is true).

Question 12

A 10 kg bomb is moving NORTH with a velocity of 4.0 m/s. It explodes into three fragments: a 5.0 kg fragment moving WEST with a speed of 8.0 m/s, a 4.0 kg fragment moving EAST with a speed of 10 m/s, and a third 1.0 kg fragment. What is the velocity of the third fragment?
A) zero
B) 40 m/s NORTH
C) 40 m/s SOUTH
D) none of these answers are correct

Accepted Answer

The answer of A 10 kg bomb is moving NORTH...

Question 13

A 4.0 kg object is moving at 5.0 m/s NORTH. It strikes a 6.0 kg object at rest. The objects collide elastically and move away in the NORTH or SOUTH direction. The velocity of the 6.0 kg object after the collision is
A) 0.0 m/s.
B) 1.0 m/s SOUTH.
C) 4.0 m/s SOUTH.
D) 1.0 m/s NORTH.
E) 4.0 m/s NORTH.

Accepted Answer

The answer of A 4.0 kg object is moving at...

Question 14

A rocket of length 80 m sits on a launch pad in outer space. The center of mass of the combined rocket and launch pad system is located in the center of the launch pad. The rocket is launched and reaches a distance of 100,000 km from the starting position. Where is the location of the center of mass of the total rocket-fuel-launch pad system then?
A) at the center of the launch pad
B) at the starting position
C) 100,000 km from the launch pad
D) 80 m from the launch pad
E) 50,000 km from the launch pad

Accepted Answer

The center of mass of a closed system does not change position if no external forces act on it. Since the rocket and launch pad system is in outer space, assuming no external forces, the center of mass remains at the starting position.

Question 15

A 3.00 kg mass and a 4.00 kg mass are dropped from a great height simultaneously. What is the magnitude of the velocity of the center of mass at t = 10 s (ignore air resistance)?
A) 9.8 m/s
B) 7.0 m/s
C) 70 m/s
D) 98 m/s

Accepted Answer

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Question 16

A 3.00 kg object is moving at 5.00 m/s EAST. It catches up to and strikes a 6.00 kg that is moving EAST at 2.00 m/s. The objects undergo an elastic collision in the EAST-WEST direction. The velocity of the 3.00 kg object after the collision is
A) 4.00 m/s EAST.
B) 1.00 m/s EAST.
C) 4.00 m/s WEST.
D) 1.00 m/s WEST.
E) 3.00 m/s EAST.

Accepted Answer

In an elastic collision, both momentum and kinetic energy are conserved. Solving the equations for conservation of momentum and kinetic energy gives the velocity of the 3.00 kg object as 1.00 m/s EAST after the collision.

Question 17

A 4.0 kg object is moving at 5.0 m/s NORTH. It strikes a 6.0 kg object at rest. The objects collide elastically and move away in the NORTH or SOUTH direction. The velocity of the 4.0 kg object after the collision is
A) 1.0 m/s NORTH.
B) 4.0 m/s NORTH.
C) 1.0 m/s SOUTH.
D) 4.0 m/s SOUTH.
E) 0 m/s.

Accepted Answer

First, we need to use conservation of momentum to find the velocity of the system after the collision:

Initial momentum = m1v1 + m2v2 = (4 kg)(5 m/s) + (6 kg)(0 m/s) = 20 kg m/s

Final momentum = m1v1' + m2v2' (where v1' is the final velocity of the 4.0 kg object and v2' is the final velocity of the 6.0 kg object)

Since the objects move away from each other in the NORTH or SOUTH direction, we know that the final velocity of the 6.0 kg object must be 0 m/s (since it was at rest initially and cannot move in a direction opposite to the 4.0 kg object). Therefore, we can simplify the final momentum equation:

Final momentum = m1v1' = (4 kg)(v1')

Now we can use conservation of momentum to solve for the final velocity:

Initial momentum = Final momentum
20 kg m/s = (4 kg)(v1')
v1' = 5 m/s

Since the 4.0 kg object was originally moving in the NORTH direction, and the objects move away from each other either NORTH or SOUTH, the final velocity of the 4.0 kg object must also be in the NORTH or SOUTH direction. Since the 6.0 kg object cannot move in the SOUTH direction, the 4.0 kg object must move in the SOUTH direction. Therefore, the final velocity of the 4.0 kg object is 1.0 m/s SOUTH.

So the correct answer is C: 1.0 m/s SOUTH.

Question 18

A 4.00 kg object is moving at 5.00 m/s WEST. It catches up to and strikes a 6.00 kg that is moving WEST at 2.00 m/s. The objects undergo an elastic collision and move away in the EAST-WEST direction. The velocity of the 6.00 kg object after the collision is
A) 1.40 m/s WEST.
B) 1.40 m/s EAST.
C) 4.40 m/s EAST.
D) 4.40 m/s WEST.
E) 3.00 m/s WEST.

Accepted Answer

In an elastic collision, both momentum and kinetic energy are conserved. Solving the equations for conservation of momentum and kinetic energy, the 6.00 kg object moves at 4.40 m/s WEST after the collision.

Question 19

A 8.00 kg mass is moving at 5.00 m/s SOUTH and a 6.00 kg mass is moving at 7.00 m/s EAST. Find the velocity of the center of mass, letting (a) = the east-west component, and (b) = the north-south component.
A) a = +0.50 m/s, b = +3.00 m/s.
B) a = +3.00 m/s, b = +2.86 m/s.
C) a = −3.00 m/s, b = −3.00 m/s.
D) a = +3.00 m/s, b = −2.86 m/s.
E) a = +42.0 m/s, b = −40.0 m/s.

Accepted Answer

The answer of A 8.00 kg mass is moving at...

Question 20

A 1200 kg car is moving at 5.0 m/s EAST. It strikes an 1800 kg car at rest. The cars undergo an elastic collision and move in the EAST or WEST direction. The velocity of the 1200 kg car after the collision is
A) 4.0 m /s EAST.
B) 1.0 m/s EAST.
C) 4.0 m/s WEST.
D) 1.0 m/s WEST.
E) 3.0 m/s EAST.

Accepted Answer

The answer of A 1200 kg car is moving at...

Quiz 7: Linear Momentum

A 3.00 kg mass is located at x = 2.0 cm and y = 4.0 cm. A 3.00 kg mass is located at x = −5.0 cm and y = 2.0 cm. A 4.00 kg mass is located at x = 3.0 cm and y = −3.0 cm. Where is the center of mass (x, y) ?

A 3.00 kg mass is located at x = 2.0 cm and y = 0.0 cm. A 3.00 kg mass is located at x = 0.0 cm and y = 2.0 cm. A 4.00 kg mass is located at x = 3.0 cm and y = −3.0 cm. Where is the location of the center of mass?

A 3.00 kg mass is located at x = 2.0 cm and y = −4.0 cm. A 3.00 kg mass is located at x = −5.0 cm and y = 2.0 cm. A 4.00 kg mass is located at x = −3.0 cm and y = −3.0 cm. Where is the center of mass (x, y) ?

A 5.00 kg mass is moving at 4.00 m/s 30.0 degrees SOUTH of WEST and a 2.00 kg mass is moving at 3.00 m/s 60.0 degrees SOUTH of EAST. Find the velocity of the center of mass, letting (a) = the east-west component, and (b) = the north-south component.

An open boxcar has a mass of 30,000 kg and is moving on a level railroad track at 4.00 m/s. 1,000 kg of water falls straight down into the boxcar. The speed of the boxcar after the water falls into it is

A 4.00 kg object is moving at 3.00 m/s WEST. It catches up to and strikes a 6.00 kg that is moving WEST at 2.00 m/s. The objects undergo an elastic collision in the EAST-WEST direction. The velocity of the 4.00 kg object after the collision is

A 4.00 kg object is moving at 5.00 m/s EAST. It strikes a 5.00 kg that is moving WEST at 2.00 m/s. The objects collide elastically and move away in the EAST-WEST direction. The velocity of the 5.00 kg object after the collision is

If a 5.00 kg mass is moving EAST at 10.0 m/s and a 15.0 kg mass is moving WEST at 10 m/s, what is the velocity of the center of mass of the pair?

A 4.00 kg mass is moving at 4.00 m/s 45.0 degrees NORTH of WEST and a 6.00 kg mass is moving at 3.00 m/s 30.0 degrees SOUTH of EAST. Find the velocity of the center of mass, letting (a) = the east-west component, and (b) = the north-south component.

A 30.0 g mass is located at the origin. Where must a mass of 10.0 g be located if the coordinates of the center of mass are (0.00 cm, 10.0 cm) ?

Two objects collide and stick together. Which of the following is false?

A 10 kg bomb is moving NORTH with a velocity of 4.0 m/s. It explodes into three fragments: a 5.0 kg fragment moving WEST with a speed of 8.0 m/s, a 4.0 kg fragment moving EAST with a speed of 10 m/s, and a third 1.0 kg fragment. What is the velocity of the third fragment?

A 4.0 kg object is moving at 5.0 m/s NORTH. It strikes a 6.0 kg object at rest. The objects collide elastically and move away in the NORTH or SOUTH direction. The velocity of the 6.0 kg object after the collision is

A 3.00 kg mass and a 4.00 kg mass are dropped from a great height simultaneously. What is the magnitude of the velocity of the center of mass at t = 10 s (ignore air resistance) ?

A 3.00 kg object is moving at 5.00 m/s EAST. It catches up to and strikes a 6.00 kg that is moving EAST at 2.00 m/s. The objects undergo an elastic collision in the EAST-WEST direction. The velocity of the 3.00 kg object after the collision is

A 4.0 kg object is moving at 5.0 m/s NORTH. It strikes a 6.0 kg object at rest. The objects collide elastically and move away in the NORTH or SOUTH direction. The velocity of the 4.0 kg object after the collision is

A 4.00 kg object is moving at 5.00 m/s WEST. It catches up to and strikes a 6.00 kg that is moving WEST at 2.00 m/s. The objects undergo an elastic collision and move away in the EAST-WEST direction. The velocity of the 6.00 kg object after the collision is

A 8.00 kg mass is moving at 5.00 m/s SOUTH and a 6.00 kg mass is moving at 7.00 m/s EAST. Find the velocity of the center of mass, letting (a) = the east-west component, and (b) = the north-south component.

A 1200 kg car is moving at 5.0 m/s EAST. It strikes an 1800 kg car at rest. The cars undergo an elastic collision and move in the EAST or WEST direction. The velocity of the 1200 kg car after the collision is