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Chapter 9: Problem 42

(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height \(h\) of the pendulum equal to \(2.6 \mathrm{~cm} .\) A second projectile (of the same mass) causes the the pendulum to swing twice as high, \(h_{2}=5.2 \mathrm{~cm} .\) The second projectile was how many times faster than the first?

Short Answer

Expert verified
The second projectile was \( \sqrt{2} \) or about 1.414 times faster than the first.

Step by step solution

01

Understanding the scenario

In a ballistic pendulum experiment, a projectile hits a pendulum and causes it to rise to a certain height. The height attained by the pendulum is related to the kinetic energy transferred by the projectile.
02

Relating potential energy to velocity

When the pendulum reaches a maximum height, its kinetic energy has been fully converted into gravitational potential energy. The potential energy at height \( h \) is given by \( mgh \), where \( m \) is the mass and \( g \) is the acceleration due to gravity. This potential energy originally came from the projectile's kinetic energy.
03

Establishing the velocity-energy relationship

The initial kinetic energy \( KE \) is \( \frac{1}{2}mv^2 \). At maximum height \( h \), this kinetic energy is equal to the gravitational potential energy \( mgh \). Therefore, \( \frac{1}{2}mv^2 = mgh \). Simplifying gives \( v^2 = 2gh \), which leads to \( v = \sqrt{2gh} \).
04

Comparing the velocities

For the first projectile, the velocity \( v_1 = \sqrt{2gh_1} \). For the second, \( v_2 = \sqrt{2gh_2} \). To compare the velocities, we find the ratio \( \frac{v_2}{v_1} \).
05

Calculating the velocity ratio

Plug in the heights \( h_1 = 0.026 \) meters and \( h_2 = 0.052 \) meters into the velocity formula from Step 3:\[ \frac{v_2}{v_1} = \sqrt{\frac{2g \cdot 0.052}{2g \cdot 0.026}} = \sqrt{\frac{0.052}{0.026}} = \sqrt{2}. \]
06

Concluding the comparison

The second projectile was \( \sqrt{2} \) times faster than the first projectile, which is approximately 1.414 times faster.

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Key Concepts

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

Projectile Motion
Projectile motion deals with objects thrown or launched near the Earth's surface. These objects move along a curved path under the influence of gravity without additional propulsion. In a ballistic pendulum experiment, we observe projectile motion initially as the projectile travels forward and hits the pendulum. This contact is temporary, but critical, as it transfers energy to the pendulum.

Key characteristics of projectile motion include:
  • Horizontal and vertical motion components, which occur simultaneously.
  • The influence of gravity, which only affects the vertical motion.
  • A parabolic trajectory for the projectile, due to the combination of linear motion and gravitational pull.
Understanding projectile motion helps us predict the projectile's motion characteristics, like velocity at impact. Here, the motion of the first projectile led to the pendulum reaching a height that reflects the energy transferred during this short-lived, swift interaction. As the projectile transfers its energy, we shift focus from its path to how this energy affects the pendulum.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy possessed by an object due to its position in a gravitational field. In the context of the ballistic pendulum, as the pendulum swings upward after being hit, it gains gravitational potential energy. This energy state change signifies the conversion from kinetic energy of the projectile to potential energy stored at the pendulum's highest point.

The formula used to calculate GPE is:
  • \[GPE = mgh\]where:
    • \(m\) is the mass of the object (pendulum in this case),
    • \(g\) is the acceleration due to gravity, approximately \(9.8 \, \text{m/s}^2\),
    • \(h\) is the height of the object above the baseline position.
In the ballistic pendulum scenario, the height \(h\) is crucial because it correlates directly with the speed of the projectile that caused the pendulum to rise. A double height implies a greater energy input, leading us to solve how much faster the second projectile traveled compared to the first.
Energy Conservation
Energy conservation is a principle stating that the total energy of an isolated system remains constant. It plays a pivotal role in understanding the ballistic pendulum problem. The kinetic energy of the incoming projectile is initially transferred to the pendulum. As the pendulum rises, this kinetic energy is gradually transformed into gravitational potential energy.

In a mathematical expression, this principle can be shown as:
  • \[\frac{1}{2}mv^2 = mgh\]Here, all terms on both sides signify energy, ensuring the total energy remains unchanged during the pendulum's motion.
  • Initially, the kinetic energy \(\frac{1}{2}mv^2\) signifies the energy due to the motion of the projectile.
  • As the pendulum reaches its peak height \(h\), this energy is fully converted into potential energy \(mgh\).
The exercise demonstrates energy conservation by comparing the height achieved by the pendulum after being struck by two different projectiles. By applying conservation laws, we calculated that the second projectile was \(\sqrt{2}\) times faster, showing how energy can seamlessly convert across different forms while adhering to conservation principles.

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Most popular questions from this chapter

(II) A radioactive nucleus at rest decays into a second nucleus, an electron, and a neutrino. The electron and neutrino are emitted at right angles and have momenta of \(9.6 \times 10^{-23} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\) and \(6.2 \times 10^{-23} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s},\) respectively. Determine the magnitude and the direction of the momentum of the second (recoiling) nucleus.

(III) ( \(a\) ) Calculate the impulse experienced when a \(65-\mathrm{kg}\) person lands on firm ground after jumping from a height of \(3.0 \mathrm{~m} .\) (b) Estimate the average force exerted on the person's feet by the ground if the landing is stiff-legged, and again \((c)\) with bent legs. With stiff legs, assume the body moves \(1.0 \mathrm{~cm}\) during impact, and when the legs are bent, about \(50 \mathrm{~cm} .\) [Hint: The average net force on her which is related to impulse, is the vector sum of gravity and the force exerted by the ground.

(III) A particle of mass \(m_{\mathrm{A}}\) traveling with speed \(v_{\mathrm{A}}\) collides elastically head-on with a stationary particle of smaller mass \(m_{\mathrm{B}} \cdot(a)\) Show that the speed of \(m_{\mathrm{B}}\) after the collision is $$ v_{\mathrm{B}}^{\prime}=\frac{2 v_{\mathrm{A}}}{1+m_{\mathrm{B}} / m_{\mathrm{A}}} $$ (b) Consider now a third particle of mass \(m_{\mathrm{C}}\) at rest between \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}}\) so that \(m_{\mathrm{A}}\) first collides head on with \(m_{\mathrm{C}}\) and then \(m_{\mathrm{C}}\) collides head on with \(m_{\mathrm{B}}\). Both collisions are elastic. Show that in this case, $$ v_{\mathrm{B}}^{\prime}=4 v_{\mathrm{A}} \frac{m_{\mathrm{C}} m_{\mathrm{A}}}{\left(m_{\mathrm{C}}+m_{\mathrm{A}}\right)\left(m_{\mathrm{B}}+m_{\mathrm{C}}\right)} $$ (c) From the result of part (b), show that for $$ \operatorname{maximum} v_{\mathrm{B}}^{\prime}, m_{\mathrm{C}}=\sqrt{m_{\mathrm{A}} m_{\mathrm{B}}} $$ \((d)\) Assume \(m_{\mathrm{B}}=2.0 \mathrm{~kg}\) \(m_{\mathrm{A}}=18.0 \mathrm{~kg}\) and \(v_{\mathrm{A}}=2.0 \mathrm{~m} / \mathrm{s}\). Use a spreadsheet to calculate and graph the values of \(v_{\mathrm{B}}^{\prime}\) from \(m_{\mathrm{C}}=0.0 \mathrm{~kg}\) to \(m_{\mathrm{C}}=50.0 \mathrm{~kg}\) in steps of \(1.0 \mathrm{~kg} .\) For what value of \(m_{\mathrm{C}}\) is the value of \(v_{\mathrm{B}}^{\prime}\) maximum? Does your numerical result agree with your result in part \((c) ?\)

(II) Car A hits car B (initially at rest and of equal mass) from behind while going \(35 \mathrm{~m} / \mathrm{s}\). Immediately after the collision, car B moves forward at \(25 \mathrm{~m} / \mathrm{s}\) and car \(\mathrm{A}\) is at rest. What fraction of the initial kinetic energy is lost in the collision?

Suppose two asteroids strike head on. Asteroid A \(\left(m_{\mathrm{A}}=7.5 \times 10^{12} \mathrm{~kg}\right)\) has velocity \(3.3 \mathrm{~km} / \mathrm{s}\) before the collision, and asteroid B \(\left(m_{\mathrm{B}}=1.45 \times 10^{13} \mathrm{~kg}\right)\) has velocity \(1.4 \mathrm{~km} / \mathrm{s}\) before the collision in the opposite direction. If the asteroids stick together, what is the velocity (magnitude and direction) of the new asteroid after the collision?
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