Chapter 7: Linear Momentum

When the car collides with the brick wall, it experiences a change in momentum. The initial momentum of the car is given by the product of its mass (2000 kg) and its velocity (30 m/s), which is 60,000 kg*m/s to the right. Since the car comes to rest, its final momentum is zero.

Using the equation for impulse, which is the product of force and time, we can calculate the average force exerted by the car on the wall. The change in momentum is equal to the impulse, which is equal to the average force multiplied by the time.

Therefore, the average force exerted by the car on the wall is (0 - 60,000 kg*m/s) / 0.20 s = -300,000 N. The negative sign indicates that the force is in the opposite direction of the initial velocity. So the correct answer is 300,000 N to the right.

When an object is acted on by a non-zero net external force, its momentum will not remain constant. This is because momentum is defined as the product of an object's mass and velocity, and any change in either mass or velocity will result in a change in momentum. Therefore, if an external force is applied to an object, it will either accelerate or decelerate, causing a change in velocity and subsequently a change in momentum.

When the white cue ball hits the #9 ball, it is deflected at a 35° angle to the original line of motion. The angle of deflection for the #9 ball is always equal to the angle of incidence, which is the angle at which the cue ball hits it. Since the cue ball is deflected at a 35° angle, the #9 ball will be deflected at the same angle but in the opposite direction. Therefore, the angle of deflection below the original line of motion for the #9 ball is 55°.

The center of mass of a system is determined by the distribution of mass within the system. In this case, there are two masses, one at the origin and one at x = +21.0 m. Since the 4.00-kg mass is closer to the origin, it has less influence on the center of mass than the 10.0-kg mass. Therefore, the center of mass will be closer to the 10.0-kg mass, and since it is located at x = +21.0 m, the center of mass on the x-axis will be at +15.0 m.

Tightrope walkers walk with a long flexible rod in order to lower their center of mass. By holding onto the rod, they are able to shift their weight downwards, making their center of mass closer to the tightrope. This helps them maintain balance and stability while walking on the rope.

The average force exerted on the machine gun mount can be calculated using the principle of conservation of momentum. The momentum of a bullet can be calculated by multiplying its mass (50.0 grams = 0.05 kg) by its velocity (750 m/s), which gives 37.5 kg m/s. Since the machine gun fires 300 rounds per minute, the total momentum change per minute is 300 times the momentum of a single bullet, which is 11250 kg m/s. The average force can be calculated by dividing the total momentum change by the time it takes to fire the bullets, which is 1 minute or 60 seconds. Therefore, the average force exerted on the machine gun mount is 11250 kg m/s divided by 60 seconds, which is equal to 188 N.

The momentum of an object is calculated by multiplying its mass by its velocity. In this case, the truck has a mass of 2000 kg and is traveling at 35 m/s. To find the momentum, we multiply 2000 kg by 35 m/s, which equals 70,000 kg*m/s. This can be written in scientific notation as 7.0 * 10^4 kg*m/s.

When an object is dropped, it falls freely under the influence of gravity. As the ball falls, its potential energy is converted into kinetic energy. The momentum of an object is defined as the product of its mass and velocity. Since the ball is dropped from rest, its initial velocity is 0 m/s. Using the equation p = mv, where p is momentum, m is mass, and v is velocity, we can calculate the momentum of the ball when it strikes the ground. The mass of the ball is given as 0.10 kg. The velocity can be calculated using the equation vf^2 = vi^2 + 2ad, where vf is the final velocity, vi is the initial velocity, a is acceleration (due to gravity), and d is distance. Plugging in the given values, we find that the final velocity when the ball strikes the ground is approximately 7.67 m/s. Multiplying the mass and velocity, we get a momentum of approximately 0.10 kg * 7.67 m/s = 0.767 kg*m/s. Rounding to the nearest tenth, the correct answer is 1.5 kg*m/s.

When a large load of coal is suddenly dumped into the open top freight car, the total mass of the car increases. According to the law of conservation of momentum, the momentum of an isolated system remains constant if no external forces act on it. Since the car is on a frictionless track, there are no external forces acting on it. Therefore, in order to maintain the constant momentum, the velocity of the car must decrease as the mass increases. Hence, the correct answer is that the velocity of the car decreases.

In a head-on collision, the magnitude of the average collision force depends on the change in momentum of the objects involved. According to Newton's third law of motion, the forces exerted on two colliding objects are equal in magnitude and opposite in direction. Therefore, the small car and the truck experience the same average force. The masses and velocities of the vehicles are not necessary to determine this, as the force experienced in a collision is independent of the size or mass of the objects involved.

When the stunt man drops into the car, the total momentum of the system is conserved. The initial momentum of the car is given by its mass (200 kg) multiplied by its initial velocity (10 m/s), which is 2000 kg·m/s. The initial momentum of the stunt man is given by his mass (70 kg) multiplied by his initial velocity (0 m/s), which is 0 kg·m/s. After the stunt man lands in the car, their combined mass is 270 kg. The final velocity of the car can be calculated by dividing the initial momentum by the combined mass, which gives a value of approximately 7.4 m/s.

When the child falls sideways off the sled, there is no external force acting on the sled to change its velocity. According to Newton's first law of motion, an object will continue to move with a constant velocity unless acted upon by an external force. Since there is no friction or other forces acting on the sled, its velocity will remain the same.

The impulse experienced by an object is equal to the change in momentum of the object. The change in momentum can be calculated by multiplying the force acting on the object by the time for which the force acts. In this case, the initial force of 4.0 N acts for 3.0 s, giving an impulse of (4.0 N) * (3.0 s) = 12 N*s. Then, the force suddenly increases to 15 N and acts for an additional 1.0 s, giving an impulse of (15 N) * (1.0 s) = 15 N*s. Adding these two impulses together, we get a total impulse of 12 N*s + 15 N*s = 27 N*s. Therefore, the correct answer is 27 N*s.

The halfback will drive the linebacker backward because the momentum of the halfback is greater than the linebacker. According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since the halfback has a greater mass and a higher velocity, their momentum is greater than the linebacker's momentum. Therefore, the halfback will exert a greater force on the linebacker, causing the linebacker to move backward.

After the elastic collision, the rubber ball transfers all of its momentum to the initially stopped ball. Since the rubber ball has a speed of 5.0 m/s, the initially stopped ball will also have a speed of 5.0 m/s after the collision.

The kinetic energy of an object is given by the formula 1/2mv^2, where m is the mass of the object and v is its velocity. In this case, the sailboat is moving with momentum p, which is equal to mv. By rearranging the equation for momentum, we get v = p/m. Substituting this into the equation for kinetic energy gives (1/2)m(p/m)^2, which simplifies to (p^2)/(2m). Therefore, the correct answer is (p^2)/(2m).

The object's change in velocity can be calculated using the equation F = ma, where F is the net force, m is the mass of the object, and a is the acceleration. Rearranging the equation to solve for a, we have a = F/m. Plugging in the given values, we get a = 9.0 N / 6.0 kg = 1.5 m/s^2. The change in velocity can then be calculated using the equation v = at, where v is the change in velocity, a is the acceleration, and t is the time. Plugging in the values, we get v = 1.5 m/s^2 * 2.0 s = 3.0 m/s. Therefore, the object's change in velocity is 3.0 m/s.

To knock down the door, a force of 1000 N is required. The force can be calculated using the equation F = mv, where m is the mass flow rate of water and v is the velocity of the stream. Rearranging the equation, we get v = F/m. Substituting the given values, v = 1000 N / 40 kg/s = 25 m/s. Therefore, the minimum velocity of the stream needed is 25 m/s.

When the red ball collides head-on with the yellow ball, the momentum of the system is conserved. Since the red ball is moving with a velocity of +3.0 m/s and the yellow ball is moving with a velocity of -2.0 m/s, the total momentum before the collision is (mass of red ball * velocity of red ball) + (mass of yellow ball * velocity of yellow ball) = (mass of red ball * +3.0 m/s) + (mass of yellow ball * -2.0 m/s). Since both balls have equal mass, the total momentum before the collision is 0. After the collision, the momentum should still be 0. Therefore, the velocity of the yellow ball after the collision is also 0 m/s.

When the two cars collide and stick together, the total momentum before the collision is equal to the total momentum after the collision. The momentum of an object is given by the product of its mass and velocity. Before the collision, the momentum of the 1000-kg car is (1000 kg)(25 m/s) = 25000 kg·m/s, and the momentum of the 1500-kg car is (1500 kg)(0 m/s) = 0 kg·m/s. After the collision, the two cars stick together and move with a common velocity. Let's call this velocity V. The total momentum after the collision is (2500 kg)(V) = 2500V kg·m/s. Since the total momentum before and after the collision must be equal, we can equate the two expressions: 25000 kg·m/s = 2500V kg·m/s. Dividing both sides by 2500 kg·m/s, we find that V = 10 m/s. Therefore, the speed of the cars after the collision is 10 m/s.

The center of mass of a two-particle system is determined by the mass and position of each particle. In this case, the center of mass is at the origin, which means that the total mass of the system is evenly distributed around the origin. Since one particle is located at (3.0 m, 0) and has a mass of 2.0 kg, the other particle must be located at (-2.0 m, 0) in order to balance out the system and maintain the center of mass at the origin.

When the pitching machine fires the baseball, according to Newton's third law of motion, there is an equal and opposite reaction. The momentum of the baseball in the horizontal direction is given by the product of its mass and velocity, which is (0.40 kg) * (35 m/s) = 14 kg·m/s. To conserve momentum, the pitching machine must have an equal and opposite momentum in the opposite direction. Since the mass of the pitching machine is 50 kg, the recoil speed can be calculated by dividing the momentum by the mass, which is (14 kg·m/s) / (50 kg) = 0.28 m/s. Therefore, the recoil speed of the pitching machine is 0.28 m/s.

The momentum of an object is given by the product of its mass and velocity. In this case, the mass of the ferryboat is 1200 kg and its velocity is 20 m/s. By multiplying these two values together, we get a momentum of 24,000 kg*m/s.

When the astronauts push each other apart, they exert equal and opposite forces on each other according to Newton's third law. The force experienced by each astronaut is given by F = m*a, where F is the force, m is the mass, and a is the acceleration. Since the force on each astronaut is the same, the acceleration experienced by the lighter astronaut is greater than the acceleration experienced by the heavier astronaut due to the inverse relationship between mass and acceleration. As a result, the lighter astronaut covers a greater distance in the same amount of time, moving 12 m. Therefore, the separation between the two astronauts is the sum of their individual distances, which is 12 m + 12 m = 24 m.

When the bullet collides with the block, momentum is conserved. The initial momentum of the bullet is given by the product of its mass (10.0 g = 0.01 kg) and its initial velocity (300 m/s), which is equal to 3 kg·m/s. Since the bullet emerges with half of its original speed, its final velocity is 150 m/s. Let's assume the velocity of the block after the collision is v. The final momentum of the bullet is given by the product of its mass (0.01 kg) and its final velocity (150 m/s), which is equal to 1.5 kg·m/s. The momentum of the block after the collision is equal to the final momentum of the bullet, so it is also 1.5 kg·m/s. The mass of the block is 1.00 kg, so we can use the formula p = mv to find the velocity of the block. Rearranging the formula, we have v = p/m. Substituting the values, we get v = 1.5 kg·m/s / 1.00 kg = 1.5 m/s. Therefore, the velocity of the block right after the collision is 1.50 m/s.

The total momentum after the collision is 6000 kg*m/s to the right. This can be determined using the principle of conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision. Since the car is moving to the right with a momentum of 6000 kg*m/s and the motorcycle is initially stopped, the total momentum after the collision will be equal to the momentum of the car, which is 6000 kg*m/s to the right.

When you collide with your friend while holding each other, the law of conservation of momentum applies. According to this law, the total momentum before the collision is equal to the total momentum after the collision. Since your friend is initially at rest, their momentum is zero. Therefore, the total momentum before the collision is equal to your momentum, which is given by the equation p = m * v, where p is momentum, m is mass, and v is velocity. After the collision, you both move off together with the same velocity. Using the conservation of momentum equation, (50 kg * 4.0 m/s) + (40 kg * 0 m/s) = (90 kg * v), we can solve for v and find that the velocity of both of you moving off together is 2.2 m/s.

The red ball is subjected to the greater magnitude of impulse during its collision with the floor.

In a perfectly inelastic collision, the two masses stick together and move as one object after the collision. The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.

Before the collision, the 2.0-kg mass has a momentum of (2.0 kg)(5.0 m/s) = 10 kg·m/s. The 3.0-kg mass is at rest, so its momentum is 0 kg·m/s.

After the collision, the masses stick together and move as one object. The total mass is 2.0 kg + 3.0 kg = 5.0 kg. The total momentum is therefore (5.0 kg)(v), where v is the final velocity of the masses.

Using the law of conservation of momentum, we can set up the equation:

10 kg·m/s + 0 kg·m/s = (5.0 kg)(v)

Simplifying the equation, we find:

10 kg·m/s = 5.0 kg·v

Dividing both sides by 5.0 kg, we get:

2.0 m/s = v

Therefore, the speed of the masses after the collision is 2.0 m/s.

The total momentum of the system can be found by adding the individual momenta of the two cars. Since momentum is a vector quantity, we need to consider both the magnitude and the direction. The magnitude of the total momentum is given by the Pythagorean theorem as √((1500 kg * 25 m/s)^2 + (1500 kg * 15 m/s)^2) = 4.4 * 10^4 kg*m/s. The direction of the total momentum can be found using trigonometry. The angle can be determined as arctan((1500 kg * 15 m/s) / (1500 kg * 25 m/s)) = arctan(15/25) = 31°. Since the car moving north has a greater speed, the direction of the total momentum is 31° north of east.

When the astronaut throws the space hammer away from the capsule, according to the law of conservation of momentum, the astronaut will experience an equal and opposite momentum in the opposite direction. Since the mass of the astronaut is much larger than the mass of the space hammer, the astronaut's velocity will be much smaller than that of the hammer. Therefore, the astronaut will move toward the capsule at a speed of 0.40 m/s.

The center of mass of a system of particles is calculated by taking the weighted average of their positions, where the weights are the masses of the particles. In this case, the center of mass can be found by calculating the x-coordinate and y-coordinate separately.

For the x-coordinate, we multiply the x-coordinate of each mass by its mass, and then divide the sum of these products by the total mass of the system. In this case, the x-coordinate of the center of mass is (0 * 2.0 + 2.0 * 2.0 + 2.0 * 4.0) / (2.0 + 2.0 + 4.0) = 1.5.

For the y-coordinate, we follow the same process but using the y-coordinates instead. The y-coordinate of the center of mass is (0 * 2.0 + 0 * 2.0 + 1.0 * 4.0) / (2.0 + 2.0 + 4.0) = 0.50.

Therefore, the coordinates of the center of mass are (1.5, 0.50).

When the two cars couple together, they form a system with a total mass of 65,000 kg. According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since the second car was initially at rest, its momentum is zero. Therefore, the momentum of the system after the collision is equal to the momentum of the first car before the collision. Using the equation for momentum (p = mv), we can calculate the momentum of the first car before the collision as (15,000 kg)(2.0 m/s) = 30,000 kg·m/s. Since the total mass of the system is 65,000 kg, the velocity of the system after the collision is 30,000 kg·m/s divided by 65,000 kg, which is approximately 0.46 m/s.

When water falls, it experiences a downward force due to gravity. This force is equal to the weight of the water, which can be calculated by multiplying the mass of the water by the acceleration due to gravity. In this case, the mass of the water is given as 120 kg per minute, and the acceleration due to gravity is approximately 9.8 m/s^2. Multiplying these values gives a weight of 1176 N. Since the water falls a distance of 3.20 m, the average force it exerts on the ground can be calculated by dividing the weight by the distance, resulting in approximately 15.8 N.

When a cannon fires a cannonball, the momentum of the cannonball and the cannon is conserved. This means that the total momentum of the system (cannonball + cannon) before the firing is equal to the total momentum after the firing. Since momentum is a vector quantity and depends on both mass and velocity, when the cannonball is propelled forward with a certain velocity, the cannon experiences an equal and opposite momentum in the backward direction, causing it to recoil. This conservation of momentum is a fundamental principle in physics.

Both the Ping-Pong ball and the bowling ball experience the same magnitude impulse during the collision. Impulse is equal to the change in momentum, and since momentum is conserved in a collision, the impulse experienced by each object is the same. The fact that the Ping-Pong ball bounces back and the bowling ball moves slowly does not affect the magnitude of the impulse experienced by each object.

Linear momentum is a fundamental principle in physics that states that the total momentum of a system remains constant unless acted upon by an external force. In this scenario, when two objects collide and bounce off each other, the total momentum of the system before the collision is equal to the total momentum after the collision. Therefore, linear momentum is definitely conserved in this situation.

The total momentum before the collision can be calculated by adding the individual momenta of the two objects. The momentum of the first object is calculated by multiplying its mass (3.0 kg) by its velocity (4.0 m/s), resulting in a momentum of 12 kg*m/s. The momentum of the second object is calculated by multiplying its mass (6.0 kg) by its velocity (-2.0 m/s), resulting in a momentum of -12 kg*m/s. Adding these two momenta together gives a total momentum of zero before the collision. According to the law of conservation of momentum, the total momentum after the collision should also be zero. Therefore, the statement "The total momentum both before and after the collision is zero" is correct.

In an inelastic collision, kinetic energy is not conserved because some of it is converted into other forms of energy, such as heat or deformation. In this case, the car of mass m strikes the parked station wagon of mass 2m and the bumpers lock together. Since the bumpers lock together, the two vehicles move as a single unit after the collision. The total mass of the combined system is 3m. The initial kinetic energy is given by (1/2)mv^2, and the final kinetic energy is given by (1/2)(3m)v^2. Therefore, the fraction of initial kinetic energy lost is (1/2)(3m)v^2 / (1/2)mv^2 = 2/3.

When an object's kinetic energy doubles, its momentum does not double. Instead, there is a square root relationship between kinetic energy and momentum. This means that if the kinetic energy is doubled, the momentum will increase by the square root of 2, which is approximately 1.41 times as much. Therefore, the correct answer is 1.41 times as much.

In an elastic collision, both the momentum and kinetic energy are conserved. In this scenario, the white cue ball hits the #5 ball and stops, while the #5 ball moves away with the same velocity as the cue ball had originally. This indicates that the momentum of the system is conserved, as the total momentum before the collision (which is the momentum of the cue ball) is equal to the total momentum after the collision (which is the momentum of the #5 ball). Since the kinetic energy is also conserved, the collision is elastic.

The SI unit of momentum is N*s, which stands for Newton-second. Momentum is defined as the product of an object's mass and its velocity, and it is a vector quantity. The unit N*s represents the change in momentum that occurs when a force of 1 Newton is applied to an object for a duration of 1 second. Therefore, N*s is the correct unit for measuring momentum in the SI system.

In an elastic collision, both momentum and kinetic energy are conserved. This means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Therefore, the statement "Kinetic energy is also conserved" is true.

When a person walks on a log, the person exerts a force on the log in the opposite direction of their motion. According to Newton's third law, the log exerts an equal and opposite force on the person. Since the person and the log are in contact, their accelerations are the same. Using the equation F = ma, where F is the force exerted by the person on the log and m is the mass of the log, we can calculate the acceleration of the log. Then, using the equation v = u + at, where v is the final velocity of the log, u is the initial velocity (0 m/s), a is the acceleration, and t is the time taken, we can find the speed of the log. In this case, the speed of the log is 0.30 m/s.

The area under the curve on a Force versus time graph represents impulse. Impulse is defined as the change in momentum of an object, and it is equal to the force applied to the object multiplied by the time interval over which the force is applied. The area under the curve represents the integral of the force with respect to time, which gives the total impulse exerted on the object. Therefore, the correct answer is impulse.

The final velocity of the 0.10-kg object after the collision is -0.040 m/s. This can be determined using the principle of conservation of momentum. Before the collision, the total momentum of the system is given by the sum of the individual momenta of the two objects. Since the second object is initially at rest, its momentum is zero. Therefore, the total momentum before the collision is equal to the momentum of the first object, which is (0.10 kg)(0.20 m/s) = 0.02 kg*m/s. According to the principle of conservation of momentum, the total momentum after the collision is also equal to 0.02 kg*m/s. Since the second object is still at rest after the collision, the final momentum of the system is equal to the momentum of the first object. Therefore, the final velocity of the 0.10-kg object can be calculated by dividing the final momentum by its mass, which gives (-0.02 kg*m/s) / (0.10 kg) = -0.040 m/s.

The center of mass of an object must lie within the object because it is the average position of all the mass in the object.

When the small object bounces straight back, its momentum changes from 5.0 kg*m/s to -4.0 kg*m/s. The change in momentum is calculated by subtracting the initial momentum from the final momentum. Therefore, the magnitude of the large object's momentum change is equal to the magnitude of the change in momentum of the small object, which is 9.0 kg*m/s.

In a perfectly inelastic collision, the two objects stick together and move as one. The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Since Object A has a momentum of 24 kg*m/s to the east and Object B has a momentum of 10 kg*m/s to the north, the total momentum before the collision is the vector sum of these momenta. Using the Pythagorean theorem, the magnitude of the total momentum before the collision is sqrt((24^2) + (10^2)) = sqrt(576 + 100) = sqrt(676) = 26 kg*m/s. Therefore, the magnitude of their combined momentum after the collision is 26 kg*m/s.