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Chapter 9: Problem 55

Block 1 of mass \(3.0 \mathrm{~kg}\) is sliding across a floor with speed \(v_{1}=2.0 \mathrm{~m} / \mathrm{s}\) when it makes a head-on, one-dimensional, elastic collision with initially stationary block 2 of mass \(2.0 \mathrm{~kg}\). The coefficient of kinetic friction between the blocks and the floor is \(\mu_{k}=\) \(0.30\). Find the speeds of (a) block 1 and (b) block 2 just after the collision. Also find (c) their final separation after friction has stopped them and \((\mathrm{d})\) the energy lost to thermal energy because of the friction.

Short Answer

Expert verified
Block 1 speed is 0.4 m/s, block 2 speed is 2.4 m/s, total separation is the sum of distances traveled by each block, and energy lost is the initial kinetic energy.

Step by step solution

01

Conservation of Momentum

For an elastic collision, the total momentum before and after the collision is conserved. The formula for momentum conservation is: \[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \] Substituting the given values: \[ 3.0 \, \mathrm{kg} \times 2.0 \, \mathrm{m/s} + 2.0 \, \mathrm{kg} \times 0 \, \mathrm{m/s} = 3.0 \, \mathrm{kg} \times v_1' + 2.0 \, \mathrm{kg} \times v_2' \] which simplifies to:\[ 6.0 = 3.0 v_1' + 2.0 v_2' \] This is our first equation.
02

Conservation of Kinetic Energy

In an elastic collision, the total kinetic energy is also conserved. Thus, \[ \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1{v_1'}^2 + \frac{1}{2}m_2{v_2'}^2 \] Substituting the known values results in: \[ \frac{1}{2} \times 3.0 \times (2.0)^2 + 0 = \frac{1}{2} \times 3.0 \times (v_1')^2 + \frac{1}{2} \times 2.0 \times (v_2')^2 \] which simplifies to:\[ 6.0 = 1.5(v_1')^2 + 1.0 (v_2')^2 \] This is our second equation.
03

Solve the System of Equations

Use the equations from Steps 1 and 2 to solve for \(v_1'\) and \(v_2'\). From momentum, we have:\[ 6 = 3v_1' + 2v_2' \] And from energy conservation: \[ 6 = 1.5(v_1')^2 + 1.0(v_2')^2 \] Solve this system of equations to find that \(v_1' = 0.4\,\mathrm{m/s}\) and \(v_2' = 2.4\,\mathrm{m/s}\).
04

Calculate Friction Force and Work Done

The friction force \(f_k\) acting on each block is given by: \[ f_k = \mu_k \cdot m \cdot g \] For block 1, it is \( f_{k1} = 0.3 \times 3.0 \times 9.8 = 8.82 \, \mathrm{N} \) and for block 2, \( f_{k2} = 0.3 \times 2.0 \times 9.8 = 5.88 \, \mathrm{N} \). The work done by friction is equal to the loss in kinetic energy.
05

Determine Distance Traveled Until Stopped

Using the work-energy principle, where the work done by friction stops the blocks, \( W = \Delta KE = \frac{1}{2} m v'^2 \). To find the distance (d): \[ f_k \cdot d = \frac{1}{2}m{v_1'}^2 + \frac{1}{2}m{v_2'}^2 - 0 \] For block 1:\[ 8.82 \cdot d_1 = \frac{1}{2} \times 3.0 \times (0.4)^2 \]Solve for \(d_1\) and then do likewise for block 2 using its variables.
06

Calculate Final Separation Distance

The total separation after both blocks come to rest can be found by adding the distances each traveled: \[ d_{total} = d_1 + d_2 \] where \(d_1\) and \(d_2\) are the distances traveled by block 1 and block 2 until they stop.
07

Calculate Energy Loss Due to Friction

Energy lost to thermal energy can be calculated as the difference in initial kinetic energy and final kinetic energy (which is zero as they come to rest):\[ \Delta E = KE_{initial} - KE_{final} = \text{(Initial Kinetic Energy)} \]

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Key Concepts

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

Momentum Conservation
Momentum conservation is a fundamental principle in physics, especially relevant in collision scenarios. It states that the total momentum of a closed system before and after a collision remains the same, provided no external forces act on it. When considering an elastic collision, this principle plays a crucial role.

For a collision between two objects, each with mass and velocity, the total momentum before the collision (\[ m_1 v_1 + m_2 v_2 \]) must equal the total momentum after the collision (\[ m_1 v_1' + m_2 v_2' \]). In the provided exercise:
  • Block 1 with mass 3.0 kg is moving at 2.0 m/s initially.
  • Block 2 is stationary with mass 2.0 kg.
The momentum equation simplifies to \[ 6.0 = 3.0 v_1' + 2.0 v_2' \]. Using this relation alongside other equations helps us solve for the velocities of both blocks post-collision.
Kinetic Energy Conservation
In an elastic collision, not only is momentum conserved, but kinetic energy is as well. The conservation of kinetic energy dictates that the total kinetic energy of a system remains constant if only conservative forces are acting. This comes into play directly in calculations involving moving bodies before and after collisions.

The exercise showcases kinetic energy conservation through the equation:\[ \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1{v_1'}^2 + \frac{1}{2}m_2{v_2'}^2 \]. Given:
  • Initial velocities lead to a kinetic energy of 6.0 J for the system.
  • This energy splits post-collision as indicated by the new velocities solved through simultaneous equations.
The formula simplifies with known variables to another equation \[ 6.0 = 1.5(v_1')^2 + 1.0 (v_2')^2 \], crucial for finding post-collision speeds.
Kinetic Friction
Kinetic friction, the force that opposes the relative motion of surfaces in contact during movement, plays a significant role in the block deceleration. It influences how far each block eventually travels before coming to rest.

The force of kinetic friction can be calculated using the formula: \[ f_k = \mu_k \cdot m \cdot g \], where
  • \(\mu_k\) is the coefficient of kinetic friction, 0.30 in this problem.
  • The gravitational force (\(g\)) is approximately 9.8 m/s\(^2\).
  • Different masses yield different friction forces (8.82 N for block 1 and 5.88 N for block 2).
Understanding kinetic friction helps predict how both blocks slow down and stop due to frictional forces, affecting their ultimate separation distance.
Work-Energy Principle
The work-energy principle establishes a relationship between the work done by forces on an object and the change in its kinetic energy. This principle is very useful in calculating how forces like friction affect the motion of sliding objects.

For our exercise, when blocks slide and stop due to friction, the work done by friction (work-energy) equals the initial kinetic energy lost by the blocks. The principle is illustrated through the equation\[ f_k \cdot d = \frac{1}{2}m{v_1'}^2 + \frac{1}{2}m{v_2'}^2 - 0 \]. Solving this,
  • We determine the stopping distances for both blocks.
  • The distances achieved before the blocks stop can be added to yield the final separation between them.
  • The overall energy lost to friction manifests as thermal energy, a form of dissipated work.
The work-energy principle not only guides us to determine distances but is also key in calculating the energy transformations during the motion.

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Most popular questions from this chapter

A \(0.25 \mathrm{~kg}\) puck is initially stationary on an ice surface with negligible friction. At time \(t=0\), a horizontal force begins to move the puck. The force is given by \(\vec{F}=\left(12.0-3.00 t^{2}\right) \hat{i}\), with \(\vec{F}\) in newtons and \(t\) in seconds, and it acts until its magnitude is zero. (a) What is the magnitude of the impulse on the puck from the force between \(t=0.750 \mathrm{~s}\) and \(t=1.25 \mathrm{~s}\) ? (b) What is the change in momentum of the puck between \(t=0\) and the instant at which \(F=0\) ?

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by \(0.50 \mathrm{~m}\), the mass that moves downward is \(75 \mathrm{~kg}\), and the collision on the floor lasts \(0.088 \mathrm{~s}\). What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

After a completely inelastic collision, two objects of the same mass and same initial speed move away together at half their initial speed. Find the angle between the initial velocities of the objects.

A \(5.20 \mathrm{~g}\) bullet moving at \(700 \mathrm{~m} / \mathrm{s}\) strikes a \(700 \mathrm{~g}\) wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to \(450 \mathrm{~m} / \mathrm{s}\). (a) What is the resulting speed of the block? (b) What is the speed of the bullet-block center of mass?

A ball of mass \(50 \mathrm{~g}\) moving with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\) strikes a wall at an angle of incidence \(45^{\circ}\) and is reflected from the wall at the same angle and with the same speed. See the overhead view in Fig. 9-38. Calculate (a) the magnitude of the change \(\Delta \vec{p}\) in the momentum of the ball, (b) the change in the magnitude of the momentum \(\vec{p}\) of the ball, and (c) the change in the magnitude of the momentum of the wall.
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