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Chapter 7: Problem 28

A projectile (mass \(=0.20 \mathrm{~kg}\) ) is fired at and embeds itself in a target (mass \(=2.50 \mathrm{~kg}\) ). The target (with the projectile in it) flies off after being struck. What percentage of the projectile's incident kinetic energy does the target (with the projectile in it) carry off after being struck?

Short Answer

Expert verified
The target carries off 7.4% of the projectile's initial kinetic energy.

Step by step solution

01

Identify the Initial and Final Systems

Initially, we have a projectile of mass \(m_1 = 0.20 \text{ kg}\) and a stationary target of mass \(m_2 = 2.50 \text{ kg}\). After the collision, the target and the projectile move together as a single mass of \(m_1 + m_2\).
02

Calculate the Initial Kinetic Energy

The initial kinetic energy is due only to the projectile, since the target is stationary. The kinetic energy of the projectile is:\[ KE_i = \frac{1}{2} m_1 v_1^2 \]where \( v_1 \) is the speed of the projectile before impact.
03

Apply Conservation of Momentum

Since no external forces act on the system, momentum is conserved. Thus, the initial momentum (just the projectile) equals the final momentum (the projectile plus target):\[ m_1 v_1 = (m_1 + m_2) v_f \]Solve this equation for \(v_f\), the final velocity of the combined system.
04

Calculate the Final Kinetic Energy

The final kinetic energy is:\[ KE_f = \frac{1}{2} (m_1 + m_2) v_f^2 \]Substitute \(v_f\) from Step 3 into this formula to get the expression for \(KE_f\) in terms of \(m_1\), \(m_2\), and \(v_1\).
05

Find the Percentage of Initial Kinetic Energy Carried Off

Use the formula for percentage to compare the initial and final kinetic energies:\[ \text{Percentage} = \left(\frac{KE_f}{KE_i}\right) \times 100\% \]Substitute the expressions for \(KE_f\) and \(KE_i\) into the formula.

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Key Concepts

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

Projectile Motion
Projectile motion refers to the path that an object follows when it is thrown or propelled into the air, influenced by gravity. This motion is typically two-dimensional, involving both horizontal and vertical components. Understanding projectile motion involves:
  • Initial Velocity: The speed at which the projectile is launched.
  • Angle of Launch: The angle relative to the ground at which the projectile is fired.
  • Air Resistance: While often neglected in simple calculations, it can impact the projectile's path.
  • Acceleration Due to Gravity: The constant downwards force acting on the projectile.
Although the projectile in this problem embeds itself in the target, initially, it travels in a manner similar to typical projectile motion. The initial speed is essential for calculating kinetic energy, as the motion is partly horizontal before the collision.
Understanding projectile motion can help in figuring out the velocity components necessary for solving collision-type problems like the one in the exercise.
Kinetic Energy
Kinetic energy is the energy of motion. For an object of mass \(m\) traveling at speed \(v\), it can be calculated using the formula:\[ KE = \frac{1}{2} mv^2 \]The kinetic energy before and after the collision in this exercise is crucial because it allows us to compare how much energy is retained in different states:
  • Initial Kinetic Energy: This is only due to the projectile since the target is stationary before the collision.
  • Final Kinetic Energy: After the collision, the projectile and target move together. This energy is shared.

The objective in this problem is to determine what fraction of the projectile's initial kinetic energy is retained as kinetic energy in the combined mass of the target and projectile. This demonstrates energy conservation by showing loss or retention.
Elastic Collision
An elastic collision means no kinetic energy is lost during the interaction. However, the problem we have solved involves an inelastic collision, where the projectile embeds in the target.

While not elastic, it is vital to note:
  • Inelastic Collisions: These often result in objects sticking together, with some kinetic energy converted to other forms of energy (e.g., heat).
  • Momentum Conservation: Even in inelastic collisions, momentum before the collision equals momentum after.
  • Energy Transformation: Some kinetic energy becomes internal energy due to deformation or heat.
Understanding elastic collisions helps highlight the differences between that scenario and our inelastic collision problem. In such exercises, observing where energy "leaks" to other forms is essential to grasp the inefficiencies of the interaction.

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Most popular questions from this chapter

An 85 -kg jogger is heading due east at a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). A \(55-\mathrm{kg}\) jogger is heading \(32^{\circ}\) north of east at a speed of \(3.0 \mathrm{~m} / \mathrm{s}\). Find the magnitude and direction of the sum of the momenta of the two joggers.

Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions (a) John has a larger mass than Barbara has. He is standing on the \(x\) axis at \(x_{\mathrm{J}}=+9.0 \mathrm{~m},\) while she is standing on the \(x\) axis at \(x_{\mathrm{B}}=+2.0 \mathrm{~m} .\) Is their centerof-mass point closer to the 9.0 -m point or the 2.0 -m point? (b) They switch positions. Is their center- of-mass point now closer to the 9.0 -m point or the 2.0 -m point? (c) In which direction, toward or away from the origin, does their center of mass move as a result of the switch? Problem John's mass is \(86 \mathrm{~kg}\), and Barbara's is \(55 \mathrm{~kg}\). How far and in which direction does their center of mass move as a result of the switch? Verify that your answer is consistent with your answers to the Concept Questions.

An \(85-\mathrm{kg}\) jogger is heading due east at a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). A \(55-\mathrm{kg}\) jogger is heading \(32^{\circ}\) north of east at a speed of \(3.0 \mathrm{~m} / \mathrm{s}\). Find the magnitude and direction of the sum of the momenta of the two joggers.

Object \(\mathrm{A}\) is moving due east, while object \(\mathrm{B}\) is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. (a) Is it possible that the two-object system has a final total momentum of zero after the collision? (b) Roughly, what is the direction of the final total momentum of the two-object system after the collision?

When jumping straight down, you can be seriously injured if you land stiff- legged One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 75 -kg man just before contact with the ground has a speed of \(6.4 \mathrm{~m} / \mathrm{s}\). (a) In a stiff-legged landing he comes to a halt in \(2.0 \mathrm{~ms}\). Find the average net force that acts on him during this time. (b) When he bends his knees, he comes to a halt in \(0.10 \mathrm{~s}\). Find the average net force now. (c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of the forces, find the force of the ground on the man in parts (a) and (b).
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