Math Problem Statement

Find the final velocities of two bumper cars after a collision. Treat the problem as an elastic collision using the masses (300 kg and 100 kg) and initial velocities (2 m/s and -4 m/s).

Solution

To solve the given problem, where the two bumper cars collide elastically, we need to use the principles of conservation of momentum and conservation of kinetic energy.

Step 1: Write the Given Information

  • Mass of car A: m1=300kgm_1 = 300 \, \text{kg}
  • Initial velocity of car A: v1=2m/sv_1 = 2 \, \text{m/s}
  • Mass of car B: m2=100kgm_2 = 100 \, \text{kg}
  • Initial velocity of car B: v2=4m/sv_2 = -4 \, \text{m/s}

Step 2: Conservation of Momentum

m1v1i+m2v2i=m1v1f+m2v2fm_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} Substituting values: 300(2)+100(4)=300v1f+100v2f300(2) + 100(-4) = 300v_{1f} + 100v_{2f} 600400=300v1f+100v2f600 - 400 = 300v_{1f} + 100v_{2f} 200 = 300v_{1f} + 100v_{2f} \tag{1}

Step 3: Conservation of Kinetic Energy

12m1v1i2+12m2v2i2=12m1v1f2+12m2v2f2\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2 Substituting values: 12(300)(22)+12(100)(42)=12(300)v1f2+12(100)v2f2\frac{1}{2}(300)(2^2) + \frac{1}{2}(100)(-4^2) = \frac{1}{2}(300)v_{1f}^2 + \frac{1}{2}(100)v_{2f}^2 600+800=150v1f2+50v2f2600 + 800 = 150v_{1f}^2 + 50v_{2f}^2 1400 = 150v_{1f}^2 + 50v_{2f}^2 \tag{2}

Step 4: Solve the System of Equations

We have two equations:

  1. 200=300v1f+100v2f200 = 300v_{1f} + 100v_{2f}
  2. 1400=150v1f2+50v2f21400 = 150v_{1f}^2 + 50v_{2f}^2

Let me solve these equations. sympy import symbols, Eq, solve

Define the variables

v1f, v2f = symbols('v1f v2f')

Define the equations

eq1 = Eq(300v1f + 100v2f, 200) eq2 = Eq(150v1f**2 + 50v2f**2, 1400)

Solve the system of equations

solutions = solve((eq1, eq2), (v1f, v2f)) solutionsThe solutions to the system of equations are:

  1. v1f=1m/s,v2f=5m/sv_{1f} = -1 \, \text{m/s}, \, v_{2f} = 5 \, \text{m/s}
  2. v1f=2m/s,v2f=4m/sv_{1f} = 2 \, \text{m/s}, \, v_{2f} = -4 \, \text{m/s}

Since this is an elastic collision, the initial and final velocities must differ. The second solution, v1f=2m/s,v2f=4m/sv_{1f} = 2 \, \text{m/s}, v_{2f} = -4 \, \text{m/s}, represents the trivial case where nothing changes. Therefore, the physically meaningful solution is:

v1f=1m/s,v2f=5m/s.v_{1f} = -1 \, \text{m/s}, \, v_{2f} = 5 \, \text{m/s}.

Final Velocities:

  • Final velocity of car A: v1f=1m/sv_{1f} = -1 \, \text{m/s}
  • Final velocity of car B: v2f=5m/sv_{2f} = 5 \, \text{m/s}

Would you like a step-by-step breakdown of how the equations were solved or have additional questions?

Related Questions:

  1. How are the equations for momentum and energy conservation derived?
  2. Why is the second solution considered trivial in this problem?
  3. Can this method be applied to inelastic collisions? If not, how does the approach change?
  4. What happens if both cars have the same mass?
  5. How does the relative velocity between the two cars change before and after the collision?

Tip:

When solving collision problems, always verify the solutions by substituting them back into both conservation equations to ensure consistency.

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Math Problem Analysis

Mathematical Concepts

Conservation of Momentum
Conservation of Kinetic Energy
Elastic Collisions

Formulas

Conservation of Momentum: m1*v1i + m2*v2i = m1*v1f + m2*v2f
Conservation of Kinetic Energy: (1/2)*m1*v1i^2 + (1/2)*m2*v2i^2 = (1/2)*m1*v1f^2 + (1/2)*m2*v2f^2

Theorems

Law of Conservation of Momentum
Law of Conservation of Kinetic Energy

Suitable Grade Level

Grades 10-12