50. what does the law of conservation of momentum state?
51. how can a roller skater have more momentum than a large truck?
52. why will an egg break when thrown against a wall, but less likely to break when thrown against a sheet?
53. what happens to the momentum of an object when its velocity doubles?
54. what are the differences between elastic and inelastic collisions?
55. when an object falls on the ground, the force of that object depends on what?
unit 4:
chapter 9 - work, power, energy
56. how much work is done by a crane that lowers 3000 n of material a distance of 200 meters?

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Accepted Answer

### Question 50
# Brief Explanations:
The Law of Conservation of Momentum states that in a closed system (a system with no external forces acting on it), the total momentum before an event (like a collision or explosion) is equal to the total momentum after the event. Mathematically, if we have a system of objects with initial momenta $ p_{1i}, p_{2i}, \dots, p_{ni} $ and final momenta $ p_{1f}, p_{2f}, \dots, p_{nf} $, then $ \sum_{i = 1}^{n} p_{i} = \text{constant} $, or more specifically for two - object collisions, $ m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f} $, where $ m $ is mass and $ v $ is velocity.
# Answer:
In a closed system (no external net force), the total momentum before an interaction (e.g., collision, explosion) equals the total momentum after the interaction. Mathematically, for a system of objects, $\sum p_{\text{initial}}=\sum p_{\text{final}}$ (or for two objects, $m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$).

### Question 51
# Brief Explanations:
Momentum is calculated using the formula $ p = mv $, where $ p $ is momentum, $ m $ is mass, and $ v $ is velocity. A large truck has a large mass ($ m $) but a roller - skater has a small mass. However, if the roller - skater's velocity ($ v $) is large enough such that the product $ mv $ (momentum) of the roller - skater is greater than the product $ mv $ of the truck (where the truck's velocity is likely to be small, maybe close to zero if it's at rest), then the roller - skater can have more momentum. For example, if the truck has a mass $ m_t = 10000\space kg $ and velocity $ v_t=0.1\space m/s $, its momentum $ p_t=m_tv_t = 10000\times0.1 = 1000\space kg\cdot m/s $. If a roller - skater has a mass $ m_s = 50\space kg $ and velocity $ v_s = 25\space m/s $, the momentum $ p_s=m_sv_s=50\times25 = 1250\space kg\cdot m/s $, and $ p_s>p_t $.
# Answer:
Momentum $ p = mv $. A roller skater (small $ m $) can have more momentum than a large truck (large $ m $) if the skater’s velocity ($ v $) is sufficiently large so that $ m_{\text{skater}}v_{\text{skater}}>m_{\text{truck}}v_{\text{truck}} $ (e.g., truck is slow/at rest, skater is very fast).

### Question 52
# Brief Explanations:
When the egg hits the wall, the time of impact ($ \Delta t $) is very short. According to the impulse - momentum theorem, $ J = F\Delta t=\Delta p $, where $ J $ is impulse, $ F $ is force, and $ \Delta p $ is the change in momentum. The change in momentum of the egg is the same in both cases (since it comes to rest from its initial velocity). When hitting the sheet, the sheet stretches, increasing the time of impact ($ \Delta t $). Since $ F=\frac{\Delta p}{\Delta t} $, a larger $ \Delta t $ (when hitting the sheet) results in a smaller force $ F $ acting on the egg. A smaller force is less likely to break the egg.
# Answer:
Using the impulse - momentum theorem ($ F\Delta t=\Delta p $): The egg’s $ \Delta p $ is the same in both cases. Hitting the wall gives a short $ \Delta t $, so $ F $ is large (breaks egg). Hitting the sheet stretches it, increasing $ \Delta t $, so $ F $ is small (less likely to break egg).

### Question 53
# Brief Explanations:
The formula for momentum is $ p = mv $. If the velocity $ v $ doubles (new velocity $ v' = 2v $) and the mass $ m $ of the object remains constant (mass is a property of the object and doesn't change with velocity in non - relativistic situations, which is the case for most everyday situations), then the new momentum $ p'=m\times v'=m\times(2v) = 2(mv)=2p $. So the momentum of the object also doubles.
# Answer:
Momentum $ p = mv $. If velocity doubles ($ v' = 2v $) and mass $ m $ is constant, new momentum $ p'=m(2v)=2mv = 2p $. So momentum doubles.

### Question 54
# Brief Explanations:
- **Elastic Collisions**: In an elastic collision, both momentum and kinetic energy are conserved. That is, $ m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f} $ (momentum conservation) and $ \frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2=\frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2 $ (kinetic energy conservation). Examples include collisions between billiard balls (approximately) or particles in atomic - level collisions.
- **Inelastic Collisions**: In an inelastic collision, only momentum is conserved, and kinetic energy is not conserved (some kinetic energy is transformed into other forms of energy like heat, sound, or used to deform the objects). A special case of inelastic collisions is a perfectly inelastic collision where the two objects stick together after the collision, and the formula becomes $ m_1v_{1i}+m_2v_{2i}=(m_1 + m_2)v_f $, where $ v_f $ is the common final velocity. Examples include a car crashing into a tree (the car deforms, kinetic energy is lost) or a bullet embedding in a block of wood.
# Answer:
- **Elastic Collisions**: Momentum ($ \sum p $) and kinetic energy ($ \sum KE $) are conserved. E.g., billiard ball collisions (approx.). Formula: $ m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f} $ and $ \frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2=\frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2 $.
- **Inelastic Collisions**: Only momentum is conserved; kinetic energy is not (lost to heat, deformation, etc.). In a perfectly inelastic collision, objects stick together: $ m_1v_{1i}+m_2v_{2i}=(m_1 + m_2)v_f $. E.g., car - tree collision, bullet embedding in wood.

### Question 55
# Brief Explanations:
When an object falls on the ground, we can use the impulse - momentum theorem $ F\Delta t=\Delta p $. The force $ F $ that the object exerts on the ground (and vice - versa, by Newton's third law) depends on two main factors:
1. The change in momentum ($ \Delta p $) of the object: $ \Delta p=m\Delta v $, where $ m $ is the mass of the object and $ \Delta v $ is the change in velocity (from the velocity just before impact to zero, usually). So a larger mass or a larger velocity before impact will lead to a larger $ \Delta p $.
2. The time of impact ($ \Delta t $): As $ F=\frac{\Delta p}{\Delta t} $, a shorter time of impact (e.g., hitting a hard surface) will lead to a larger force, and a longer time of impact (e.g., hitting a soft surface) will lead to a smaller force. Additionally, the nature of the collision (elastic or inelastic) and the properties of the surface (hardness, elasticity) also affect the force by influencing $ \Delta t $ and $ \Delta p $.
# Answer:
Using $ F\Delta t=\Delta p $ (impulse - momentum), the force depends on: 1. $ \Delta p=m\Delta v $ (mass $ m $, velocity change $ \Delta v $ before impact). 2. Time of impact $ \Delta t $ (shorter $ \Delta t $ → larger $ F $, longer $ \Delta t $ → smaller $ F $). Surface properties (hard/soft) also affect $ \Delta t $.

### Question 56
# Explanation:
## Step 1: Recall the work formula
The formula for work is $ W = Fd\cos\theta $, where $ F $ is the force, $ d $ is the displacement, and $ \theta $ is the angle between the force and the displacement. When the crane lowers the material, the force exerted by the crane on the material is in the upward direction (to support the weight of the material), and the displacement of the material is in the downward direction. So the angle $ \theta $ between the force ($ F $) and the displacement ($ d $) is $ 180^{\circ} $, and $ \cos(180^{\circ})=- 1 $. The magnitude of the force $ F $ is equal to the weight of the material, so $ F = 3000\space N $, and the displacement $ d = 200\space m $.
## Step 2: Calculate the work
Substitute the values into the work formula: $ W=Fd\cos\theta=(3000\space N)\times(200\space m)\times\cos(180^{\circ}) $. Since $ \cos(180^{\circ})=- 1 $, we have $ W = 3000\times200\times(- 1)=- 600000\space J $. The negative sign indicates that the force applied by the crane is opposite to the direction of the displacement (the crane is doing negative work as the gravitational force is doing positive work in this case, but we are calculating the work done by the crane). If we consider the magnitude of the work done (the amount of energy transferred), we can also note that when the force and displacement are in opposite directions ( $ \theta = 180^{\circ} $ ), $ W=-Fd $ (for the force of the crane).
# Answer:
The work done by the crane is $- 600000\space J$ (or $600000\space J$ of negative work). If we consider the magnitude, it is $6\times10^{5}\space J$ of work done by the crane (with the negative sign indicating the force - displacement direction relationship).

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