/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 214 Two barges, each with a displace... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 214

Two barges, each with a displacement (mass) of \(500 \mathrm{Mg},\) are loosely moored in calm water. A stunt driver starts his 1500 -kg car from rest at \(A\), drives along the deck, and leaves the end of the \(15^{\circ}\) ramp at a speed of \(50 \mathrm{km} / \mathrm{h}\) relative to the barge and ramp. The driver successfully jumps the gap and brings his car to rest relative to barge 2 at \(B\). Calculate the velocity \(v_{2}\) imparted to barge 2 just after the car has come to rest on the barge. Neglect the resistance of the water to motion at the low velocities involved.

Short Answer

Expert verified
The velocity imparted to barge 2 is approximately \(0.0415\) m/s.

Step by step solution

01

Identify and Define Greek Variables

Let the velocity of barge 2 be \( v_2 \), the initial velocity of the car relative to the barge and ramp be \( v_{c, rel} = 50 \text{ km/h} \) (which is \( \frac{50 \times 1000}{3600} \text{ m/s} \)), and the mass of the car be \( m_c = 1500 \text{ kg} \). Convert \( v_{c, rel} \) to meters per second for ease of calculation.
02

Initial Momentum Considerations

Initially, only the car is moving relative to the Earth as it leaves the ramp. Convert its relative speed to absolute speed by noting that the barge is initially stationary and doesn't contribute to initial momentum except for the car. The initial velocity of the car in meters per second \( v_{c, abs} = v_{c, rel} \).
03

Conservation of Momentum

Momentum before and after the car comes to rest must be conserved. Only linear motion, as water resistance is negligible, is considered. The initial momentum of the system:\[m_i = m_c \cdot v_{c, abs}\]After coming to rest on barge 2, the car and barge 2's total mass is \( m_c + m_b \), and the final velocity of the car-barge 2 system is \( v_2 \). The final momentum is:\[m_f = (m_c + m_b) \cdot v_2 \]Equating initial and final momentum:\[m_c \cdot v_{c, abs} = (m_c + m_b) \cdot v_2\]
04

Solving for Barge Velocity

Rearrange to solve for \( v_2 \):\[v_2 = \frac{m_c \cdot v_{c, abs}}{m_c + m_b}\]Substitute \( m_b = 500 \times 10^3 \text{ kg} \):\[v_2 = \frac{1500 \text{ kg} \times \frac{50000 \text{ m/s}}{3600}}{1500 \text{ kg} + 500000 \text{ kg}}\]
05

Calculate Final Velocity

Compute \( v_2 \):1. Convert \( v_{c, rel} \) to \( v_{c, abs} \): \( \approx 13.89 \text{ m/s} \)2. Solve using the formula in Step 4 for \( v_2 \):\[v_2 = \frac{1500 \times 13.89}{1500 + 500000} \approx 0.0415 \text{ m/s}\]

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Key Concepts

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

Velocity Calculation
Velocity is essentially the speed of something in a specific direction and is an important concept when dealing with motion. For our physics problem, we need to find the velocity of barge 2 after a car lands on it. To do this, we make use of the conservation of momentum. Start by understanding that the initial velocity of the car relative to the barge is given as 50 km/h. This needs to be converted to meters per second, which requires the formula: \[ v_{c, rel} = \frac{50 \times 1000}{3600} \approx 13.89 \text{ m/s} \]Once the car lands on barge 2 and comes to a halt, barge 2 will have gained velocity. By using the principle of conservation of momentum, we can set up the equation and solve for this velocity. Calculating that velocity involves plugging into the equation derived from equating initial and final momentum.
Linear Motion
Linear motion refers to movement in a straight line. In this exercise, we consider both the car and the barges as moving linearly. Initially, the car moves across a flat ramp and eventually changes direction slightly due to the 15掳 angle. However, we focus on the linear component of its velocity. Given the car and barge are treated as moving along a straight path, we apply linear momentum concepts. After the car comes to rest on barge 2, its motion has transferred to the barge leading it to move in a linear motion as well. This is calculated using the conservation of linear momentum where movement in only one direction is considered since the problem specifically neglects lateral and water resistance.
Physics Problem-Solving
Solving physics problems effectively involves breaking them down into manageable steps. This problem requires applying the conservation of momentum, a key physics principle utilized to predict changes during interactions like collisions or separations between bodies. Let's discuss some practical tips for solving such exercises:
  • Identify all given data clearly and convert units when necessary to ensure consistency.
  • Determine which principles or laws apply, such as momentum conservation when addressing collision or transfer of motion.
  • Set up equations based on these principles and write each step carefully to keep track of thoughts and solutions.
  • Verify results against the context of the problem; ensure they make physical sense.
By following these steps and ensuring a thorough understanding of fundamental concepts, solving similar physics problems becomes a more structured and less daunting task.

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