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(I) A \(7700-\mathrm{kg}\) boxcar traveling \(18 \mathrm{~m} / \mathrm{s}\) strikes a second car. The two stick together and move off with a speed of \(5.0 \mathrm{~m} / \mathrm{s} .\) What is the mass of the second car?

Short Answer

Expert verified
The mass of the second car is 20020 kg.

Step by step solution

01

Understand the Conservation of Momentum Principle

The law of conservation of momentum states that the total momentum of a closed system is constant if no external forces act on it. In this scenario, the total momentum before the collision is equal to the total momentum after the collision.
02

Write the Momentum Equation Before Collision

The momentum of the system before the collision consists only of the boxcar since the second car is stationary initially. The momentum is the product of mass and velocity: \[ p_{ ext{initial}} = m_1 imes v_1 = 7700 ext{ kg} imes 18 ext{ m/s} \]
03

Calculate Initial Momentum

Substitute the given values to calculate the initial momentum:\[ p_{ ext{initial}} = 7700 imes 18 = 138600 ext{ kg m/s} \]
04

Write the Momentum Equation After Collision

After the collision, both the boxcar and the second car are moving together with a speed of 5.0 m/s. The combined mass is \(m_1 + m_2\), and the momentum of the system is:\[ p_{ ext{final}} = (m_1 + m_2) imes 5 \]
05

Set Initial and Final Momentum Equal

According to the conservation of momentum:\[ p_{ ext{initial}} = p_{ ext{final}} \]So, we have:\[ 138600 = (7700 + m_2) imes 5 \]
06

Solve for the Mass of the Second Car

To find \(m_2\), the mass of the second car, divide both sides by 5:\[ 27720 = 7700 + m_2 \]Now, solve for \(m_2\):\[ m_2 = 27720 - 7700 = 20020 \]
07

Conclusion

The mass of the second car is 20020 kg based on the conservation of momentum in the system.

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics that describes the quantity of motion an object has. It is calculated as the product of an object's mass and its velocity. The formula for momentum \( p \) is given by:\[ p = m \times v \]where \( m \) is the mass and \( v \) is the velocity of the object.

Momentum is a vector quantity, meaning it has both magnitude and direction. This is important for understanding how objects move and interact with each other. For example, in collisions, momentum plays a critical role in determining the final velocities of objects. It's important for students to distinguish momentum from simple speed or kinetic energy; while related, these concepts are not interchangeable. Understanding momentum helps predict the results of interactions in a variety of physical systems.
Collision
A collision occurs when two or more objects come into contact in such a way that they exert forces on each other. In this exercise, we specifically focus on an inelastic collision, where the objects stick together after colliding.

During a collision:
  • Momentum is transferred between the colliding bodies.
  • Some energy might be lost to factors like sound, heat, or deformation.
The key principle for solving collision problems is the conservation of momentum. This principle states that in the absence of external forces, the total momentum of a system remains unchanged through the collision. Therefore, knowing the initial and final momentum allows us to solve for unknown properties, such as the mass of an unknown object in our example.

In real-world applications, understanding collisions helps in fields like automotive safety, where minimizing the impact on passengers during collisions is crucial.
Mass Calculation
Calculating mass in the context of conservation of momentum involves using the momentum equations before and after a collision. The exercise demonstrates this with two colliding boxcars.

Steps to Calculate Mass in a Collision:
  • Determine the initial momentum using the formula \( p = m_1 \times v_1 \).
  • Express the final combined momentum as \( (m_1 + m_2) \times v_f \).
  • Set initial and final momentum equal due to conservation of momentum: \( m_1 \times v_1 = (m_1 + m_2) \times v_f \).
  • Solve for the unknown mass \( m_2 \) by rearranging the equation.
In this specific problem, the calculation reveals that the mass of the second car is 20020 kg after considering the initial momentum of the moving boxcar and the final velocity after collision.

Understanding these calculations helps reinforce the concept of momentum conservation and illustrates how theoretical physics applies to practical problems, such as calculating the effects of vehicle collisions.

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Most popular questions from this chapter

A massless spring with spring constant \(k\) is placed between a block of mass \(m\) and a block of mass \(3 m .\) Initially the blocks are at rest on a frictionless surface and they are held together so that the spring between them is compressed by an amount \(D\) from its equilibrium length. The blocks are then released and the spring pushes them off in opposite directions. Find the speeds of the two blocks when they detach from the spring.

(II) Determine the fraction of kinetic energy lost by a neutron \(\left(m_{1}=1.01 \mathrm{u}\right)\) when it collides head-on and elastically with a target particle at rest which is \((a){ }_{1}^{1} \mathrm{H}(m=1.01 \mathrm{u})\) (b) \({ }_{1}^{2} \mathrm{H}\) (heavy hydrogen, \(\left.m=2.01 \mathrm{u}\right) ;(c){ }_{6}^{12} \mathrm{C}(m=12.00 \mathrm{u})\) (d) \({ }^{208} \mathrm{~Pb}\) (lead, \(m=208 \mathrm{u}\) ).

(III) A huge balloon and its gondola, of mass \(M,\) are in the air and stationary with respect to the ground. A passenger, of mass \(m\), then climbs out and slides down a rope with speed \(v\), measured with respect to the balloon. With what speed and direction (relative to Earth) does the balloon then move? What happens if the passenger stops?

A 4800 -kg open railroad car coasts along with a constant speed of 8.60 \(\mathrm{m} / \mathrm{s}\) on a level track. Snow begins to fall vertically and fills the car at a rate of 3.80 \(\mathrm{kg} / \mathrm{min}\) . Ignoring friction with the tracks, what is the speed of the car after 60.0 \(\mathrm{min}\) ? (See Section 2 of "Linear Momentum.")

(II) The force on a bullet is given by the formula \(F=\left[740-\left(2.3 \times 10^{5} \mathrm{~s}^{-1}\right) t\right] \mathrm{N}\) over the time interval \(t=0\) to \(t=3.0 \times 10^{-3} \mathrm{~s}\). ( \(a\) ) Plot a graph of \(F\) versus \(t\) for \(t=0\) to \(t=3.0 \mathrm{~ms} .\) (b) Use the graph to estimate the impulse given the bullet. ( \(c\) ) Determine the impulse by integration. ( \(d\) ) If the bullet achieves a speed of \(260 \mathrm{~m} / \mathrm{s}\) as a result of this impulse, given to it in the barrel of a gun, what must the bullet's mass be? \((e)\) What is the recoil speed of the \(4.5-\mathrm{kg}\) gun?

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