Sure, let’s look into the detailed, step-by-step solution for question 3 about the linear momentum of a 3 g rocket moving at a constant speed of 3 m/s and at an angle of 100° above the horizontal.
### Step-by-Step Solution:
1. Understand the Given Data:
- Mass of the rocket [tex]\( m = 3 \)[/tex] grams.
- Speed of the rocket [tex]\( v = 3 \)[/tex] meters per second.
- Angle of the rocket's motion relative to the horizontal [tex]\( \theta = 100^\circ \)[/tex].
2. Convert Units:
- The mass needs to be converted from grams to kilograms.
[tex]\[
1 \text{ gram} = 0.001 \text{ kilograms}
\][/tex]
So,
[tex]\[
3 \text{ g} = 3 \times 0.001 \text{ kg} = 0.003 \text{ kg}
\][/tex]
3. Understand the Concept:
- Linear momentum ([tex]\( p \)[/tex]) is the product of the mass ([tex]\( m \)[/tex]) and velocity ([tex]\( v \)[/tex]) of an object.
[tex]\[
p = m \times v
\][/tex]
4. Calculate the Linear Momentum:
[tex]\[
p = 0.003 \text{ kg} \times 3 \text{ m/s}
\][/tex]
[tex]\[
p = 0.009 \text{ kg m/s}
\][/tex]
### Conclusion:
The linear momentum [tex]\( p \)[/tex] of the 3 g rocket moving at a constant speed of 3 m/s is:
[tex]\[
p = 0.009 \text{ kg m/s}
\][/tex]
There you have it. The linear momentum of the rocket is [tex]\( 0.009 \text{ kg m/s} \)[/tex].
