Some students conduct an experiment to prove conservation of momentum they use 2 objects that collide measurements are taken before and after the collision which 2 qualities will the students multiply together before and after the collision

Answer 1

In an experiment to prove the conservation of momentum during a collision between two objects, the students will typically multiply two quantities: mass and velocity.

Before the collision, they will measure the mass and velocity of each object and calculate their momentum using the formula:

\[ \text{Momentum} (p) = \text{mass} (m) \times \text{velocity} (v) \]

They will do the same after the collision to confirm that the total momentum before the collision is equal to the total momentum after the collision:

\[ \text{Total Momentum Before} = m_1 v_1 + m_2 v_2 \] \[ \text{Total Momentum After} = m_1 v_1' + m_2 v_2' \]

where \(m_1\) and \(m_2\) are the masses of the two objects, \(v_1\) and \(v_2\) are their velocities before the collision, and \(v_1'\) and \(v_2'\) are their velocities after the collision. The principle of conservation of momentum states that the total momentum before the collision should equal the total momentum after the collision, provided no external forces are acting on the system.

Answer 2

In a closed system 2 moving bodies collide there are no external forces acting on the system how does the total momentum of the system Before the collision compare with the total momentum of the system after the collision

Answer 3

In a closed system where two moving bodies collide and no external forces are acting on the system, the total momentum of the system before the collision is equal to the total momentum of the system after the collision. This is a consequence of the principle of conservation of momentum.

Mathematically, this can be expressed as:

\[ \text{Total Momentum Before Collision} = \text{Total Momentum After Collision} \]

If we denote the masses and velocities of the two bodies before and after the collision as follows:

• \(m_1\) and \(m_2\) are the masses of the two bodies.
• \(v_1\) and \(v_2\) are their velocities before the collision.
• \(v_1'\) and \(v_2'\) are their velocities after the collision.

Then the conservation of momentum can be stated as:

\[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \]

This equation demonstrates that the total momentum remains constant throughout the interaction as long as the system is closed and no external forces are influencing the motion of the bodies.

Answer 4

As part of an experiment on momentum a billiard ball with a mass of 0.180 kg travels 1.5 in 0.5 seconds what is the momentum of the ball

Answer 5

To find the momentum of the billiard ball, you can use the formula for momentum:

\[ \text{Momentum} (p) = \text{mass} (m) \times \text{velocity} (v) \]

First, we need to calculate the velocity of the billiard ball. Velocity is defined as the displacement (distance traveled) divided by the time taken.

Given:

• Mass of the billiard ball, \(m = 0.180 , \text{kg}\)
• Distance traveled, \(d = 1.5 , \text{m}\)
• Time taken, \(t = 0.5 , \text{s}\)

Now, we can calculate the velocity:

\[ v = \frac{d}{t} = \frac{1.5 , \text{m}}{0.5 , \text{s}} = 3 , \text{m/s} \]

Next, we can calculate the momentum:

\[ p = m \times v = 0.180 , \text{kg} \times 3 , \text{m/s} = 0.540 , \text{kg} \cdot \text{m/s} \]

Therefore, the momentum of the billiard ball is 0.540 kg·m/s.

Answer 6

2 students performance experiment with soccer balls they kick 2 soccer balls so that they collide they measure the total momentum of the 2 soccer balls before and after the collision which statement explains why the total momentum of the soccer balls before and after the collision may not be the same

Answer 7

The total momentum of the two soccer balls before and after the collision may not be the same due to the presence of external forces acting on the system, or due to factors such as:

1. Friction and Air Resistance: If the collision occurs on a surface where friction is present, or if there is significant air resistance, these external forces can affect the motion of the soccer balls, causing a loss of momentum. For instance, friction between the soccer balls and the ground can slow them down after the collision, resulting in a smaller total momentum after the collision compared to before.

2. Inelastic Collision: If the collision is not perfectly elastic (which is often the case in real-life scenarios), some kinetic energy may be converted into other forms of energy (like sound, heat, or deformation of the balls), which can lead to a difference in momentum. Inelastic collisions result in loss of kinetic energy and can change the velocities of the balls in such a way that the total momentum might not equal the initial momentum.

3. Measurement Errors: There could be inaccuracies in the measurements of mass or velocity, which can lead to discrepancies in calculating the total momentum before and after the collision.

In summary, if the total momentum of the soccer balls before and after the collision is not the same, it can be attributed to external forces (like friction or air resistance), the nature of the collision (inelastic), or experimental errors in measurement.

Answer 8

To students of upper cars want to prove conservation of momentum they collide their bumper cars. The 1st bumper car has a mass of 120 kg and was moving with a velocity of 4.0 M/S Before the collision and with the velocity of -2.0 m/s After the collision. Ⅱ bumper car has a mass of 90 kg it was moving at a velocity of -5.0 m/s Before the collision. To prove conservation of momentum what must the velocity of the second bumper car be after the collision

Answer 9

To prove conservation of momentum, we need to ensure that the total momentum of the system before the collision is equal to the total momentum after the collision.

Let's denote the quantities as follows:

• Mass of the first bumper car, \(m_1 = 120 , \text{kg}\)

• Initial velocity of the first bumper car, \(v_1 = 4.0 , \text{m/s}\)

• Final velocity of the first bumper car, \(v_1' = -2.0 , \text{m/s}\)

• Mass of the second bumper car, \(m_2 = 90 , \text{kg}\)

• Initial velocity of the second bumper car, \(v_2 = -5.0 , \text{m/s}\)

• Final velocity of the second bumper car, \(v_2'\) (this is what we need to find)

Step 1: Calculate the total momentum before the collision.

\[ \text{Total Momentum Before} = m_1 v_1 + m_2 v_2 \]

Substituting in the known values:

\[ \text{Total Momentum Before} = (120 , \text{kg} \times 4.0 , \text{m/s}) + (90 , \text{kg} \times -5.0 , \text{m/s}) \]

Calculating each term:

\[ = 480 , \text{kg} \cdot \text{m/s} - 450 , \text{kg} \cdot \text{m/s} = 30 , \text{kg} \cdot \text{m/s} \]

Step 2: Calculate the total momentum after the collision.

\[ \text{Total Momentum After} = m_1 v_1' + m_2 v_2' \]

Substituting in the known values and letting \(v_2'\) be the unknown:

\[ 30 , \text{kg} \cdot \text{m/s} = (120 , \text{kg} \times -2.0 , \text{m/s}) + (90 , \text{kg} \times v_2') \]

Calculating the first term:

\[ 30 , \text{kg} \cdot \text{m/s} = -240 , \text{kg} \cdot \text{m/s} + 90 , \text{kg} \times v_2' \]

Step 3: Solve for \(v_2'\).

Now we can rearrange the equation to isolate \(v_2'\):

\[ 30 + 240 = 90 , v_2' \] \[ 270 = 90 , v_2' \] \[ v_2' = \frac{270}{90} = 3.0 , \text{m/s} \]

Conclusion:

The velocity of the second bumper car after the collision should be 3.0 m/s to satisfy the conservation of momentum.