Question

A billiard ball moves with 3 kg⋅m/s of momentum and strikes three other billiard balls. What is the total momentum of the balls after the collision?(1 point)
Responses

less than 3 kg⋅m/s

3 kg⋅m/s

It depends on the mass of each ball.

more than 3 kg⋅m/s

Answer 1

According to the law of conservation of momentum, the total momentum of a closed system remains constant if no external forces act on it. In this case, if a billiard ball with a momentum of 3 kg⋅m/s strikes three other billiard balls, the total momentum of the system after the collision will still equal the initial momentum, provided there are no external forces.

Therefore, the total momentum of the balls after the collision will be:

**3 kg⋅m/s**.

Answer 2

an object has a mass of 5 kg and a velocity of 20 m/s. What is the momentum of the object?
Responses

100 kg⋅m/s

4 kg⋅m/s

0.25 kg⋅m/s

25 kg⋅m/s

Answer 3

Momentum (p) is calculated using the formula:

\[ p = m \times v \]

where:
- \( m \) is the mass of the object,
- \( v \) is the velocity of the object.

For the given object:
- Mass (\( m \)) = 5 kg
- Velocity (\( v \)) = 20 m/s

Substituting the values into the formula:

\[ p = 5 \, \text{kg} \times 20 \, \text{m/s} = 100 \, \text{kg⋅m/s} \]

Thus, the momentum of the object is:

**100 kg⋅m/s**.

Answer 4

Two objects in a closed system collide while moving directly toward each other. Each has a mass of 1.00 kg and is moving at 1.80 m/s. What is the total momentum after the collision?
Responses

0.00 kg·m/s

0.90 kg·m/s

1.80 kg·m/s

3.60 kg·m/s

Answer 5

To determine the total momentum after the collision of two objects in a closed system, we first need to calculate their individual momenta and then sum them up.

**Object 1:**
- Mass \( m_1 = 1.00 \, \text{kg} \)
- Velocity \( v_1 = 1.80 \, \text{m/s} \) (let's consider this positive)

**Object 2:**
- Mass \( m_2 = 1.00 \, \text{kg} \)
- Velocity \( v_2 = -1.80 \, \text{m/s} \) (since it is moving in the opposite direction, we consider this negative)

Now we calculate the total momentum before the collision:

\[
p_{\text{total}} = p_1 + p_2
\]

Calculating \( p_1 \) and \( p_2 \):

\[
p_1 = m_1 \times v_1 = 1.00 \, \text{kg} \times 1.80 \, \text{m/s} = 1.80 \, \text{kg⋅m/s}
\]

\[
p_2 = m_2 \times v_2 = 1.00 \, \text{kg} \times (-1.80 \, \text{m/s}) = -1.80 \, \text{kg⋅m/s}
\]

Now, summing the momenta:

\[
p_{\text{total}} = 1.80 \, \text{kg⋅m/s} + (-1.80 \, \text{kg⋅m/s}) = 0.00 \, \text{kg⋅m/s}
\]

Thus, the total momentum after the collision is:

**0.00 kg·m/s**.

Answer 6

A small rock bounces off the windshield of a moving car and experiences a large change in velocity. How does the car’s momentum change?
Responses:

It does not change.

It has a change in momentum greater than that of the rock, in the opposite direction.

It has a change in momentum equal to that of the rock, in the opposite direction.

It has a much smaller change in momentum than that of the rock, in the opposite direction.

Answer 7

According to the law of conservation of momentum, in an isolated system, when two objects interact, the total momentum before the interaction equals the total momentum after the interaction.

When the small rock bounces off the windshield of the moving car, both the rock and the car experience a change in momentum. The key points are:

1. The **change in momentum** of the rock and the car is related by Newton's third law (for every action, there is an equal and opposite reaction).
2. The force exerted by the rock on the car during the collision will result in an equal and opposite force exerted by the car on the rock.

Since momentum is a vector quantity, the car will experience a change in momentum equal in magnitude to that of the rock but in the opposite direction.

Therefore, the correct response is:

**It has a change in momentum equal to that of the rock, in the opposite direction.**