Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Let's break this question down step-by-step:
### Part 5. a)
i) Momentum:
Momentum is the product of the mass and velocity of an object. It is a vector quantity, meaning it has both magnitude and direction. The mathematical expression for momentum ([tex]\( p \)[/tex]) is given by:
[tex]\[ p = m \times v \][/tex]
where [tex]\( m \)[/tex] is the mass of the object and [tex]\( v \)[/tex] is the velocity.
ii) Impulse:
Impulse is the change in momentum of an object when a force is applied over a period of time. It is also a vector quantity. The relationship between impulse ([tex]\( J \)[/tex]) and the change in momentum can be expressed as:
[tex]\[ J = \Delta p = F \times \Delta t \][/tex]
where [tex]\( F \)[/tex] is the applied force and [tex]\( \Delta t \)[/tex] is the time period over which the force acts.
### Part 5. b)
#### Given Data:
- Mass of the bullet ([tex]\( m_{\text{bullet}} \)[/tex]) = [tex]\( 50 \, \text{g} \)[/tex] = [tex]\( 0.05 \, \text{kg} \)[/tex]
- Velocity of the bullet ([tex]\( v_{\text{bullet}} \)[/tex]) = [tex]\( 200 \, \text{m/s} \)[/tex]
- Mass of the lead block ([tex]\( m_{\text{block}} \)[/tex]) = [tex]\( 950 \, \text{g} \)[/tex] = [tex]\( 0.95 \, \text{kg} \)[/tex]
#### Conservation of Momentum:
Since the lead can move freely, the law of conservation of momentum applies. The total momentum before and after the impact must be equal.
The initial momentum is primarily due to the bullet since the block is initially at rest:
[tex]\[ p_{\text{initial}} = m_{\text{bullet}} \times v_{\text{bullet}} = 0.05 \, \text{kg} \times 200 \, \text{m/s} = 10 \, \text{kg} \cdot \text{m/s} \][/tex]
After the impact, the bullet and block move together with a common velocity ([tex]\( v_{\text{final}} \)[/tex]). The total mass after impact is:
[tex]\[ m_{\text{total}} = m_{\text{bullet}} + m_{\text{block}} = 0.05 \, \text{kg} + 0.95 \, \text{kg} = 1 \, \text{kg} \][/tex]
The final velocity ([tex]\( v_{\text{final}} \)[/tex]) can be calculated by equating the initial and final momentum:
[tex]\[ 10 \, \text{kg} \cdot \text{m/s} = 1 \, \text{kg} \times v_{\text{final}} \][/tex]
[tex]\[ v_{\text{final}} = \frac{10 \, \text{kg} \cdot \text{m/s}}{1 \, \text{kg}} = 10 \, \text{m/s} \][/tex]
#### i) Kinetic Energy After Impact:
The kinetic energy ([tex]\( KE \)[/tex]) of the combined system after impact is given by:
[tex]\[ KE_{\text{after}} = \frac{1}{2} m_{\text{total}} v_{\text{final}}^2 \][/tex]
Substituting the values:
[tex]\[ KE_{\text{after}} = \frac{1}{2} \times 1 \, \text{kg} \times (10 \, \text{m/s})^2 = \frac{1}{2} \times 1 \times 100 = 50 \, \text{J} \][/tex]
Therefore, the kinetic energy after impact is [tex]\( 50 \, \text{J} \)[/tex].
#### ii) Loss of Energy:
To find the loss of energy, we first determine the initial kinetic energy of the bullet:
[tex]\[ KE_{\text{initial}} = \frac{1}{2} m_{\text{bullet}} v_{\text{bullet}}^2 \][/tex]
Substituting the values:
[tex]\[ KE_{\text{initial}} = \frac{1}{2} \times 0.05 \, \text{kg} \times ( 200 \, \text{m/s})^2 = \frac{1}{2} \times 0.05 \times 40000 = 1000 \, \text{J} \][/tex]
The loss of energy ([tex]\( \Delta KE \)[/tex]) is the difference between the initial and final kinetic energy:
[tex]\[ \Delta KE = KE_{\text{initial}} - KE_{\text{after}} = 1000 \, \text{J} - 50 \, \text{J} = 950 \, \text{J} \][/tex]
Therefore, the loss of energy is [tex]\( 950 \, \text{J} \)[/tex].
### Part 5. a)
i) Momentum:
Momentum is the product of the mass and velocity of an object. It is a vector quantity, meaning it has both magnitude and direction. The mathematical expression for momentum ([tex]\( p \)[/tex]) is given by:
[tex]\[ p = m \times v \][/tex]
where [tex]\( m \)[/tex] is the mass of the object and [tex]\( v \)[/tex] is the velocity.
ii) Impulse:
Impulse is the change in momentum of an object when a force is applied over a period of time. It is also a vector quantity. The relationship between impulse ([tex]\( J \)[/tex]) and the change in momentum can be expressed as:
[tex]\[ J = \Delta p = F \times \Delta t \][/tex]
where [tex]\( F \)[/tex] is the applied force and [tex]\( \Delta t \)[/tex] is the time period over which the force acts.
### Part 5. b)
#### Given Data:
- Mass of the bullet ([tex]\( m_{\text{bullet}} \)[/tex]) = [tex]\( 50 \, \text{g} \)[/tex] = [tex]\( 0.05 \, \text{kg} \)[/tex]
- Velocity of the bullet ([tex]\( v_{\text{bullet}} \)[/tex]) = [tex]\( 200 \, \text{m/s} \)[/tex]
- Mass of the lead block ([tex]\( m_{\text{block}} \)[/tex]) = [tex]\( 950 \, \text{g} \)[/tex] = [tex]\( 0.95 \, \text{kg} \)[/tex]
#### Conservation of Momentum:
Since the lead can move freely, the law of conservation of momentum applies. The total momentum before and after the impact must be equal.
The initial momentum is primarily due to the bullet since the block is initially at rest:
[tex]\[ p_{\text{initial}} = m_{\text{bullet}} \times v_{\text{bullet}} = 0.05 \, \text{kg} \times 200 \, \text{m/s} = 10 \, \text{kg} \cdot \text{m/s} \][/tex]
After the impact, the bullet and block move together with a common velocity ([tex]\( v_{\text{final}} \)[/tex]). The total mass after impact is:
[tex]\[ m_{\text{total}} = m_{\text{bullet}} + m_{\text{block}} = 0.05 \, \text{kg} + 0.95 \, \text{kg} = 1 \, \text{kg} \][/tex]
The final velocity ([tex]\( v_{\text{final}} \)[/tex]) can be calculated by equating the initial and final momentum:
[tex]\[ 10 \, \text{kg} \cdot \text{m/s} = 1 \, \text{kg} \times v_{\text{final}} \][/tex]
[tex]\[ v_{\text{final}} = \frac{10 \, \text{kg} \cdot \text{m/s}}{1 \, \text{kg}} = 10 \, \text{m/s} \][/tex]
#### i) Kinetic Energy After Impact:
The kinetic energy ([tex]\( KE \)[/tex]) of the combined system after impact is given by:
[tex]\[ KE_{\text{after}} = \frac{1}{2} m_{\text{total}} v_{\text{final}}^2 \][/tex]
Substituting the values:
[tex]\[ KE_{\text{after}} = \frac{1}{2} \times 1 \, \text{kg} \times (10 \, \text{m/s})^2 = \frac{1}{2} \times 1 \times 100 = 50 \, \text{J} \][/tex]
Therefore, the kinetic energy after impact is [tex]\( 50 \, \text{J} \)[/tex].
#### ii) Loss of Energy:
To find the loss of energy, we first determine the initial kinetic energy of the bullet:
[tex]\[ KE_{\text{initial}} = \frac{1}{2} m_{\text{bullet}} v_{\text{bullet}}^2 \][/tex]
Substituting the values:
[tex]\[ KE_{\text{initial}} = \frac{1}{2} \times 0.05 \, \text{kg} \times ( 200 \, \text{m/s})^2 = \frac{1}{2} \times 0.05 \times 40000 = 1000 \, \text{J} \][/tex]
The loss of energy ([tex]\( \Delta KE \)[/tex]) is the difference between the initial and final kinetic energy:
[tex]\[ \Delta KE = KE_{\text{initial}} - KE_{\text{after}} = 1000 \, \text{J} - 50 \, \text{J} = 950 \, \text{J} \][/tex]
Therefore, the loss of energy is [tex]\( 950 \, \text{J} \)[/tex].
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
