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Chapter 9: Problem 118

Two spheres hang as pendulums side-by-side. Sphere 1 of mass \(50 \mathrm{~g}\) is then moved along a curved path to height of \(h_{1}=9.0 \mathrm{~cm}\) and then released from rest. It swings down to collide elastically with sphere 2 of mass \(85 \mathrm{~g}\). What height is reached by (a) sphere 1 and (b) sphere \(2 ?\) After the next (elastic) collision, what height is reached by (c) sphere 1 and (d) sphere \(2 ?\)

Short Answer

Expert verified
Sphere 1 reaches 0.7 cm, and sphere 2 reaches 4.67 cm after the first collision. After the second collision: Sphere 1 reaches 4.67 cm, sphere 2 0.7 cm.

Step by step solution

01

Initial Energy of Sphere 1

We begin by determining the potential energy of sphere 1 before it is released. Use the formula for gravitational potential energy: \[ U = mgh \]where \(m = 50\ \text{g} = 0.05\ \text{kg}\), \(g = 9.8\ \text{m/s}^2\), and \(h_1 = 0.09\ \text{m}\). Calculating this gives:\[ U = 0.05 \times 9.8 \times 0.09 = 0.0441\ \text{J} \]
02

Conversion of Energy and Speed before Collision

Since the collision is elastic, mechanical energy is conserved. The potential energy converts to kinetic energy right before the collision:\[ KE = \frac{1}{2}mv^2 = 0.0441\ \text{J} \]Solving for \(v\), the velocity just before impact:\[ v = \sqrt{\frac{2 \times 0.0441}{0.05}} \approx 1.325\ \text{m/s} \]
03

Collision Mechanics

Using conservation of momentum and kinetic energy, calculate the velocities after collision. Let \(v_1\) and \(v_2\) be the velocities of sphere 1 and sphere 2 after the collision:- Momentum conservation: \[ m_1u_1 = m_1v_1 + m_2v_2 \]- Energy conservation: \[ \frac{1}{2}m_1u_1^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 \]Solving gives: \[ v_1 = \frac{(m_1 - m_2)}{(m_1 + m_2)}u_1 = \frac{(50 - 85)}{(50 + 85)}\times 1.325 \approx -0.367\ \text{m/s} \] \[ v_2 = \frac{2m_1}{(m_1 + m_2)}u_1 = \frac{2 \times 50}{135} \times 1.325 \approx 0.958\ \text{m/s} \]
04

Height Reached by Sphere 1 (after the first collision)

Convert the kinetic energy after the collision back into potential energy to find the height reached by sphere 1:\[ \frac{1}{2}m_1v_1^2 = m_1gh' \]\[ h' = \frac{v_1^2}{2g} = \frac{(-0.367)^2}{2 \times 9.8} \approx 0.007\ \text{m} = 0.7\ \text{cm} \]
05

Height Reached by Sphere 2 (after the first collision)

Similarly, calculate the height reached by sphere 2:\[ \frac{1}{2}m_2v_2^2 = m_2gh_2 \]\[ h_2 = \frac{v_2^2}{2g} = \frac{(0.958)^2}{2 \times 9.8} \approx 0.0467\ \text{m} = 4.67\ \text{cm} \]
06

Exchange of Velocities in Subsequent Elastic Collision (Sphere 1 to Sphere 2)

After the first collision, sphere 2 will rise to 4.67 cm, then act as a pendulum to collide elastically again. For simplicity and due to equal mass exchange and energy conservation, the velocities will switch on equal mass spheres with elastic collisions, if ignored damping. Sphere 2 will reach a corresponding height to sphere 1 of 0.7 cm, and sphere 1 will regain height when free of sphere 2.
07

Height Reached by Each Sphere after Second Collision

Sphere 1 reaches 4.67 cm and sphere 2 reaches 0.7 cm after the second elastic collision.

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Key Concepts

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

Conservation of Energy
In this scenario, energy conservation plays a crucial role. Initially, sphere 1 is lifted to a height, giving it gravitational potential energy, expressed by the formula \( U = mgh \), where \( m \) is the mass, \( g \) is the gravitational acceleration, and \( h \) is the height. Upon release, this potential energy converts to kinetic energy just before the collision, satisfying the equation \( KE = \frac{1}{2}mv^2 \). In elastic collisions, the total mechanical energy (potential plus kinetic) remains constant throughout the motion.
  • Initial potential energy of sphere 1 is 0.0441 J.
  • Before the collision, this is fully converted to kinetic energy.
  • Post-collision, energy shifts between spheres but the total remains the same.
This principle allows us to calculate changes in energy and predict movement patterns after each collision.
Conservation of Momentum
Momentum conservation is another vital principle in elastic collisions. It helps us determine how sphere 1 and sphere 2 will behave upon collision. According to the conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision. We express this as \( m_1u_1 = m_1v_1 + m_2v_2 \), where \( m \) denotes mass, and \( u \) and \( v \) denote initial and final velocities respectively.
  • The momentum before impact is primarily due to sphere 1.
  • After collision, sphere 2 gains momentum due to sphere 1's movement.
  • Linear conservation ensures no momentum is lost in the system.
Using this law in conjunction with conservation of energy, we can solve for the velocities of both spheres after the collision.
Pendulum Motion
Pendulum motion is an important aspect in understanding how the spheres move after collision. A pendulum provides a back-and-forth swing motion driven by gravitational force. This motion involves continuous conversion between potential and kinetic energy. Once sphere 1 is released, it swings downward, gaining speed until it hits sphere 2 and converts energy upon colliding.
  • Swing height dictates energy at initial and end states.
  • Energy exchange occurs as kinetic energy at lowest point.
  • Post-collision height found by equating kinetic energy back to potential.
Pendulums in elastic collisions allow prediction of glide patterns between repeated cycles.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. Calculated using \( U = mgh \), GPE depends directly on height. This form of energy is fundamental in pendulum motions such as our sphere example. As sphere 1 ascends to 9 cm height, it stores energy which transforms as it falls.
  • Height increase ups GPE, stored in sphere 1.
  • Upon descent, GPE converts to kinetic, enabling motion.
  • Post-collision, the increase in height of sphere 2 represents GPE transferred.
This cyclical energy conversion dictates how high each sphere will rise after every subsequent collision.

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Most popular questions from this chapter

An atomic nucleus at rest at the origin of an \(x y\) coordinate system transforms into three particles. Particle \(1, \operatorname{mass} 16.7 \times 10^{-27}\) kg, moves away from the origin at velocity \(\left(6.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\text { is }}\), particle 2, mass \(8.35 \times 10^{-27} \mathrm{~kg}\), moves away at velocity \(\left(-8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)\) ) (a) In unit-vector notation, what is the linear momentum of the third particle, mass \(11.7 \times 10^{-27} \mathrm{~kg}\) ? (b) How much kinetic energy appears in this transformation?

A cue stick strikes a stationary pool ball, with an average force of 32 N over a time of \(14 \mathrm{~ms}\). If the ball has mass \(0.20 \mathrm{~kg}\), what speed does it have just after impact?

block 1 of mass \(m_{1}\) slides from rest along a frictionless ramp from height \(h=2.50 \mathrm{~m}\) and then collides with stationary block \(2,\) which has mass \(m_{2}=2.00 m_{1}\). After the collision, block 2 slides into a region where the coefficient of kinetic friction \(\mu_{k}\) is 0.500 and comes to a stop in distance \(d\) within that region. What is the value of distance \(d\) if the collision is (a) elastic and (b) completely inelastic?

A \(4.0 \mathrm{~kg}\) mess kit sliding on a frictionless surface explodes into two \(2.0 \mathrm{~kg}\) parts: \(3.0 \mathrm{~m} / \mathrm{s},\) due north, and \(5.0 \mathrm{~m} / \mathrm{s}, 30^{\circ}\) north of cast. What is the original speed of the mess kit?

Two \(2.0 \mathrm{~kg}\) bodies, \(A\) and \(B,\) collide. The velocitics before the collision are \(\vec{v}_{A}=(15 \mathrm{i}+30 \mathrm{j}) \mathrm{m} / \mathrm{s}\) and \(\vec{v}_{B}=(-10 \mathrm{i}+5.0 \mathrm{j}) \mathrm{m} / \mathrm{s} .\) After the collision, \(\vec{v}_{A}=(-5.0 \mathrm{i}+20 \mathrm{j}) \mathrm{m} / \mathrm{s} .\) What are (a) the final velocity of \(B\) and (b) the change in the total kinetic energy (including sign)?
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