Illustration of a two-body collision process, depicting initial conditions, interaction, and final motion.
Diagram showing three stages of a collision between two objects: before collision, at collision, and after collision, with mass and velocity labels.
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Step-by-step solution for: Mass and Momentum: Law of Conservation of Momentum
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Show Answer Key & Explanations
Step-by-step solution for: Mass and Momentum: Law of Conservation of Momentum
The image depicts a physics problem concerning the collision of two masses, $m_1$ and $m_2$. The goal is to determine the final velocities, $v_1$ and $v_2$, after the collision, given their initial velocities, $u_1$ and $u_2$.
To solve this, we apply the fundamental conservation laws of physics for collisions:
1. Conservation of Momentum: The total momentum of the system before the collision equals the total momentum after the collision.
2. Conservation of Kinetic Energy: For an elastic collision (implied by the "before" and "after" states shown without energy loss), the total kinetic energy is also conserved.
These two principles provide a system of two equations with two unknowns ($v_1$ and $v_2$), which can be solved simultaneously.
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Step 1: Apply Conservation of Momentum
The total momentum before the collision is $m_1 u_1 + m_2 u_2$.
The total momentum after the collision is $m_1 v_1 + m_2 v_2$.
Setting them equal gives the first equation:
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad \text{(Equation 1)}$$
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Step 2: Apply Conservation of Kinetic Energy
The total kinetic energy before the collision is $\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2$.
The total kinetic energy after the collision is $\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$.
Setting them equal gives the second equation:
$$\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \quad \text{(Equation 2)}$$
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Step 3: Solve the System of Equations
By algebraically manipulating these two equations, we can derive the standard formulas for the final velocities in a one-dimensional elastic collision.
- Rearranging Equation 1 to isolate terms involving $m_1$ and $m_2$:
$$m_1(u_1 - v_1) = m_2(v_2 - u_2)$$
- Rearranging Equation 2 (multiplying through by 2 to eliminate the 1/2):
$$m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2)$$
- Using the difference of squares identity ($a^2 - b^2 = (a-b)(a+b)$) on both sides:
$$m_1(u_1 - v_1)(u_1 + v_1) = m_2(v_2 - u_2)(v_2 + u_2)$$
- Dividing the kinetic energy equation by the momentum equation (assuming $u_1 \neq v_1$ and $v_2 \neq u_2$):
$$u_1 + v_1 = v_2 + u_2$$
This simplifies to:
$$u_1 - u_2 = v_2 - v_1 \quad \text{(Equation 3)}$$
This equation states that the relative velocity of approach equals the relative velocity of separation.
- Solving Equation 3 for $v_2$:
$$v_2 = u_1 - u_2 + v_1$$
- Substituting this expression for $v_2$ into Equation 1:
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2(u_1 - u_2 + v_1)$$
- Expanding and solving for $v_1$:
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 u_1 - m_2 u_2 + m_2 v_1$$
$$m_1 u_1 + m_2 u_2 - m_2 u_1 + m_2 u_2 = v_1(m_1 + m_2)$$
$$u_1(m_1 - m_2) + 2m_2 u_2 = v_1(m_1 + m_2)$$
Therefore:
$$v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1 + \frac{2m_2}{m_1 + m_2}u_2$$
- Similarly, substituting the expression for $v_1$ back into Equation 3 or using symmetry, we find $v_2$:
$$v_2 = \frac{2m_1}{m_1 + m_2}u_1 + \frac{m_2 - m_1}{m_1 + m_2}u_2$$
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Final Solution
The final velocities of the two masses after a one-dimensional elastic collision are given by:
- $v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u_1 + \left( \frac{2m_2}{m_1 + m_2} \right) u_2$
- $v_2 = \left( \frac{2m_1}{m_1 + m_2} \right) u_1 + \left( \frac{m_2 - m_1}{m_1 + m_2} \right) u_2$
These formulas allow you to calculate the post-collision velocities for any given values of $m_1$, $m_2$, $u_1$, and $u_2$.
To solve this, we apply the fundamental conservation laws of physics for collisions:
1. Conservation of Momentum: The total momentum of the system before the collision equals the total momentum after the collision.
2. Conservation of Kinetic Energy: For an elastic collision (implied by the "before" and "after" states shown without energy loss), the total kinetic energy is also conserved.
These two principles provide a system of two equations with two unknowns ($v_1$ and $v_2$), which can be solved simultaneously.
---
Step 1: Apply Conservation of Momentum
The total momentum before the collision is $m_1 u_1 + m_2 u_2$.
The total momentum after the collision is $m_1 v_1 + m_2 v_2$.
Setting them equal gives the first equation:
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad \text{(Equation 1)}$$
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Step 2: Apply Conservation of Kinetic Energy
The total kinetic energy before the collision is $\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2$.
The total kinetic energy after the collision is $\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$.
Setting them equal gives the second equation:
$$\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \quad \text{(Equation 2)}$$
---
Step 3: Solve the System of Equations
By algebraically manipulating these two equations, we can derive the standard formulas for the final velocities in a one-dimensional elastic collision.
- Rearranging Equation 1 to isolate terms involving $m_1$ and $m_2$:
$$m_1(u_1 - v_1) = m_2(v_2 - u_2)$$
- Rearranging Equation 2 (multiplying through by 2 to eliminate the 1/2):
$$m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2)$$
- Using the difference of squares identity ($a^2 - b^2 = (a-b)(a+b)$) on both sides:
$$m_1(u_1 - v_1)(u_1 + v_1) = m_2(v_2 - u_2)(v_2 + u_2)$$
- Dividing the kinetic energy equation by the momentum equation (assuming $u_1 \neq v_1$ and $v_2 \neq u_2$):
$$u_1 + v_1 = v_2 + u_2$$
This simplifies to:
$$u_1 - u_2 = v_2 - v_1 \quad \text{(Equation 3)}$$
This equation states that the relative velocity of approach equals the relative velocity of separation.
- Solving Equation 3 for $v_2$:
$$v_2 = u_1 - u_2 + v_1$$
- Substituting this expression for $v_2$ into Equation 1:
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2(u_1 - u_2 + v_1)$$
- Expanding and solving for $v_1$:
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 u_1 - m_2 u_2 + m_2 v_1$$
$$m_1 u_1 + m_2 u_2 - m_2 u_1 + m_2 u_2 = v_1(m_1 + m_2)$$
$$u_1(m_1 - m_2) + 2m_2 u_2 = v_1(m_1 + m_2)$$
Therefore:
$$v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1 + \frac{2m_2}{m_1 + m_2}u_2$$
- Similarly, substituting the expression for $v_1$ back into Equation 3 or using symmetry, we find $v_2$:
$$v_2 = \frac{2m_1}{m_1 + m_2}u_1 + \frac{m_2 - m_1}{m_1 + m_2}u_2$$
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Final Solution
The final velocities of the two masses after a one-dimensional elastic collision are given by:
- $v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u_1 + \left( \frac{2m_2}{m_1 + m_2} \right) u_2$
- $v_2 = \left( \frac{2m_1}{m_1 + m_2} \right) u_1 + \left( \frac{m_2 - m_1}{m_1 + m_2} \right) u_2$
These formulas allow you to calculate the post-collision velocities for any given values of $m_1$, $m_2$, $u_1$, and $u_2$.
Parent Tip: Review the logic above to help your child master the concept of law of conservation of momentum.
