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A projectile experiences a force of \(2.0 \mathrm{kN}\) for a time of \(3.6 \mathrm{~ms}\). What is the magnitude of the impulse it received? [Hint: ms means millisecond.]

Short Answer

Expert verified
The impulse is \( 7.2 \mathrm{~N} \cdot \mathrm{s} \).

Step by step solution

01

Understand the Concept of Impulse

Impulse is the product of force and the time duration over which the force acts. The formula for impulse \( J \) is given by \( J = F \cdot \Delta t \) where \( F \) is the force and \( \Delta t \) is the time duration.
02

Convert Units as Necessary

The force is given as \( 2.0 \mathrm{kN} \), which is equivalent to \( 2000 \mathrm{~N} \), and the time is given as \( 3.6 \mathrm{~ms} \), which must be converted to seconds for consistency in standard units. Thus, \( 3.6 \mathrm{~ms} = 3.6 \times 10^{-3} \mathrm{~s} \).
03

Calculate the Impulse

Substitute the force and the time into the formula: \[ J = 2000 \mathrm{~N} \times 3.6 \times 10^{-3} \mathrm{~s} = 7.2 \mathrm{~N} \cdot \mathrm{s}. \]
04

Conclusion

The magnitude of the impulse received by the projectile is \( 7.2 \mathrm{~N} \cdot \mathrm{s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Force and Time Relationship
The concept of impulse revolves around the relationship between force and time. In physics, when a force is applied to an object, the impact it makes is not solely dependent on the force's magnitude. It is also about how long the force acts. This force-time connection is crucial in determining the impulse an object receives. Impulse, denoted by the symbol \( J \), is calculated as the product of force \( F \) and the time period \( \Delta t \) over which the force is exerted:
  • \( J = F \times \Delta t \)
Imagine pushing a door gently; a small force over a long time might result in the same effect as a large push for a short time. This is similar to how impulses work. The longer or stronger the force applied, the greater the impulse.
Unit Conversion
When dealing with physics equations, consistency in units is vital. In the impulse calculation, both force and time need to be in standard units:
  • Force: typically measured in Newtons (N)
  • Time: usually measured in seconds (s)
For our exercise:- The force is given as \( 2.0 \text{ kN} \). You convert kilonewtons into newtons since \( 1 \text{ kN} = 1000 \text{ N} \). Therefore, \( 2.0 \text{ kN} = 2000 \text{ N} \).
- Time is provided in milliseconds (ms). There are 1000 milliseconds in a second, hence, \( 3.6 \text{ ms} = 3.6 \times 10^{-3} \text{ s} \).
This method of unit conversion ensures that your calculations are correct and result in meaningful, standardized answers.
Impulse Calculation
After understanding the relationship and ensuring correct unit conversion, the next step is calculating the impulse. With the figures obtained:
  • Force \( F = 2000 \text{ N} \)
  • Time \( \Delta t = 3.6 \times 10^{-3} \text{ s} \)
Substitute these values into the impulse formula:\[ J = 2000 \text{ N} \times 3.6 \times 10^{-3} \text{ s} = 7.2 \text{ N} \cdot \text{s} \]The impulse is a measure of how much momentum is transferred to an object, and in this case, the calculation shows that the projectile received an impulse of \( 7.2 \text{ N} \cdot \text{s} \). Impulse can change an object's velocity, demonstrating the effective force applied over time.
Projectile Motion
Projectile motion is typically associated with an object launched into motion, subject to gravity, and often ignored in horizontal movement. When calculating impulse on a projectile, it essentially influences its velocity, either adding to it or reducing it, depending on the direction of the force. The provided scenario doesn't include gravity or air resistance, but focuses merely on the force applied and the corresponding impulse. In real-world applications, consider:
  • The initial velocity, which changes due to impulse.
  • The effect of air resistance affecting the projectile's path.
  • Gravity, primarily affecting vertical motion.
By analyzing these additional forces, one can predict the projectile's motion more accurately. In our study case though, only force and time were focused, emphasizing impulse’s direct effect on motion.

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Most popular questions from this chapter

A ball of mass \(m\) at rest at the coordinate origin explodes into three equal pieces. At some instant, one piece is on the \(x\) -axis at \(x=40\) \(\mathrm{cm}\) and another is at \(x=20 \mathrm{~cm}, y=-60 \mathrm{~cm}\). Where is the third piece at that instant?

Imagine that a \(1.20-\mathrm{kg}\) hard-rubber ball traveling at \(10.0 \mathrm{~m} / \mathrm{s}\) bounces off a brick wall in an essentially elastic collision. Determine the change in the momentum of the ball. [Hint: What change in momentum will just stop the ball?]

What force is exerted on a stationary flat plate held perpendicular to a jet of water as shown in Fig. 8-6? The horizontal speed of the water is \(80 \mathrm{~cm} / \mathrm{s}\), and \(30 \mathrm{~mL}\) of the water hit the plate each second. Assume the water moves parallel to the plate after striking it. One milliliter (mL) of water has a mass of \(1.00 \mathrm{~g}\). This question deals with speed, mass, time, and force, and that suggests impulse-momentum and Newton's Second Law. The plate exerts an impulse on the water and changes its horizontal momentum. The water exerts a counterforce on the plate. Taking the direction to the right as positive, (Impulse) \(_{x}=\) Change in \(x\) -directed momentum $$ F_{x} \Delta t=\left(m v_{x}\right)_{\text {final }}-\left(m v_{x}\right)_{\text {initial }} $$ Let \(t\) be \(1.00 \mathrm{~s}\) so that \(m\) will be the mass that strikes in \(1.00 \mathrm{~s}\), namely \(30 \mathrm{~g}\). Then the above equation becomes $$ F_{x}(1.00 \mathrm{~s})=(0.030 \mathrm{~kg})(0 \mathrm{~m} / \mathrm{s})-(0.030 \mathrm{~kg})(0.80 \mathrm{~m} / \mathrm{s}) $$ from which \(F_{x}=-0.024 \mathrm{~N}\). This is the force exerted by the plate on the water. The law of action and reaction tells us that the jet exerts an equal but opposite force on the plate.

Four masses are positioned in the \(x y\) -plane as follows: \(300 \mathrm{~g}\) at \((x=\) \(0, y=2.0 \mathrm{~m}), 500 \mathrm{~g}\) at \((-20 \mathrm{~m},-3.0 \mathrm{~m}), 700 \mathrm{~g}\) at \((50 \mathrm{~cm}, 30 \mathrm{~cm})\) and \(900 \mathrm{~g}\) at \((-80 \mathrm{~cm}, 150 \mathrm{~cm})\). Find their center of mass.

A 90 -g ball moving at \(100 \mathrm{~cm} / \mathrm{s}\) collides head-on with a stationary 10-g ball. Determine the speed of each after impact if \((a)\) they stick together, \((b)\) the collision is perfectly elastic, \((c)\) the coefficient of restitution is \(0.90\).

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