To solve the problem, we need to find the ratio of the final momentum \( P \) when a force of 1 N acts on a 1 kg mass at rest for 1 second, to the final momentum \( P_1 \) when the same force acts on the mass through a distance of 1 m.
### Step 1: Calculate \( P \)
Using the formula for change in momentum:
\[
\Delta P = F \cdot t
\]
Given:
- Force \( F = 1 \, \text{N} \)
- Time \( t = 1 \, \text{s} \)
Substituting the values:
\[
\Delta P = 1 \, \text{N} \cdot 1 \, \text{s} = 1 \, \text{kg m/s}
\]
Thus, the final momentum \( P \) is:
\[
P = 1 \, \text{kg m/s}
\]
### Step 2: Calculate \( P_1 \)
Now, we need to calculate the final momentum \( P_1 \) when the same force acts through a distance of 1 m. We can use the work-energy principle, which states that the work done on an object is equal to the change in kinetic energy.
The work done \( W \) is given by:
\[
W = F \cdot s
\]
Where:
- \( F = 1 \, \text{N} \)
- \( s = 1 \, \text{m} \)
Substituting the values:
\[
W = 1 \, \text{N} \cdot 1 \, \text{m} = 1 \, \text{J}
\]
Since the object starts from rest, the work done is equal to the kinetic energy gained:
\[
\frac{1}{2} mv^2 = W
\]
Substituting \( W = 1 \, \text{J} \) and \( m = 1 \, \text{kg} \):
\[
\frac{1}{2} \cdot 1 \cdot v^2 = 1
\]
Solving for \( v^2 \):
\[
v^2 = 2 \, \text{m}^2/\text{s}^2
\]
Thus, the final velocity \( v \) is:
\[
v = \sqrt{2} \, \text{m/s}
\]
Now, we can find the final momentum \( P_1 \):
\[
P_1 = mv = 1 \cdot \sqrt{2} = \sqrt{2} \, \text{kg m/s}
\]
### Step 3: Calculate the ratio \( \frac{P}{P_1} \)
Now, we can find the ratio of \( P \) to \( P_1 \):
\[
\frac{P}{P_1} = \frac{1 \, \text{kg m/s}}{\sqrt{2} \, \text{kg m/s}} = \frac{1}{\sqrt{2}}
\]
### Step 4: Simplify the ratio
To express the ratio in a more understandable form:
\[
\frac{P}{P_1} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}
\]
Thus, the ratio \( P : P_1 \) can be expressed as:
\[
1 : \sqrt{2}
\]
### Final Answer
The ratio of \( P \) to \( P_1 \) is \( 1 : \sqrt{2} \).
