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Chapter 10: Problem 87

Elastic Collision A body of mass \(2.0 \mathrm{~kg}\) makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed of the \(2.0 \mathrm{~kg}\) body was \(4.0 \mathrm{~m} / \mathrm{s}\) ?

Short Answer

Expert verified
The mass of the other body is 1.2 kg. The speed of the center of mass is 2.5 m/s.

Step by step solution

01

- Understanding Elastic Collision

In an elastic collision, both momentum and kinetic energy are conserved. Denote the mass of the initially moving body as \( m_1 = 2.0 \, \mathrm{kg} \), its initial velocity \( v_1 = 4.0 \, \mathrm{m/s} \), and its final velocity \( v_1' = \frac{v_1}{4} = 1.0 \, \mathrm{m/s} \). Let the mass of the second body be \( m_2 \) and its final velocity be \( v_2' \). The second body is initially at rest, so \( v_2 = 0 \).
02

- Apply Conservation of Momentum

Using the conservation of momentum: \[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \] Substituting the known values: \[ 2.0 \, \mathrm{kg} \times 4.0 \, \mathrm{m/s} = 2.0 \, \mathrm{kg} \times 1.0 \, \mathrm{m/s} + m_2 v_2' \] Simplify to find: \[ 8.0 \, \mathrm{kg \cdot m/s} = 2.0 \, \mathrm{kg \cdot m/s} + m_2 v_2' \] Therefore: \[ m_2 v_2' = 6.0 \, \mathrm{kg \cdot m/s} \]
03

- Apply Conservation of Kinetic Energy

Using the conservation of kinetic energy: \[ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} m_2 v_2'^2 \] Substituting the known values: \[ \frac{1}{2} \times 2.0 \, \mathrm{kg} \times (4.0 \, \mathrm{m/s})^2 = \frac{1}{2} \times 2.0 \, \mathrm{kg} \times (1.0 \, \mathrm{m/s})^2 + \frac{1}{2} \times m_2 v_2'^2 \] Simplify to find: \[ 16 \, \mathrm{J} = 1 \, \mathrm{J} + \frac{1}{2} m_2 v_2'^2 \] Therefore: \[ \frac{1}{2} m_2 v_2'^2 = 15 \, \mathrm{J} \] And: \[ m_2 v_2'^2 = 30 \, \mathrm{J} \]
04

- Solving for Mass of Second Body

From step 2, we have \[ m_2 v_2' = 6.0 \, \mathrm{kg \cdot m/s} \] and from step 3, we have \[ m_2 v_2'^2 = 30 \, \mathrm{J} \] Dividing the kinetic energy equation by the momentum equation: \[ v_2' = \frac{30 \, \mathrm{J}}{6.0 \, \mathrm{kg \cdot m/s}} = 5.0 \, \mathrm{m/s} \] Then substituting back to find \( m_2 \): \[ m_2 = \frac{6.0 \, \mathrm{kg \cdot m/s}}{5.0 \, \mathrm{m/s}} = 1.2 \, \mathrm{kg} \]
05

- Center of Mass Speed

To find the speed of the center of mass, use: \[ v_{\text{cm}} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \] Initially: \[ v_{\text{cm}} = \frac{2.0 \, \mathrm{kg} \times 4.0 \, \mathrm{m/s} + 1.2 \, \mathrm{kg} \times 0.0 \, \mathrm{m/s}}{2.0 \, \mathrm{kg} + 1.2 \, \mathrm{kg}} = \frac{8.0 \, \mathrm{kg \cdot m/s}}{3.2 \, \mathrm{kg}} = 2.5 \, \mathrm{m/s} \]

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Key Concepts

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

Conservation of Momentum
In physics, the principle of conservation of momentum states that in a closed system, the total momentum before and after a collision remains the same. Momentum is the product of mass and velocity. In this exercise, we have two bodies: the first body has a mass of ewline 2.0 kg and an initial velocity of 4.0 m/s, while the second body is initially at rest, hence has a velocity of 0 m/s.
During an elastic collision, the momentum equation is:
equation here: \[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \]By substituting the given values, we calculate the momentum of each body before and after collision. From the simplified equation, we can solve for unknowns such as the final velocities or masses of the bodies, ensuring the total momentum stays constant.
Let's remember:
  • Momentum = mass 脳 velocity
  • Closed system means no external forces
Conservation of Kinetic Energy
Kinetic energy of an object is given by the formula:\[ K.E. = \frac{1}{2} m v^2 \]In an elastic collision, the total kinetic energy before and after the collision remains constant. This is demonstrated by showing the addition of kinetic energies of both bodies before and after collision.
In the given exercise, the initial kinetic energy is:\[ \frac{1}{2} \times 2.0 \mathrm{~kg} \times (4.0 \mathrm{~m/s})^2 = 16 \text{ J} \]After the collision, the kinetic energy of the first body is:\[\frac{1}{2} \times 2.0 \mathrm{~kg} \times (1.0 \mathrm{~m/s})^2 = 1 \text{ J} \]Using the conservation of kinetic energy, we set up the equation:\[ 16 \text{ J} = 1 \text{ J} + \frac{1}{2} m_2 v_2'^2 \]By rearranging and solving, we find the kinetic energy of the second body.
  • Kinetic Energy = \[ \frac{1}{2} m v^2 \]
  • Total K.E. before collision = Total K.E. after collision
Center of Mass
The center of mass of a system is a point representing the average location of the total mass of the system. It can be found using the formula:\[ v_\text{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \]For the two bodies in the problem, we initially calculate the center of mass by taking into account their respective masses and velocities. Since one of the bodies is initially at rest, its contribution to the momentum is zero.
The calculation would be:
First, the numerator (combined momentum):
\[2.0 \mathrm{~kg} \times 4.0 \mathrm{~m/s} + 1.2 \mathrm{~kg} \times 0.0 \mathrm{~m/s} = 8.0 \text{ kg路m/s}\]Then, the denominator (total mass):
\[2.0 \text{ kg} + 1.2 \text{ kg} = 3.2 \text{ kg}\]So, the speed of the center of mass is:
\[v_\text{cm} = \frac{8.0 \mathrm{~kg \cdot m/s}}{3.2 \mathrm{~kg}} = 2.5 \mathrm{~m/s} \]This speed represents the velocity of the entire mass system moving together.
  • Center of Mass = average position of system's mass
  • Uses combined mass and momentum

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