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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 \(0.90\).

Short Answer

Expert verified
(a) Both at 0.9 m/s; (b) 90 g at 0 m/s, 10 g at 1 m/s; (c) 90 g at 0.1 m/s, 10 g at 1 m/s.

Step by step solution

01

Convert units

First, let's convert the mass and velocity units to standard SI units. The mass of the balls: \(90\, \text{g} = 0.09\, \text{kg}\) and \(10\, \text{g} = 0.01\, \text{kg}\). The initial velocity of the moving ball: \(100\, \text{cm/s} = 1\, \text{m/s}\). The stationary ball has an initial velocity of \(0\, \text{m/s}\).
02

Case (a) - Inelastic collision (stick together)

In an inelastic collision where they stick together, use the conservation of momentum: \[ m_1v_1 + m_2v_2 = (m_1 + m_2)v_f \]Plugging in the values: \[ 0.09 \times 1 + 0.01 \times 0 = (0.09 + 0.01)v_f \]\[ v_f = \frac{0.09}{0.10} = 0.9\, \text{m/s} \]Both balls move together at \(0.9\, \text{m/s}\).
03

Case (b) - Perfectly elastic collision

For a perfectly elastic collision, both momentum and kinetic energy are conserved. Solve the following system of equations:1. \(m_1v_1 + m_2v_2 = m_1v_{1f} + m_2v_{2f}\)2. \(\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2\)Substituting the values into the momentum equation:\[ 0.09 \times 1 = 0.09v_{1f} + 0.01v_{2f} \]And into the kinetic energy equation:\[ 0.045 = 0.045v_{1f}^2 + 0.005v_{2f}^2 \]Solving these gives: \(v_{1f} = 0, v_{2f} = 1\) m/s. The heavier ball stops, and the lighter ball moves at 1 m/s.
04

Case (c) - Collision with a coefficient of restitution 0.90

The coefficient of restitution \(e\) is given by: \[ e = \frac{v_{2f} - v_{1f}}{v_1 - v_2} \] Given \(e = 0.90\), and substituting known values:\[ 0.90 = \frac{v_{2f} - v_{1f}}{1 - 0} \]\[ v_{2f} - v_{1f} = 0.90 \]Combine this with the conservation of momentum equation:\[ 0.09 = 0.09v_{1f} + 0.01v_{2f} \]Solve the two equations together:1. \(0.09 = 0.09v_{1f} + 0.01v_{2f}\)2. \(v_{2f} - v_{1f} = 0.90\)Solving this pair gives: \(v_{1f} = 0.1\, \text{m/s}, v_{2f} = 1.0\, \text{m/s}\).

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Key Concepts

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

Inelastic Collision
In an inelastic collision, two objects collide and do not bounce apart. The main characteristic of this type of collision is that kinetic energy is not conserved, though the momentum is. Typically, the objects stick together post-collision, moving as a single mass.
  • Imagine a 90 g ball moving towards a 10 g stationary ball.
  • They crash and cling together, forming a 100 g mass.
  • In this scenario, only momentum is conserved.
For our example, we calculate their shared velocity using the conservation of momentum formula: \[ m_1v_1 + m_2v_2 = (m_1 + m_2)v_f \]The initial momentum of the system is solely due to the moving 90 g ball. Thus, after substitution and solving, the combined velocity of both balls is found to be \(0.9\, \text{m/s}\). This reduced speed indicates kinetic energy was lost, typically transformed into heat or deformation energy.
Elastic Collision
An elastic collision is one where both momentum and kinetic energy are conserved. In this type of collision, after impact, the objects do not stick together but continue moving separately.
  • Picture again our 90 g moving ball colliding with the 10 g ball.
  • After the hit, they part ways without any energy loss.
  • Both momentum and energy equations must be satisfied.
When doing the math:
  • Use the momentum conservation: \(m_1v_1 + m_2v_2 = m_1v_{1f} + m_2v_{2f}\)
  • And kinetic energy: \(\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2\)
Solving these equations for the given scenario, one finds the 90 g ball stops (\(v_{1f} = 0\, \text{m/s}\)), and the 10 g ball moves ahead with the previous speed of \(1\, \text{m/s}\). No kinetic energy is lost.
Momentum Conservation
Momentum conservation is a key principle in mechanics. It states that within an isolated system, the total momentum remains constant before and after an event, such as a collision.
  • Momentum, represented as \(p\), is calculated by the formula \(p = mv\).
  • In a closed system (no external forces), the sum of all momenta of the involved bodies remains the same.
For example:
  • A 90 g and a 10 g ball interact.
  • Regardless of whether they stick or bounce apart, the initial and final total momentum will equalize.
While momentum dictates the movement of mass, remember it does not account for energy changes directly. Hence each collision type - elastic or inelastic - involves its distinct energy focus.
Kinetic Energy Conservation
Kinetic energy conservation refers to the idea that, in certain types of collisions, the total kinetic energy of the system remains unchanged pre- and post-collision. This is not universal for all collisions and primarily applies to elastic collisions.
  • Kinetic energy \(KE\) is given by \(\frac{1}{2}mv^2\).
  • Unlike elastic collisions, inelastic collisions do not return all kinetic energy in the same form post-impact.
In the scenario with our balls:
  • During a perfectly elastic collision, the kinetic energy remains consistent before and after collision.
  • In the inelastic example, energy conservation does not hold, evidenced by the reduced velocity of the combined mass.
Thus, acknowledging the collision type helps anticipate potential energy transformations - exterior forces like friction often alter kinetic pathways even when momentum is conserved.

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Most popular questions from this chapter

Three point masses are placed on the \(x\) -axis: \(200 \mathrm{~g}\) at \(x=0,500 \mathrm{~g}\) at \(x=30 \mathrm{~cm}\), and \(400 \mathrm{~g}\) at \(x=70 \mathrm{~cm} .\) Find their center of mass. We can make the calculation with respect to any point, but since all the data is measured from the \(x=0\) origin, that point will do nicely. $$ x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(0.20 \mathrm{~kg})+(0.30 \mathrm{~m})(0.50 \mathrm{~kg})+(0.70 \mathrm{~m})(0.40 \mathrm{~kg})}{(0.20+0.50+0.40) \mathrm{kg}}=0.39 \mathrm{~m} $$ The center of mass is located at a distance of \(0.39 \mathrm{~m}\), in the positive \(x\) -direction, from the origin. The \(y\) - and \(z\) -coordinates of the center of mass are zero.

A \(2.0\) -kg brick is moving at a speed of \(6.0 \mathrm{~m} / \mathrm{s}\). How large a force \(F\) is needed to stop the brick in a time of \(7.0 \times 10^{-4} \mathrm{~s}\) ? Since we have a force and the time over which it acts, that suggests using the impulse equation (i.e., Newton's Second Law): Impulse on brick \(=\) Change in momentum of brick $$ \begin{aligned} F \Delta t &=m v_{f}-m v_{i} \\ F\left(7.0 \times 10^{-4} \mathrm{~s}\right) &=0-(2.0 \mathrm{~kg})(6.0 \mathrm{~m} / \mathrm{s}) \end{aligned} $$ from which \(F=-1.7 \times 10^{4} \mathrm{~N}\). The minus sign indicates that the force opposes the motion.

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?

Two balls of equal mass, moving with speeds of \(3 \mathrm{~m} / \mathrm{s}\), collide head-on. Find the speed of each after impact if (a) they stick together, \((b)\) the collision is perfectly elastic, \((c)\) the coefficient of restitution is \(1 / 3\).

Typically, a tennis ball hit during a serve travels away at about \(51 \mathrm{~m} / \mathrm{s}\). If the ball is at rest mid-air when struck, and it has a mass of \(0.058 \mathrm{~kg}\), what is the change in its momentum on leaving the racket?

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