Problem 10
Question
An engine of the orbital maneuvering system (OMS) on a space shuttle exerts a force of \((26,700 \mathrm{N}) \hat{J}\) for 3.90 \(\mathrm{s}\) , exhausting a negligible mass of fuel relative to the \(95,000-\mathrm{kg}\) mass of the shuttle. (a) What is the impulse of the force for this 3.90 s? (b) What is the shuttle's change in momentum from this impulse? (c) What is the shuttle's change in velocity from this impulse? (d) Why can't we find the resulting change in the kinetic energy of the shuttle?
Step-by-Step Solution
Verified Answer
(a) 104,130 Ns; (b) 104,130 Ns; (c) 1.096 m/s; (d) Need initial and final velocities.
1Step 1: Calculate Impulse
Impulse is calculated by multiplying the force with the time interval during which the force acts. The formula for impulse \( J \) is given by \( J = F \, \Delta t \).Given: \[ F = 26,700 \, \mathrm{N} \quad \text{and} \quad \Delta t = 3.90 \, \mathrm{s} \]Thus, \[ J = 26,700 \, \mathrm{N} \times 3.90 \, \mathrm{s} = 104,130 \, \mathrm{Ns} \]
2Step 2: Determine Change in Momentum
The change in momentum of an object is equal to the impulse applied to it, according to the impulse-momentum theorem. Therefore, the shuttle's change in momentum \( \Delta p \) is:\[ \Delta p = J = 104,130 \, \mathrm{Ns} \]
3Step 3: Calculate Change in Velocity
Using the relation between impulse, change in momentum, and the mass of the shuttle, we find the change in velocity. Since \( \Delta p = m \Delta v \), we can solve for \( \Delta v \):Given \( m = 95,000 \, \mathrm{kg} \), then:\[ \Delta v = \frac{\Delta p}{m} = \frac{104,130 \, \mathrm{Ns}}{95,000 \, \mathrm{kg}} = 1.096 \, \mathrm{m/s} \]
4Step 4: Explain Change in Kinetic Energy
To find the change in kinetic energy, we need both the initial and final velocities of the shuttle. Without this information, we cannot calculate the kinetic energy change directly. The impulse only gives us the change in velocity, not the exact initial or final velocities needed for the kinetic energy calculations.
Key Concepts
ImpulseChange in MomentumChange in VelocityKinetic Energy Change
Impulse
Impulse is a concept in physics that measures the effect of a force applied over a certain period of time. It is represented by the symbol \( J \) and is calculated as the product of the force \( F \) and the time interval \( \Delta t \) during which the force is applied. The formula is given by:
\[ J = F \cdot \Delta t \]
In the context of the space shuttle problem, a force of 26,700 N is applied for 3.90 seconds, resulting in an impulse of \( 104,130 \, \text{Ns} \).
\[ J = F \cdot \Delta t \]
In the context of the space shuttle problem, a force of 26,700 N is applied for 3.90 seconds, resulting in an impulse of \( 104,130 \, \text{Ns} \).
- Impulse accounts for both the magnitude of the force and the duration for which it acts.
- It is a vector quantity, having both magnitude and direction.
- Impulse has the same units as momentum (Newton-seconds or \( \text{Ns} \)).
Change in Momentum
The change in momentum of an object is directly related to the impulse applied to it, as described by the impulse-momentum theorem. This theorem states that the impulse exerted on an object is equal to the change in its momentum \( \Delta p \). The formula can be expressed as:
\[ \Delta p = J \]
For the space shuttle, this means that an impulse of \( 104,130 \, \text{Ns} \) results in an identical change in momentum of \( 104,130 \, \text{Ns} \). Momentum, like impulse, is a vector and has both direction and magnitude.
\[ \Delta p = J \]
For the space shuttle, this means that an impulse of \( 104,130 \, \text{Ns} \) results in an identical change in momentum of \( 104,130 \, \text{Ns} \). Momentum, like impulse, is a vector and has both direction and magnitude.
- Momentum is defined as the mass of an object multiplied by its velocity \( p = m \cdot v \).
- A change in momentum can occur by altering either the object's velocity or its mass.
- Since the mass of the shuttle remains constant, any change in momentum comes purely from a change in velocity.
Change in Velocity
The change in velocity is derived from the impulse received by an object, considering its mass remains constant. Using the relationship between impulse and momentum, we calculate change in velocity \( \Delta v \) using the formula:
\[ \Delta v = \frac{\Delta p}{m} \]
For the shuttle, with a mass of 95,000 kg and a change in momentum of 104,130 Ns, the change in velocity is calculated as \( 1.096 \, \text{m/s} \).
\[ \Delta v = \frac{\Delta p}{m} \]
For the shuttle, with a mass of 95,000 kg and a change in momentum of 104,130 Ns, the change in velocity is calculated as \( 1.096 \, \text{m/s} \).
- Velocity changes as a consequence of external forces and impulses acting on an object.
- The equation uses constant mass, focusing solely on how speed varies due to force application.
- This increase in velocity is unidirectional, aligning with the force's direction to influence the object's momentum.
Kinetic Energy Change
Kinetic energy is related to an object's motion, defined as \( KE = \frac{1}{2}mv^2 \). The change in kinetic energy depends on the difference between the object's final and initial velocities. A challenge we face, especially in this exercise, is lacking precise values for these velocities.
This lack of initial velocity impedes a direct calculation of kinetic energy change from the impulse alone:
This lack of initial velocity impedes a direct calculation of kinetic energy change from the impulse alone:
- Kinetic energy change requires both initial and final velocities for exact results.
- Impulse provides the magnitude of velocity change, not the starting point or endpoint values.
- To fully calculate kinetic energy change, additional data on the object's velocity before and after the force's application is necessary.
Other exercises in this chapter
Problem 8
Force of a Baseball Swing. A baseball has mass 0.145 \(\mathrm{kg}\) . (a) If the velocity of a pitched ball has a magnitude of 45.0 \(\mathrm{m} / \mathrm{s}\)
View solution Problem 9
A 0.160 -kg hockey puck is moving on an icy, frictionless, horizontal surface. At \(t=0,\) the puck is moving to the right at 3.00 \(\mathrm{m} / \mathrm{s} .\)
View solution Problem 11
CALC At time \(t=0,\) a 2150 -kg rocket in outer space fires an engine that exerts an increasing force on it in the \(+x\) -direction. This force obeys the equa
View solution Problem 12
A bat strikes a 0.145 -kg baseball. Just before impact, the ball is traveling horizontally to the right at \(50.0 \mathrm{m} / \mathrm{s},\) and it leaves the b
View solution