/*! 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 33 A \(2.0 \mathrm{kg}\) block slid... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(2.0 \mathrm{kg}\) block slides along a frictionless surface at \(1.0 \mathrm{m} / \mathrm{s}\). A second block, sliding at \(4.0 \mathrm{m} / \mathrm{s},\) collides with the first from behind and sticks to it. The final velocity of the combined blocks is \(2.0 \mathrm{m} / \mathrm{s}\). What was the mass of the second block?

Short Answer

Expert verified
The mass of the second block is 1.0 kg.

Step by step solution

01

Identify the Given Variables

The known variables in this problem are the initial speed of the first block (\(v1i = 1.0 \, m/s\) and its mass \(m1 = 2.0 \, kg\)) and the initial speed of the second block (\(v2i = 4.0 \, m/s\)). The final speed of the two blocks after the collision (\(v_f = 2.0 \, m/s\)) is also known.
02

Apply the Principle of Conservation of Momentum

The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act upon it. This can be expressed as \(m1 * v1i + m2 * v2i = (m1 + m2) * v_f\), where \(m2\) is the mass of the second block, which we are trying to calculate.
03

Solve for the Unknown

We can rearrange the above equation to solve for \(m2\). Doing so gives us: \(m2 = (m1 + m2) * v_f - m1 * v1i / v2i \) Replacing all known values gives us \(m2 = ((2.0 \, kg) * (2.0 \, m/s) - (2.0 \, kg) * (1.0 \, m/s)) / (4.0 \, m/s) = 1.0 \, kg\)

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Key Concepts

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

Understanding Collisions
Collisions are common events in our day-to-day lives, whether it's during a game of pool or cars on a road. They occur when two objects come into contact and exert forces on each other over a short period.
In the context of physics, a collision is noteworthy because it's an interaction where energy and momentum can be transferred or transformed. There are different types of collisions, with two primary kinds being elastic and inelastic. During elastic collisions, both momentum and kinetic energy are conserved. In contrast, inelastic collisions, like the one in our problem, only conserve momentum.
In this exercise, we dealt with an inelastic collision because the two blocks stick together after the collision. Understanding the type of collision helps us know which physical quantities to conserve and analyze, significantly aiding in problem-solving.
Momentum Basics
The concept of momentum plays a pivotal role in understanding collisions. Momentum is a property of moving objects and can be thought of as the "quantity of motion" an object has. Mathematically, it's the product of an object's mass and velocity.
Momentum is a vector quantity, meaning it has both magnitude and direction. This characteristic is particularly important when dealing with collisions because the direction can affect how momentum is transferred between objects.
In physics problems like the one provided, the law of conservation of momentum is crucial. This law tells us that if no external forces are acting, the total momentum just before the collision is equal to the total momentum just after. It's this principle that allows us to solve for unknowns in collision problems, as seen with the mass of the second block in our problem.
Approaching Physics Problems
Tackling physics problems effectively requires a structured approach. First, it's essential to clearly define and understand all given information. In our problem, the initial velocities and masses were key starters. Organizing this data helps in laying the groundwork for applying concepts.
Next, apply relevant physics principles. For collision problems, laws such as the conservation of momentum are paramount. These laws guide the formation of equations needed to find unknowns, acting as the bridge between given data and solutions.
Finally, solve the equations step-by-step. This involves algebraic manipulation to isolate unknowns. In our case, rearranging the conservation of momentum equation was necessary to determine the second block's mass. Always double-check calculations for accuracy, ensuring a reliable and correct solution. Combining these steps makes solving physics problems more manageable and less daunting.

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Most popular questions from this chapter

One billiard ball is shot east at \(2.00 \mathrm{m} / \mathrm{s}\). A second, identical billiard ball is shot west at \(1.00 \mathrm{m} / \mathrm{s}\). The balls have a glancing collision, not a head-on collision, deflecting the second ball by \(90^{\circ}\) and sending it north at \(1.41 \mathrm{m} / \mathrm{s}\). What are the speed and direction of the first ball after the collision?

A tennis player swings her 1000 g racket with a speed of \(10 \mathrm{m} / \mathrm{s} .\) She hits a \(60 \mathrm{g}\) tennis ball that was approaching her at a speed of \(20 \mathrm{m} / \mathrm{s} .\) The ball rebounds at \(40 \mathrm{m} / \mathrm{s} .\) a. How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision. b. If the tennis ball and racket are in contact for \(10 \mathrm{ms}\), what is the average force that the racket exerts on the ball?

A manufacturer makes a golf ball that compresses more than a traditional golf ball when struck by a club. How will this affect the average force during the collision? A. The force will decrease. B. The force will not be affected. C. The force will increase.

The carbon isotope \({ }^{14} \mathrm{C}\) is used for carbon dating of archeological artifacts. \({ }^{14} \mathrm{C}\) (mass \(2.34 \times 10^{-26} \mathrm{kg}\) ) decays by the process known as beta decay in which the nucleus emits an electron (the beta particle) and a subatomic particle called a neutrino. In one such decay, the electron and the neutrino are emitted at right angles to each other. The electron (mass \(9.11 \times 10^{-31} \mathrm{kg}\) ) has a speed of \(5.00 \times 10^{7} \mathrm{m} / \mathrm{s}\) and the neutrino has a momentum of \(8.00 \times 10^{-24} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .\) What is the recoil speed of the nucleus?

\(A\) typical raindrop is much more massive than a mosquito and much faster than a mosquito flies. How does a mosquito survive the impact? Recent research has found that the collision of a falling raindrop with a mosquito is a perfectly inelastic collision. That is, the mosquito is "swept up" by the raindrop and ends up traveling along with the raindrop. Once the relative speed between the mosquito and the raindrop is zero, the mosquito is able to detach itself from the drop and fly away. a. A hovering mosquito is hit by a raindrop that is 40 times as massive and falling at \(8.2 \mathrm{m} / \mathrm{s},\) a typical raindrop speed. How fast is the raindrop, with the attached mosquito, falling immediately afterward if the collision is perfectly inelastic? b. Because a raindrop is "soft" and deformable, the collision duration is a relatively long \(8.0 \mathrm{ms}\). What is the mosquito's average acceleration, in \(g\) 's, during the collision? The peak acceleration is roughly twice the value you found, but the mosquito's rigid exoskeleton allows it to survive accelerations of this magnitude. In contrast, humans cannot survive an acceleration of more than about \(10 \mathrm{g}\).

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