A child's spring gun when fired vertically sends the ball to a height h, starting with a launch velocity. The same gun is used on the moon and the ball starts with a velocity v' and used to height h'. Then

Options

A. `h gt h', v' gt v`
B. `h' gt h, v' gt v`
C. `h'=h, v'lt v`
D. `h'=h,v'=v`

Answer

To solve the problem, we need to analyze the situation of a child's spring gun that launches a ball to a certain height on Earth and then on the Moon. We will use the principles of energy conservation to derive the relationships between the heights and velocities.

### Step 1: Understand the Energy Conservation Principle
When the spring gun is fired, the potential energy stored in the spring is converted into kinetic energy and gravitational potential energy as the ball rises.

### Step 2: Write the Energy Equation for Earth
On Earth, when the spring gun launches the ball, the initial energy (E_initial) is given by the potential energy of the spring:
\[ E_{\text{initial}} = \frac{1}{2} k x^2 \]
As the ball rises to height \( h \), the energy is converted to gravitational potential energy:
\[ E_{\text{final}} = mgh \]
By conservation of energy:
\[ \frac{1}{2} k x^2 = mgh \]

### Step 3: Write the Energy Equation for the Moon
On the Moon, the same spring gun launches the ball to height \( h' \) with a different launch velocity \( v' \). The energy conservation equation becomes:
\[ \frac{1}{2} k x^2 = mg'h' \]
where \( g' \) is the acceleration due to gravity on the Moon.

### Step 4: Relate the Two Equations
From the two equations, we have:
1. \( \frac{1}{2} k x^2 = mgh \) (for Earth)
2. \( \frac{1}{2} k x^2 = mg'h' \) (for Moon)

Setting them equal to each other gives:
\[ mgh = mg'h' \]

### Step 5: Cancel Mass and Rearrange
Since mass \( m \) is constant and can be canceled out, we get:
\[ gh = g'h' \]
Rearranging gives us:
\[ \frac{h'}{h} = \frac{g}{g'} \]

### Step 6: Analyze the Gravitational Acceleration
We know that \( g' < g \) (the gravitational acceleration on the Moon is less than that on Earth). Therefore:
\[ \frac{g}{g'} > 1 \]
This implies:
\[ h' > h \]

### Step 7: Analyze the Velocities
Using the kinetic energy at launch, we can also relate the velocities. The kinetic energy at launch for Earth is:
\[ \frac{1}{2} mv^2 = mgh \]
For the Moon:
\[ \frac{1}{2} mv'^2 = mg'h' \]

By conservation of energy, we can derive:
\[ v^2 = 2gh \]
\[ v'^2 = 2g'h' \]

### Step 8: Relate the Velocities
From the height relation we derived earlier:
\[ h' = \frac{g}{g'}h \]
Substituting this into the velocity equation gives:
\[ v'^2 = 2g' \left(\frac{g}{g'}h\right) = 2gh \]
Thus:
\[ v'^2 = \frac{g'}{g}v^2 \]
Since \( g' < g \), it follows that:
\[ v' > v \]

### Conclusion
From our analysis, we conclude that:
- The height on the Moon \( h' \) is greater than the height on Earth \( h \) (i.e., \( h' > h \)).
- The launch velocity on the Moon \( v' \) is greater than the launch velocity on Earth \( v \) (i.e., \( v' > v \)).

### Final Result
Thus, the correct relationships are:
- \( h' > h \)
- \( v' > v \)