Train car A is at rest when it is hit by train car B. The two cars, which have the same mass, are stuck together and move off after the collision. How does the final velocity of train cars A and B after the collision compare to the initial velocity of train car B before the collision?
The final velocity is double train car B’s initial velocity.
The final velocity is the same as train car B’s initial velocity.
The final velocity is half of train car B’s initial velocity.
The final velocity is zero since train car B will stop.
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum of an isolated system remains constant before and after a collision.
Momentum (p) is calculated by multiplying an object's mass (m) by its velocity (v): p = m * v.
In this case, since train car A is at rest, its initial momentum is zero (p_initial_A = 0). Train car B has an initial velocity (v_initial_B).
After the collision, the two cars stick together and move off as one unit. Let's denote the final velocity of the combined cars as v_final_AB.
To find the final velocity, we can use the equation for conservation of momentum:
Total initial momentum = Total final momentum
(m_B * v_initial_B) + (m_A * 0) = (m_A + m_B) * v_final_AB
Since mass of train car A and train car B are the same (m_A = m_B), we can simplify the equation:
m_B * v_initial_B = 2m_B * v_final_AB
Dividing both sides of the equation by m_B, we get:
v_initial_B = 2 * v_final_AB
Therefore, the final velocity of train cars A and B after the collision is half of train car B's initial velocity. So, the correct answer is: The final velocity is half of train car B's initial velocity.
The final velocity is the same as train car B’s initial velocity.
The final velocity is the same as train car B’s initial velocity.
WRONG ITS The final velocity is half of train car B’s initial velocity.
Apologies for the mistake. Indeed, the correct answer is that the final velocity is half of train car B's initial velocity. Thank you for pointing that out.
An astronaut on a space walk floats a little too far away from the space station. Without air to push against, he cannot paddle back. However, the astronaut is holding a hammer. Explain how the astronaut could use the hammer to move and why this will get him back to the space station.
The astronaut can use the principle of conservation of momentum to move and get back to the space station.
If the astronaut throws the hammer away from the space station with some force, according to Newton's third law of motion, an equal and opposite force will act on the astronaut. As a result, the astronaut will start moving in the opposite direction, towards the space station.
This phenomenon is based on the principle of conservation of momentum. Initially, the momentum of the astronaut and the hammer is zero because they are both at rest. When the astronaut throws the hammer, the hammer gains momentum in one direction. To ensure the total momentum remains zero, the astronaut gains an equal and opposite momentum in the opposite direction.
Once the astronaut starts moving back towards the space station, they can simply catch the hammer or retrieve it somehow to stop themselves from moving away further. By repeating this process of throwing the hammer and retrieving it, the astronaut can continue to change their momentum and move closer to the space station until they reach it.
It is important to note that in space, there is no significant air resistance or friction to slow down the motion, so the astronaut's movements can continue using this method until they reach the desired destination.
The astronaut can throw the hammer in a direction away from the space station. While he is holding the hammer, the total momentum of the astronaut and hammer is 0 kg • m/s. According to the law of conservation of momentum, the total momentum after he throws the hammer must still be 0 kg • m/s. In order for momentum to be conserved, the astronaut will have to move in the opposite direction of the hammer, which will be toward the space station.
What did you include in your response? Check all that apply.
The astronaut should throw the hammer away from the space station.
The total momentum before and after the astronaut throws the hammer is 0 kg • m/s.
To satisfy the law of conservation of momentum, the astronaut will move in the opposite direction of the hammer.
Based on the response given above, the following statements were included:
- The astronaut should throw the hammer away from the space station.
- The total momentum before and after the astronaut throws the hammer is 0 kg • m/s.
- To satisfy the law of conservation of momentum, the astronaut will move in the opposite direction of the hammer.
Goal posts at the ends of football fields are padded as a safety measure for players who might run into them. How does thick padding around the goal post reduce injuries to players?
Thick padding around the goal posts reduces injuries to players by providing cushioning and reducing the force of impact in case players collide with the posts. Here's how it works:
1. Absorption of Impact: The thick padding around the goal posts is usually made of soft materials like foam or rubber. When players collide with the padded posts, the padding absorbs some of the impact forces. This helps to slow down the players' momentum gradually and reduces the intensity of the collision.
2. Decreased Risk of Head Injuries: Football goal posts often have a crossbar at the top, and players may inadvertently collide with it while jumping or attempting to score. The padded goal posts help to cushion those collisions, reducing the risk of head injuries. The soft padding helps to distribute and dissipate the force over a larger area, reducing the concentration of impact on a specific point.
3. Risk of Cutting and Bruising: Without padding, the goal posts would consist of hard materials, such as metal or wood. Colliding with these hard surfaces could lead to injuries like cuts, bruises, or even bone fractures. The padding acts as a protective barrier, reducing the likelihood of injuries caused by direct contact with the posts.
4. Psychological Confidence: The presence of thick padding around the goal posts can also provide players with psychological confidence. Knowing that there is some level of protection in place can help players focus on the game without being overly concerned about potential injuries from colliding with the goal posts. This psychological assurance can promote better performance and minimize hesitation during crucial moments of the game.
Overall, the purpose of thick padding around the goal posts is to minimize the risk and severity of injuries that can occur when players make contact with the posts. By absorbing impact, reducing the risk of head injuries, preventing cuts and bruises, and providing psychological reassurance, the padding helps ensure the safety of the players on the field.
I got that wrong but I don't think it matters here is a new question: The padding around the goal post increases the time of the collision between the player and the post, which decreases the force exerted to bring the player to a stop.
What did you include in your response? Check all that apply.
The time of the collision increases.
The force on the player decreases.
Based on your new question, the included statements in the response are:
- The time of the collision increases.
- The force on the player decreases.
TEST TIME!! get me all right answers or else.
What is the momentum of a 5 kg object that has a velocity of 1.2 m/s?
3.8 kg • m/s
4.2 kg • m/s
6.0 kg • m/s
6.2 kg • m/s
The correct answer is calculated by multiplying the mass (m) of the object by its velocity (v).
Momentum (p) = mass (m) * velocity (v)
Given:
Mass (m) = 5 kg
Velocity (v) = 1.2 m/s
Momentum (p) = 5 kg * 1.2 m/s
Momentum (p) = 6 kg • m/s
Therefore, the correct answer is 6.0 kg • m/s.
A bowling ball collides with a pin and pushes the pin toward the right. Which describes the reaction force?
The pin pushes the bowling ball toward the right.
The pin pushes the bowling ball toward the left.
The bowling ball pulls the pin toward the right.
The bowling ball pulls the pin toward the left.
According to Newton's third law of motion, for every action, there is an equal and opposite reaction. Therefore, the correct answer is:
The pin pushes the bowling ball toward the left.
In this scenario, when the bowling ball collides with the pin and pushes it toward the right, the pin reacts by exerting an equal and opposite force on the bowling ball, pushing it toward the left.
A car has a momentum of 20,000 kg • m/s. What would the car’s momentum be if its velocity doubles?
10,000 kg • m/s
20,000 kg • m/s
40,000 kg • m/s
80,000 kg • m/s
The momentum of an object is given by the product of its mass and velocity. Mathematically, momentum (p) is calculated as momentum (p) = mass (m) * velocity (v).
Let's assume the initial momentum of the car is 20,000 kg • m/s, and the initial velocity is denoted as v_initial.
We can set up the equation as follows:
p_initial = m * v_initial
Now, let's say the velocity of the car doubles, so the new velocity is 2 * v_initial.
The new momentum, p_final, can be calculated as follows:
p_final = m * (2 * v_initial) = 2 * (m * v_initial)
Since the mass of the car remains the same, we can simplify the equation by factoring out 2 from the product on the right side:
p_final = 2 * p_initial
Therefore, the new momentum (p_final) is twice the initial momentum (p_initial).
In this case, the initial momentum is 20,000 kg • m/s, so doubling it would result in:
p_final = 2 * 20,000 kg • m/s = 40,000 kg • m/s
Hence, the correct answer is 40,000 kg • m/s.
Two carts collide and bounce apart. Cart 1 had a momentum of –6 kg • m/s before the collision. Cart 2 had a momentum of 10 kg • m/s before the collision.
What is the total momentum of the carts after the collision?
–16 kg • m/s
–10 kg • m/s
4 kg • m/s
10 kg • m/s
To find the total momentum of the carts after the collision, we need to add up the momenta of both carts.
Given:
Cart 1 momentum before collision = -6 kg • m/s
Cart 2 momentum before collision = 10 kg • m/s
To find the total momentum after the collision, we simply add the momenta of the two carts:
Total momentum = Cart 1 momentum + Cart 2 momentum
Total momentum = (-6 kg • m/s) + (10 kg • m/s)
Total momentum = 4 kg • m/s
Therefore, the total momentum of the carts after the collision is 4 kg • m/s.
A 50 kg pitcher throws a baseball with a mass of 0.15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is no net force on the system, what is the velocity of the pitcher?
−0.1 m/s
−0.2 m/s
−0.7 m/s
−1.4 m/s
According to the law of conservation of momentum, the total momentum before the pitcher throws the baseball should be equal to the total momentum after the throw.
Given:
Mass of the pitcher (m_p) = 50 kg
Mass of the baseball (m_b) = 0.15 kg
Initial velocity of the baseball (v_b_initial) = 35 m/s
Final velocity of the baseball (v_b_final) = ?
Final velocity of the pitcher (v_p_final) = ?
Using the conservation of momentum, we can write the equation:
(m_p * v_p_final) + (m_b * v_b_final) = (m_p * 0) + (m_b * v_b_initial)
Since there is no net force on the system, the momentum of the pitcher before and after the throw is zero.
(50 kg * v_p_final) + (0.15 kg * v_b_final) = (50 kg * 0) + (0.15 kg * 35 m/s)
Simplifying the equation:
(50 kg * v_p_final) + (0.15 kg * v_b_final) = 0 + 5.25 kg • m/s
As the question asks for the velocity of the pitcher (v_p_final), we can isolate that variable:
50 kg * v_p_final = 5.25 kg • m/s
v_p_final = 5.25 kg • m/s / 50 kg
Calculating this, we get:
v_p_final = 0.105 m/s
Therefore, the velocity of the pitcher after throwing the baseball is approximately 0.105 m/s.
Please note that the answer options listed in the question are all negative, but based on the given information, the velocity is positive.
It's not POSITIVE they are all NEGATIVE - ignore that question here is a new one I better get 90 or 80.
The chart shows masses and velocities of four objects.
A 3-column table with 4 rows titled Motion of 4 Objects. The first column labeled Object has entries W, X, Y, Z. The second column labeled Mass (kilograms) has entries 12, 15, 18, 28. The third column labeled Velocity (meters per second) has entries 5, 8, 2, 10.
Which lists the objects in order, from least to greatest momentum?
W, X, Y, Z
Z, Y, X, W
Z, X, W, Y
Y, W, X, Z
To determine the momentum of each object, we can calculate the product of mass and velocity for each row in the table. The objects will be listed in order from least to greatest momentum.
Given:
Object W: Mass = 12 kg, Velocity = 5 m/s
Object X: Mass = 15 kg, Velocity = 8 m/s
Object Y: Mass = 18 kg, Velocity = 2 m/s
Object Z: Mass = 28 kg, Velocity = 10 m/s
To calculate momentum, we use the formula: Momentum = Mass * Velocity
Momentum of W = 12 kg * 5 m/s = 60 kg • m/s
Momentum of X = 15 kg * 8 m/s = 120 kg • m/s
Momentum of Y = 18 kg * 2 m/s = 36 kg • m/s
Momentum of Z = 28 kg * 10 m/s = 280 kg • m/s
From least to greatest momentum, the correct order is: Y, W, X, Z.
Therefore, the correct answer is: Y, W, X, Z.
The law of conservation of momentum states that, if left alone, what happens to the total momentum of two interacting objects that make up a system?
it increases over time
it remains the same
it eventually becomes zero
changes based on how they collide
The law of conservation of momentum states that the total momentum of two interacting objects that make up a system remains the same if there are no external forces acting on the system. In other words, the total momentum before the interaction is equal to the total momentum after the interaction. Therefore, the correct answer is:
- it remains the same.
The diagram shows two balls before they collide.
2 balls with grey arrows pointing to them from the outside. The left ball has below it m subscript 1 = 0.6 kilograms v subscript 1 = 0.5 meters per second. The right ball has below it m subscript 2 = 0.5 kilograms v subscript 2 = negative 0.2 meters per second.
What is the momentum of the system after the collision?
0.0 kg • m/s
0.2 kg • m/s
0.3 kg • m/s
0.4 kg • m/s
To find the momentum of the system after the collision, we need to add up the individual momenta of the two balls.
Given:
Mass of ball 1 (m₁) = 0.6 kg
Velocity of ball 1 (v₁) = 0.5 m/s
Mass of ball 2 (m₂) = 0.5 kg
Velocity of ball 2 (v₂) = -0.2 m/s
The momentum of each ball is given by the product of its mass and velocity.
Momentum of ball 1 (p₁) = m₁ * v₁ = 0.6 kg * 0.5 m/s = 0.3 kg • m/s
Momentum of ball 2 (p₂) = m₂ * v₂ = 0.5 kg * (-0.2 m/s) = -0.1 kg • m/s
To find the total momentum of the system, we add up the individual momenta:
Total Momentum = Momentum of ball 1 + Momentum of ball 2
Total Momentum = 0.3 kg • m/s + (-0.1 kg • m/s) = 0.2 kg • m/s
Therefore, the momentum of the system after the collision is 0.2 kg • m/s.
Football helmets are made with padding that helps reduce head injuries when a player collides with an object. Which best explains how the padding reduces injuries?
It extends the time of the collision.
It reduces the time of the collision.
It eliminates the reaction force of the object.
It eliminates the action force of the object.
The correct answer is: It extends the time of the collision.
Padding in football helmets helps reduce head injuries by extending the time of the collision. When a player's head collides with an object, such as another player or the ground, the padding in the helmet compresses, which increases the duration of the impact. By increasing the time of the collision, the force and acceleration experienced by the player's head are reduced.
This extension of the collision time allows the force to be distributed over a longer period, reducing the intensity of the impact and lowering the risk of severe head injuries. It helps absorb and dissipate the energy of the collision, providing cushioning and protection to the player's head.
The chart show the masses and velocities of two colliding objects that stick together after a collision.
According to the law of conservation of momentum, what is the momentum of the object after the collision?
4,500 g · m/s
1,750 g · m/s
1,500 kg · m/s
3,000 kg · m/s
HELLO ITS MY LAST QUESTION BOT
The chart show the masses and velocities of two colliding objects that stick together after a collision.
According to the law of conservation of momentum, what is the momentum of the object after the collision?
4,500 g · m/s
1,750 g · m/s
1,500 kg · m/s
3,000 kg · m/s
new user MRRAGE
copy n paste this for answer hehe
The chart show the masses and velocities of two colliding objects that stick together after a collision.
According to the law of conservation of momentum, what is the momentum of the object after the collision?
4,500 g · m/s
1,750 g · m/s
1,500 kg · m/s
3,000 kg · m/s