Answer :
To solve this problem, we need to find the momentum of a 350 kg object, considering the effects of a force applied over time.
1. Understand the Given Values:
- Mass of the object, [tex]\( m = 350 \, \text{kg} \)[/tex]
- Initial velocity of the object, [tex]\( v_i = \frac{15.0 \, \text{m/s}}{3} = 5.0 \, \text{m/s} \)[/tex]
- Force applied, [tex]\( F = 235 \, \text{N} \)[/tex]
- Time for which the force is applied, [tex]\( t = 3 \, \text{seconds} \)[/tex]
2. Find the Final Velocity:
- When a force is applied to an object, it causes an acceleration. This acceleration can be calculated using Newton's second law:
[tex]\[
a = \frac{F}{m} = \frac{235 \, \text{N}}{350 \, \text{kg}} \approx 0.6714 \, \text{m/s}^2
\][/tex]
- The acceleration will change the velocity of the object over the given time. We can use the equation for final velocity:
[tex]\[
v_f = v_i + a \times t = 5.0 \, \text{m/s} + 0.6714 \, \text{m/s}^2 \times 3 \, \text{s} \approx 7.0142 \, \text{m/s}
\][/tex]
3. Calculate Initial and Final Momentum:
- Momentum ([tex]\( p \)[/tex]) is given by the formula:
[tex]\[
p = m \times v
\][/tex]
- Initial Momentum:
[tex]\[
p_i = 350 \, \text{kg} \times 5.0 \, \text{m/s} = 1750.0 \, \text{kg} \cdot \text{m/s}
\][/tex]
- Final Momentum:
[tex]\[
p_f = 350 \, \text{kg} \times 7.0142 \, \text{m/s} \approx 2455.0 \, \text{kg} \cdot \text{m/s}
\][/tex]
Therefore, the initial momentum of the object is [tex]\( 1750.0 \, \text{kg} \cdot \text{m/s} \)[/tex], and the final momentum after the force is applied is [tex]\( 2455.0 \, \text{kg} \cdot \text{m/s} \)[/tex].
1. Understand the Given Values:
- Mass of the object, [tex]\( m = 350 \, \text{kg} \)[/tex]
- Initial velocity of the object, [tex]\( v_i = \frac{15.0 \, \text{m/s}}{3} = 5.0 \, \text{m/s} \)[/tex]
- Force applied, [tex]\( F = 235 \, \text{N} \)[/tex]
- Time for which the force is applied, [tex]\( t = 3 \, \text{seconds} \)[/tex]
2. Find the Final Velocity:
- When a force is applied to an object, it causes an acceleration. This acceleration can be calculated using Newton's second law:
[tex]\[
a = \frac{F}{m} = \frac{235 \, \text{N}}{350 \, \text{kg}} \approx 0.6714 \, \text{m/s}^2
\][/tex]
- The acceleration will change the velocity of the object over the given time. We can use the equation for final velocity:
[tex]\[
v_f = v_i + a \times t = 5.0 \, \text{m/s} + 0.6714 \, \text{m/s}^2 \times 3 \, \text{s} \approx 7.0142 \, \text{m/s}
\][/tex]
3. Calculate Initial and Final Momentum:
- Momentum ([tex]\( p \)[/tex]) is given by the formula:
[tex]\[
p = m \times v
\][/tex]
- Initial Momentum:
[tex]\[
p_i = 350 \, \text{kg} \times 5.0 \, \text{m/s} = 1750.0 \, \text{kg} \cdot \text{m/s}
\][/tex]
- Final Momentum:
[tex]\[
p_f = 350 \, \text{kg} \times 7.0142 \, \text{m/s} \approx 2455.0 \, \text{kg} \cdot \text{m/s}
\][/tex]
Therefore, the initial momentum of the object is [tex]\( 1750.0 \, \text{kg} \cdot \text{m/s} \)[/tex], and the final momentum after the force is applied is [tex]\( 2455.0 \, \text{kg} \cdot \text{m/s} \)[/tex].
