/*! 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 35 The \(5-\mathrm{Mg}\) bus \(B\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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The \(5-\mathrm{Mg}\) bus \(B\) is traveling to the right at \(20 \mathrm{m} / \mathrm{s}\). Meanwhile a \(2-\mathrm{Mg}\) car \(A\) is traveling at \(15 \mathrm{m} / \mathrm{s}\) to the right. If the vehicles crash and become entangled, determine their common velocity just after the collision. Assume that the vehicles are free to roll during collision.

Short Answer

Expert verified
The common velocity of the bus and the car immediately after the collision is 18.57 m/s to the right.

Step by step solution

01

Calculate initial momentum

The initial momentum of the system is the sum of the momentum of the bus and car before the collision. Momentum is calculated as the product of mass and velocity. So for bus \(B\), which has a mass of 5 Mg (which is \(5 \times 10^6 \)g = \(5 \times 10^3 \) kg because 1Mg = 1000kg) and is moving at 20 m/s to the right, the momentum will be \(5 \times 10^3 \times 20 = 1 \times 10^5 kg \cdot m/s\). For the car \(A\), which has a mass of 2 Mg (which is \(2 \times 10^3 \) kg) and is moving at 15 m/s to the right, the momentum will be \(2 \times 10^3 \times 15 =3 \times 10^4 kg \cdot m/s\). So total initial momentum will be \(1 \times 10^5 + 3 \times 10^4 = 1.3 \times 10^5 kg \cdot m/s\) to the right.
02

Calculate final momentum

The final momentum of the system after collision is the momentum of the combined mass of the bus and the car moving with the common velocity \(v\), which we don't know yet. The total mass of the bus and the car is \(5 \times 10^3 + 2 \times 10^3 = 7 \times 10^3\) kg. Therefore, the final momentum is \(7 \times 10^3 \times v\) kg.m/s.
03

Conservation of Momentum

Now we use the principle of conservation of momentum that the initial momentum equals the final momentum to solve for the common velocity \(v\). Thus we find that \(1.3 \times 10^5 = 7 \times 10^3 \times v\). Solving for \(v\), we get \(v= 1.3 \times 10^5 / 7 \times 10^3 = 18.57 m/s\). This is the common velocity of the combined bus and car immediately after the collision.

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Key Concepts

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

Momentum
Understanding momentum is crucial in physics, especially when analyzing moving objects and collisions. Momentum, represented by the symbol p, is the product of an object’s mass (m) and its velocity (v), and is given by the formula p = m * v. In the context of the exercise, a bus and a car are traveling to the right with different velocities and masses.

We calculate the initial momentum of both individually and add them together to find the system's total momentum before the collision. The bus, with a mass of 5000 kg (5 Mg), and velocity of 20 m/s, has a momentum of 100,000 kgâ‹…m/s. The car, lighter at 2000 kg (2 Mg), moving at 15 m/s, contributes another 30,000 kgâ‹…m/s. This results in a combined momentum of 130,000 kgâ‹…m/s. Momentum is a vector quantity, meaning it has both magnitude and direction, which is why velocities are direction-specific; in this case, to the right.
Inelastic Collision
An inelastic collision, unlike a perfectly elastic collision where objects bounce off each other without losing kinetic energy, is one where colliding objects stick together after impact and move with a common velocity, losing some kinetic energy in the process. This kind of collision is exemplified by the bus and car that become entangled.

The law of conservation of momentum still applies; the total momentum before the collision equals the total momentum after the collision. However, because kinetic energy is not conserved, some is converted into other forms of energy, such as sound or heat. In practical terms, like our roadway scenario, the deformation of the vehicles during collision accounts for the energy loss.
Common Velocity
After an inelastic collision, the two objects move together with a shared or 'common' velocity. This is a single velocity value that describes the motion of the combined mass of the colliding objects. For the bus and car, we calculate it using the conservation of momentum principle.

With the total system momentum and the combined mass known, we can find their common velocity post-collision. The formula for this scenario is derived from setting the initial momentum equal to the final momentum of the combined masses, then solving for the final velocity. v = p / (mbus + mcar), where p is the total momentum before collision, and mbus and mcar are the masses of the bus and car, respectively.
Mass-Velocity Product
The mass-velocity product is essentially another term for momentum and highlights the direct contribution of an object's mass and velocity to its momentum. In our exercise, we first determine the product for both the bus and car, then use these to calculate the common velocity after their inelastic collision.

The mass-velocity product accents the idea that both mass and velocity are equally important in determining how much momentum an object has. A larger mass or a higher velocity will result in greater momentum. In the collision between the bus and car, their individual mass-velocity products add up to the system’s total momentum pre-collision, which remains conserved post-collision, even though their individual velocities no longer exist, replaced by a new common velocity.

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Most popular questions from this chapter

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