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A 5.00-kg ball, moving to the right at a velocity of \(+2.00 \mathrm{m} / \mathrm{s}\) on a frictionless table, collides head-on with a stationary \(7.50-\mathrm{kg}\) ball. Find the final velocities of the balls if the collision is (a) elastic and (b) completely inelastic.

Short Answer

Expert verified
(a) Elastic: \(v_{1f} = -0.8\, \text{m/s}\), \(v_{2f} = 1.2\, \text{m/s}\). (b) Inelastic: \(v_{1f} = v_{2f} = 0.80\, \text{m/s}\).

Step by step solution

01

Identify System Parameters

The initial velocities, masses, and conditions of the two balls need to be identified.- Mass of Ball 1, \(m_1 = 5.00\, \text{kg}\), with initial velocity \(v_{1i} = +2.00\, \mathrm{m/s}\).- Mass of Ball 2, \(m_2 = 7.50\, \text{kg}\), with initial velocity \(v_{2i} = 0\).These form the basis for both parts of the problem (elastic and completely inelastic collisions).
02

Use Conservation of Momentum

The total momentum before and after the collision must be equal. \[ m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \] Substitute the known values: \[ 5.00\, (2.00) + 7.50\, (0) = 5.00\, v_{1f} + 7.50\, v_{2f} \] Which simplifies to: \[ 10.0 = 5.00\, v_{1f} + 7.50\, v_{2f} \]
03

Solve for Elastic Collision

In an elastic collision, both momentum and kinetic energy are conserved. Combine kinetic energy conservation: \[ \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 \] This simplifies to \( 10 = v_{1f}^2 + \frac{3}{2}v_{2f}^2 \), add with the momentum equation to solve for \(v_{1f}\) and \(v_{2f}\). Solving these yields: \[ v_{1f} = -0.8\, \mathrm{m/s}, \quad v_{2f} = 1.2\, \mathrm{m/s} \]
04

Solve for Completely Inelastic Collision

In a completely inelastic collision, the balls move together after collision.Use momentum conservation equation: \[ m_1v_{1i} + m_2v_{2i} = (m_1 + m_2) v_f \] Substitute known values: \[ 10.0 = (5.00 + 7.50) v_f \] \[ v_f = \frac{10.0}{12.5} = 0.80\, \mathrm{m/s} \] So, \(v_{1f} = v_{2f} = 0.80\, \mathrm{m/s}\).

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Key Concepts

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

Elastic Collision
In an elastic collision, two important quantities remain constant: momentum and kinetic energy. This means that neither is lost during the collision. Such collisions are quite rare in everyday life, as energy transformations often occur (like sound or heat), but they are useful for theoretical models.
  • Both momentum and kinetic energy are preserved.
  • Objects bounce apart after the collision.
For instance, in the problem given, we can calculate the velocities of both balls after the collision using these principles. By solving the equations for both momentum and kinetic energy conservation, we find the final velocities: the first ball moves to the left at \(-0.8\, \mathrm{m/s}\), while the second ball moves to the right at \(1.2\, \mathrm{m/s}\). This tells us that energy remained within the system, bouncing the balls away from each other.
Inelastic Collision
Inelastic collisions, on the other hand, do not conserve kinetic energy. Instead, some kinetic energy is transformed into other forms of energy, such as heat or sound. However, momentum is still conserved.
  • Momentum is conserved, but kinetic energy is not.
  • Objects can stick together after the collision.
In a completely inelastic collision, the two objects stick together and move as one. Referring to our exercise, after the completely inelastic collision, both balls travel together with a common velocity of \(0.80\, \mathrm{m/s}\). This shared speed shows kinetic energy lost when transforming into another type, for instance, internal energy or deformation.
Momentum Conservation
Momentum conservation is one of the fundamental laws in physics. It tells us that in the absence of external forces, the total momentum of a system remains constant before and after a collision.
  • Momentum conservation applies to both elastic and inelastic collisions.
  • Total initial momentum equals total final momentum.
The momentum conservation principle is crucial for solving collision problems, like our current exercise. It involves setting the momentum before equal to the momentum after. Using the initial data, we applied this to find both the final velocities of each ball in elastic as well as completely inelastic cases. Whether energy is conserved or transformed, momentum stays the same when dealing with isolated systems.
Kinetic Energy Conservation
Kinetic energy conservation plays a significant role, especially in elastic collisions. Kinetic energy is the energy of motion, and conserving it means all energy remains in the form of motion, without becoming heat or sound.
  • Only conserved in elastic collisions.
  • Helps determine the final velocities of objects.
In our example, the conservation of kinetic energy equation helps us calculate the final velocities in an elastic collision scenario. By equating the sum of the initial kinetic energies with the sum of the final kinetic energies and combining this with the momentum equations, we identified how both objects moved post-collision. This dual conservation requirement provides ample information to solve for unknown variables in any elastic collision.

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Most popular questions from this chapter

A ball is dropped from rest from the top of a \(6.10-\mathrm{m}\) -tall building, falls straight downward, collides inelastically with the ground, and bounces back. The ball loses \(10.0 \%\) of its kinetic energy every time it collides with the ground. How many bounces can the ball make and still reach a windowsill that is \(2.44 \mathrm{m}\) above the ground?

A model rocket is constructed with a motor that can provide a total impulse of \(29.0 \mathrm{N} \cdot \mathrm{s}\) The mass of the rocket is 0.175 kg. What is the speed that this rocket achieves when launched from rest? Neglect the effects of gravity and air resistance.

A mine car (mass \(=440 \mathrm{kg}\) ) rolls at a speed of \(0.50 \mathrm{m} / \mathrm{s}\) on a horizontal track, as the drawing shows. A 150-kg chunk of coal has a speed of \(0.80 \mathrm{m} / \mathrm{s}\) when it leaves the chute. Determine the speed of the car-coal system after the coal has come to rest in the car.

An electron collides elastically with a stationary hydrogen atom. The mass of the hydrogen atom is 1837 times that of the electron. Assume that all motion, before and after the collision, occurs along the same straight line. What is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision? \(?\)

A lumberjack (mass \(=98 \mathrm{kg}\) ) is standing at rest on one end of a floating log (mass \(=230 \mathrm{kg}\) ) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of \(+3.6 \mathrm{m} / \mathrm{s}\) relative to the shore, and then hops onto an identical floating log that is initially at rest. Neglect any friction and resistance between the logs and the water. (a) What is the velocity of the first log just before the lumberjack jumps off? (b) Determine the velocity of the second log if the lumberjack comes to rest on it.

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