Math Problem Statement

Ann (mass 50 kg) is standing at the left end of a 15-m-long, 500 kg cart that has frictionless wheels and rolls on a frictionless track. Initially both Ann and the cart are at rest. Suddenly, Ann starts running along the cart at a speed of 5.0m/s relative to the cart. How far will Ann have run relative to the ground when she reaches the right end of the cart?

Solution

This problem involves the principle of conservation of momentum and relative velocity. Here's how we solve it step by step:

Key Information

  1. Mass of Ann, mA=50kgm_A = 50 \, \text{kg}
  2. Mass of the cart, mC=500kgm_C = 500 \, \text{kg}
  3. Initial velocity of Ann and cart: vA,initial=vC,initial=0v_{A, \text{initial}} = v_{C, \text{initial}} = 0
  4. Velocity of Ann relative to the cart: vA/C=5.0m/sv_{A/C} = 5.0 \, \text{m/s}
  5. Length of the cart: L=15mL = 15 \, \text{m}

Step 1: Momentum Conservation

Since the system is isolated and no external horizontal forces act on it, the total momentum of the system remains conserved. Let vCv_C be the velocity of the cart relative to the ground after Ann starts running. Ann’s velocity relative to the ground is vA=vC+5.0v_A = v_C + 5.0.

The total momentum equation is: mAvA+mCvC=0m_A v_A + m_C v_C = 0

Substitute vA=vC+5.0v_A = v_C + 5.0: mA(vC+5.0)+mCvC=0m_A (v_C + 5.0) + m_C v_C = 0

Expanding and solving for vCv_C: 50(vC+5.0)+500vC=050(v_C + 5.0) + 500v_C = 0 50vC+250+500vC=050v_C + 250 + 500v_C = 0 550vC=250550v_C = -250 vC=250550=0.4545m/sv_C = -\frac{250}{550} = -0.4545 \, \text{m/s}

The cart moves in the opposite direction of Ann’s motion with a velocity of vC=0.4545m/sv_C = -0.4545 \, \text{m/s}.

Step 2: Relative Motion and Distance

Ann runs a distance of L=15mL = 15 \, \text{m} relative to the cart. Her velocity relative to the ground is: vA=vC+5.0=0.4545+5.0=4.545m/sv_A = v_C + 5.0 = -0.4545 + 5.0 = 4.545 \, \text{m/s}

The time it takes for Ann to traverse the 15 m relative to the cart is: t=distancevelocity relative to cart=155.0=3.0secondst = \frac{\text{distance}}{\text{velocity relative to cart}} = \frac{15}{5.0} = 3.0 \, \text{seconds}

During this time, the distance Ann covers relative to the ground is: distance relative to ground=vAt=4.5453.0=13.64m\text{distance relative to ground} = v_A \cdot t = 4.545 \cdot 3.0 = 13.64 \, \text{m}

Final Answer:

Ann will have run 13.64 m relative to the ground when she reaches the right end of the cart.


Would you like further clarification or additional examples? Here are five related questions for you to explore:

  1. What happens if the cart had a different mass? How would Ann’s distance relative to the ground change?
  2. What would the result be if Ann ran at a different speed relative to the cart?
  3. How does the principle of conservation of momentum apply to other scenarios, like collisions?
  4. How would friction on the cart's wheels affect the results?
  5. How does the relative motion concept apply when Ann runs faster than the speed of the cart?

Tip: Always identify the relative frame of reference clearly in momentum problems to avoid confusion.

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Math Problem Analysis

Mathematical Concepts

Conservation of Momentum
Relative Motion
Kinematics

Formulas

Conservation of Momentum: m_A * v_A + m_C * v_C = 0
Relative Velocity: v_A = v_C + v_A/C
Distance-Time Relationship: distance = velocity * time

Theorems

Conservation of Momentum

Suitable Grade Level

Grades 10-12