Question 1

A 310-g air track cart is traveling at 1.25 m/s and a 260-g cart traveling in the opposite direction at 1.33 m/s. What is the speed of the center of mass of the two carts?

Accepted Answer

The speed of the center of mass can be calculated using the formula: $v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$, where $m_1$ and $m_2$ are the masses of the carts, and $v_1$ and $v_2$ are their velocities. Given that one cart is moving in the opposite direction, its velocity is considered negative. Thus, $v_{cm} = \frac{(0.31 \times 1.25) + (0.26 \times -1.33)}{0.31 + 0.26} = \frac{0.3875 - 0.3458}{0.57} = 0.0731 \, m/s$, which rounds to 0.0732 m/s, making option B correct.

Question 2

A curling stone slides on ice with a speed of 1.70 m/s and collides elastically with an identical, stationary curling stone. After the collision, the first stone has a velocity of 0.800 m/s in a direction that makes a counterclockwise angle of 61.9° with its original direction of travel. At what speed and what direction is the second stone traveling after the collision?

Accepted Answer

The answer of A curling stone slides on ice with...

Question 3

A 320-g air track cart traveling at 1.25 m/s collides elastically with a stationary 270-g cart. What is the speed of the 270-g cart after the collision?

Accepted Answer

In an elastic collision, both momentum and kinetic energy are conserved. Using these principles, the speed of the 270-g cart after the collision can be calculated to be 1.36 m/s.

Question 4

A 120-g rubber ball is thrown with a speed of 10 m/s at an oncoming 900-kg car which is approaching at 30 m/s, and undergoes a one-dimensional elastic collision with the car. What is the speed of the ball after the collision?

Accepted Answer

In an elastic collision, both momentum and kinetic energy are conserved. Using these principles, the final speed of the ball can be calculated to be 70 m/s.

Question 5

A 1000-kg helicopter hovers over a forest fire and starts to squirt water downward with a speed of 10 m/s at a rate of 40 kg/s. What is the initial upward acceleration on the helicopter when it first starts to release the water?

Accepted Answer

The force exerted by the water is equal to the rate of change of momentum, which is 40 kg/s * 10 m/s = 400 N. The upward acceleration is then 400 N / 1000 kg = 0.40 m/s².

Question 6

A mass of 2 kg traveling at 3 m/s undergoes a one-dimensional elastic collision with a group of four 1-kg masses that are at rest in contact with each other and lined up in the same direction as the velocity of the 2-kg mass. As a result of the collision

Accepted Answer

In an elastic collision, both momentum and kinetic energy are conserved. Initially, the momentum of the system is 2 kg x 3 m/s = 6 kg m/s, and after the collision, the momentum is still conserved. Let us assume that two of the 1-kg masses take off in one direction at a speed of v and the other two masses take off in the opposite direction at the same speed v. Then, the final momentum of the system would be (2 kg)(0 m/s) + (1 kg)(v m/s) + (1 kg)(v m/s) + (1 kg)(-v m/s) + (1 kg)(-v m/s) = 0 kg m/s. Solving for v, we get v = 3 m/s. Thus, two of the 1-kg masses take off in one direction at 3 m/s, while the other two take off in the opposite direction at 3 m/s, leaving the 2-kg mass at rest. Therefore, the correct answer is B.

Question 7

In a physics demonstration, two rubber balls are dropped down a tube towards the ground, with the lighter ball behind, and not quite in contact with, the heavier ball. The heavier ball collides with the ground and bounces, then collides with the still descending lighter ball. Assume that all the collisions are elastic. In one such demonstration, the heavier ball has a mass of 500 grams, the lighter one a mass of 100 grams, and they are released 1.00 m above the ground. How high does the lighter ball rise after the two collisions?

Accepted Answer

In an elastic collision, the lighter ball can gain a significant amount of kinetic energy from the heavier ball. Using conservation of momentum and energy, the lighter ball can rise to a height of approximately 5.44 m after the collisions.

Question 8

A curling stone slides on ice with a speed of 2.0 m/s and collides inelastically with an identical, stationary curling stone. After the collision, the first stone is deflected by a counterclockwise angle of 28° from its original direction of travel, and the second stone moves in a direction that makes a 42° clockwise angle with the original direction of travel of the first stone. What fraction of the initial energy is lost in this collision?

Accepted Answer

The answer of A curling stone slides on ice with...

Question 9

In a particle accelerator, protons are fired at a stationary target and collide elastically with the nuclei of the atoms in the target. In one such experiment, a beam of protons is fired with a speed of 2.10 × 10⁷ m/s. The protons that undergo a collision rebound with a speed of 1.68 × 10⁷ m/s. What is the mass of the nuclei that the protons are colliding against?

Accepted Answer

The answer of In a particle accelerator, protons are fired...

Question 10

A 340-g air track cart traveling at 1.25 m/s collides elastically with a 300-g cart traveling in the opposite direction at 1.33 m/s. What is the speed of the 300-g cart after the collision?

Accepted Answer

The answer of A 340-g air track cart traveling at...

Question 11

A rocket is fired in a gravity-free environment in deep space. The rocket has an initial mass of 8000 kg and ejects gas with a velocity of 2000 m/s. How much gas must be ejected in the first second to give the rocket an acceleration of 30.0 m/s²?

Accepted Answer

Explanation:The force on the rocket is equal to the rate of change of momentum of the gas:$$F = \frac{d(mv)}{dt}$$where m is the mass of the gas and v is its velocity. The acceleration of the rocket is:$$a = \frac{F}{m_{rocket}}$$where $m_{rocket}$ is the mass of the rocket. Substituting the first equation into the second equation, we get:$$a = \frac{d(mv)}{dt} \frac{1}{m_{rocket}}$$$$a = \frac{v}{m_{rocket}} \frac{dm}{dt}$$Solving for $\frac{dm}{dt}$, we get:$$\frac{dm}{dt} = \frac{am_{rocket}}{v}$$Substituting the given values, we get:$$\frac{dm}{dt} = \frac{(30.0 \ m/s^2)(8000 \ kg)}{2000 \ m/s} = 120 \ kg/s$$Therefore, the correct answer is A.

Question 12

A 320-g air track cart traveling at 1.25 m/s collides elastically with a stationary 270-g cart. What is the speed of the 320-g cart after the collision?

Accepted Answer

The answer of A 320-g air track cart traveling at...

Question 13

A puck with a mass m₁ = 28.0 g moving at 1.00 m/s approaches a stationary puck with a mass m₂ = 102 g on an air table and they undergo a two-dimensional elastic collision. As a result of their interaction, the incident puck moves away with a speed v₁ = 0.785 m/s and the other puck moves away with a speed v₂ in a different direction. What is the angle between the velocities v₁ and v₂?

Accepted Answer

The answer of A puck with a mass m₁ =...

Question 14

A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the center of mass of the bent wire?

Accepted Answer

The center of mass of each segment of the wire (10 cm long) is at its midpoint, 5 cm from the bend. Since the wire is uniform and the segments are equal in length, the center of mass of the L-shape will be located at the point where the medians of the right triangle (formed by the two segments and the line connecting their midpoints) intersect. This point is 1/3 of the way along the median from the right angle, which is $\sqrt{5^2 + 5^2} = \sqrt{50}$ cm long. Therefore, the distance from the bend to the center of mass is $\frac{1}{3} \times \sqrt{50} \approx 3.5$ cm.

Question 15

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

Accepted Answer

To find the center of mass, we can use the formula: 
x_cm = (m1*x1 + m2*x2 + m3*x3)/(m1+m2+m3) 
y_cm = (m1*y1 + m2*y2 + m3*y3)/(m1+m2+m3) 
Plugging in the values given: 
x_cm = (6*0 + 4*3 + 2*0)/(6+4+2) = 1 
y_cm = (6*0 + 4*0 + 2*3)/(6+4+2) = 0.5 
Therefore, the center of mass is at (1 m, 0.5 m). The correct choice is D.

Question 16

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

Accepted Answer

The center of mass of a uniform rope is at its midpoint. When half of the rope has lifted off the table, the center of mass is 0.25 m above the table.

Question 17

An 80.0-g puck moving at 1.20 m/s on an air table collides with a stationary 100-g puck. After the collision, the 80.0-g puck moves in a direction that makes an angle of 45.0° with the original direction of motion and the 100-g puck moves at an angle of 12.0° in the opposite direction. What is the speed of the 80.0-g puck after the collision?

Accepted Answer

The answer of An 80.0-g puck moving at 1.20 m/s...

Question 18

A 550-g mass traveling at 8.0 m/s undergoes a head-on elastic collision with a 250-g mass traveling toward it also at 8.0 m/s. What is the speed of the 250-g mass after the collision?

Accepted Answer

Explanation:Momentum is conserved in the collision, so the total momentum before the collision is equal to the total momentum after the collision. The 550-g mass has a momentum of 550 g * 8.0 m/s = 4400 g m/s, and the 250-g mass has a momentum of 250 g * 8.0 m/s = 2000 g m/s. The total momentum before the collision is therefore 4400 g m/s + 2000 g m/s = 6400 g m/s. After the collision, the 550-g mass has a momentum of 550 g * v = 550 v g m/s, and the 250-g mass has a momentum of 250 g * v = 250 v g m/s. The total momentum after the collision is therefore 550 v g m/s + 250 v g m/s = 800 v g m/s. Since momentum is conserved, we have 6400 g m/s = 800 v g m/s, so v = 8.0 m/s. The speed of the 250-g mass after the collision is therefore 8.0 m/s.

Question 19

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 of the man-board system?

Accepted Answer

The center of mass is 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 20

A 1200-kg pick-up truck traveling south at 15.0 m/s collides with a 750-kg car traveling east. The two vehicles stick together. A patrolman investigating the accident determines that the final position of the wreckage after the collision is 25.0 m, at an angle of 50.0° south of east, from the point of impact. He also determines that the coefficient of friction between the tires and the road at that location was 0.400. What was the speed of the car before the collision?

Accepted Answer

We can use conservation of momentum in two dimensions to solve the problem. The total momentum in the x-direction before the collision is:

Px = (1200 kg)(0 m/s) + (750 kg)(v) = 750v

The total momentum in the y-direction before the collision is:

Py = (1200 kg)(-15.0 m/s) + (750 kg)(0 m/s) = -18000 kg*m/s

After the collision, the two vehicles stick together and move at an angle of 50.0° south of east. We can resolve this momentum vector into its x- and y-components:

Pf,x = (1200 kg + 750 kg)(vf,x) = 1950 vf,x

Pf,y = 0

We can use the Pythagorean theorem to find the magnitude of the final momentum vector:

|Pf| = sqrt[(Pf,x)^2 + (Pf,y)^2] = 1950 vf,x

We also know that the final position of the wreckage is 25.0 m at an angle of 50.0° south of east, which means that the displacement in the x-direction is:

xf = 25.0 m cos(50.0°) = 16.0 m

We can use the coefficient of friction and the displacement in the x-direction to find the work done by friction:

Wf = f * d * cos(180°) = -f * d = -0.400 * 16.0 m = -6.40 J

Since there are no external forces acting on the system in the x-direction, the work done by friction must be equal to the change in kinetic energy:

Wf = Kf,x - Ki,x

-6.40 J = (1/2)(1950 kg)(vf,x)^2 - (1/2)(750 kg)(v)^2

We can solve this equation for vf,x:

vf,x = sqrt[((0.400)(16.0 m)(750 kg)(v)^2 - 6.40 J)/(1950 kg)]

vf,x = sqrt[(120000 kg*m^2/s^2)(v^2) - 20.8 J)/(1950 kg)]

Now we can use the fact that the magnitude of the final momentum vector is equal to the magnitude of the initial momentum vector, since there are no external forces acting in the y-direction:

|Pf| = |Pi| = sqrt[(750v)^2 + (-18000 kg*m/s)^2] = 18525 kg*m/s

|Pf| = sqrt[(1950 kg)^2 (vf,x)^2] = (1950 kg)(vf,x)

Combining both equations gives:

18525 kg*m/s = (1950 kg)(vf,x) = 1950 sqrt[(120000 kg*m^2/s^2)(v^2) - 20.8 J)/(1950 kg)]

18525 = sqrt[(120000v^2 - 20.8)]

342219375 = 120000v^2 - 20.8

v^2 = 2851.864

v = 53.4 m/s

However, we should check if this is reasonable based on the initial speeds of the two vehicles. The pick-up truck was traveling south at 15.0 m/s, which means that the car must have been traveling at least 43.4 m/s east to give the resultant velocity of 23.4 m/s southeast. Therefore, the correct answer is:

v = 43.4 m/s = 15.6 m/s

The answer in the given format is:

Answer: D

Quiz 9: Linear Momentum and Collisions

A 310-g air track cart is traveling at 1.25 m/s and a 260-g cart traveling in the opposite direction at 1.33 m/s. What is the speed of the center of mass of the two carts?

A 320-g air track cart traveling at 1.25 m/s collides elastically with a stationary 270-g cart. What is the speed of the 270-g cart after the collision?

A 120-g rubber ball is thrown with a speed of 10 m/s at an oncoming 900-kg car which is approaching at 30 m/s, and undergoes a one-dimensional elastic collision with the car. What is the speed of the ball after the collision?

A 1000-kg helicopter hovers over a forest fire and starts to squirt water downward with a speed of 10 m/s at a rate of 40 kg/s. What is the initial upward acceleration on the helicopter when it first starts to release the water?

A mass of 2 kg traveling at 3 m/s undergoes a one-dimensional elastic collision with a group of four 1-kg masses that are at rest in contact with each other and lined up in the same direction as the velocity of the 2-kg mass. As a result of the collision

A 340-g air track cart traveling at 1.25 m/s collides elastically with a 300-g cart traveling in the opposite direction at 1.33 m/s. What is the speed of the 300-g cart after the collision?

A rocket is fired in a gravity-free environment in deep space. The rocket has an initial mass of 8000 kg and ejects gas with a velocity of 2000 m/s. How much gas must be ejected in the first second to give the rocket an acceleration of 30.0 m/s²?

A 320-g air track cart traveling at 1.25 m/s collides elastically with a stationary 270-g cart. What is the speed of the 320-g cart after the collision?

A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the center of mass of the bent wire?

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

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

A 550-g mass traveling at 8.0 m/s undergoes a head-on elastic collision with a 250-g mass traveling toward it also at 8.0 m/s. What is the speed of the 250-g mass after the collision?

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 of the man-board system?

Quiz 9: Linear Momentum and Collisions

A 310-g air track cart is traveling at 1.25 m/s and a 260-g cart traveling in the opposite direction at 1.33 m/s. What is the speed of the center of mass of the two carts?

A 320-g air track cart traveling at 1.25 m/s collides elastically with a stationary 270-g cart. What is the speed of the 270-g cart after the collision?

A 120-g rubber ball is thrown with a speed of 10 m/s at an oncoming 900-kg car which is approaching at 30 m/s, and undergoes a one-dimensional elastic collision with the car. What is the speed of the ball after the collision?

A 1000-kg helicopter hovers over a forest fire and starts to squirt water downward with a speed of 10 m/s at a rate of 40 kg/s. What is the initial upward acceleration on the helicopter when it first starts to release the water?

A mass of 2 kg traveling at 3 m/s undergoes a one-dimensional elastic collision with a group of four 1-kg masses that are at rest in contact with each other and lined up in the same direction as the velocity of the 2-kg mass. As a result of the collision

A 340-g air track cart traveling at 1.25 m/s collides elastically with a 300-g cart traveling in the opposite direction at 1.33 m/s. What is the speed of the 300-g cart after the collision?

A rocket is fired in a gravity-free environment in deep space. The rocket has an initial mass of 8000 kg and ejects gas with a velocity of 2000 m/s. How much gas must be ejected in the first second to give the rocket an acceleration of 30.0 m/s2?

A 320-g air track cart traveling at 1.25 m/s collides elastically with a stationary 270-g cart. What is the speed of the 320-g cart after the collision?

A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the center of mass of the bent wire?

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

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

A 550-g mass traveling at 8.0 m/s undergoes a head-on elastic collision with a 250-g mass traveling toward it also at 8.0 m/s. What is the speed of the 250-g mass after the collision?

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 of the man-board system?

A rocket is fired in a gravity-free environment in deep space. The rocket has an initial mass of 8000 kg and ejects gas with a velocity of 2000 m/s. How much gas must be ejected in the first second to give the rocket an acceleration of 30.0 m/s²?