/*! 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 Two large blocks of wood are sli... [FREE SOLUTION] | 91Ó°ÊÓ

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Two large blocks of wood are sliding toward cach other on the frictionless surface of a frozen pond. Block \(A\) has mass \(4.00 \mathrm{~kg}\) and is initially sliding cast at \(2.00 \mathrm{~m} / \mathrm{s}\). Block \(B\) has mass \(6.00 \mathrm{~kg}\) and is initially sliding west at \(2.50 \mathrm{~m} / \mathrm{s}\). The blocks collide head-on. Aner the collision block \(B\) is sliding east at \(0.50 \mathrm{~m} / \mathrm{s}\). What is the decrease in the total kinctic encrgy of the two blocks as a result of the collision?

Short Answer

Expert verified
The decrease in the total kinetic energy of the two blocks as a result of the collision is 13.50 Joules.

Step by step solution

01

Find Initial Momenta

Calculate the initial momentum of each block. Remember that momentum (p) is given by the formula \( p = mv \) where m is the mass and v is the velocity. The total initial momentum is the sum of the momentum of block A and block B. Because they are moving in opposite directions, the momenta will have opposite signs. Block A momentum: \( p_{A1} = 4.00 kg * 2.00 m/s = 8.00 kg*m/s \). Block B momentum: \( p_{B1} = 6.00 kg * -2.50 m/s = -15.00 kg*m/s \). So total initial momentum, \( p_{1} = p_{A1} + p_{B1} = 8.00 kg*m/s -15.00 kg*m/s = -7.00 kg*m/s \).
02

Find Final Momenta

Now, find the final momentum of block B and use the conservation of momentum principle to find the final momentum of block A. After the collision, the momentum must still equal to the total initial momentum. We know Block B is sliding east at \(0.50 \mathrm{~m} / \mathrm{s}\), so its momentum is: \( p_{B2} = 6.00 kg * 0.50 m/s = 3.00 kg*m/s \). The final momentum of block A must be: \( p_{A2} = p_{1} - p_{B2} = -7.00 kg*m/s - 3.00 kg*m/s = -10.00 kg*m/s \). Since it's negative, block A is moving west. To find its velocity, divide its momenta by its mass: \( V_{A2} = p_{A2} / m_{A} = -10.00 kg*m/s / 4.00 kg = -2.50 m/s \).
03

Find Initial and Final Kinetic Energy

Calculate the initial and final kinetic energy (KE), which is given by the formula \(\frac{1}{2} * m * v^2\). For block A initial Kinetic energy: \( KE_{A1} = \frac{1}{2} * 4.00 kg * (2.00 m/s)^2 = 8.00 Joules \), for block B initial Kinetic energy: \( KE_{B1} = \frac{1}{2} * 6.00 kg * (-2.50 m/s)^2 = 18.75 Joules \). So the total initial kinetic energy \( KE_{1} = KE_{A1} + KE_{B1} = 8.00 Joules + 18.75 Joules = 26.75 Joules \). Now, the final kinetic energy for block A: \( KE_{A2} = \frac{1}{2} * 4.00 kg * (-2.50 m/s)^2 = 12.50 Joules \), for block B: \( KE_{B2} = \frac{1}{2} * 6.00 kg * (0.50 m/s)^2 = 0.75 Joules \). So the total final kinetic energy \( KE_{2} = KE_{A2} + KE_{B2} = 12.50 Joules + 0.75 Joules = 13.25 Joules \).
04

Find the Decrease in Kinetic Energy

The decrease in the total kinetic energy is given by the difference between the initial and the final total kinetic energy. So, the decrease in total kinetic energy: \( \Delta KE = KE_{1} - KE_{2} =26.75 Joules - 13.25 Joules = 13.50 Joules \).

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Key Concepts

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

Kinetic Energy
Kinetic Energy (KE) is the energy possessed by an object due to its motion. It's calculated using the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. For example, if a skater glides across an ice rink, the skater has kinetic energy proportional to their mass and the square of their speed. This energy can be transformed into other types, like potential energy, or can be transferred to other objects upon collision, which may lead to changes in the objects' velocities and thus their kinetic energies.

In our textbook problem, we analyze the kinetic energy of two sliding blocks before and after they collide. We notice that the total kinetic energy decreases as a result of the collision. This decrease is a key characteristic of inelastic collisions, where some kinetic energy is transformed into other forms of energy, such as sound, heat, or deformation energy. It is important to understand how kinetic energy is calculated and how it changes in different scenarios, as it provides valuable insights into the behavior of moving objects.
Inelastic Collision
An inelastic collision is a type of collision where the colliding objects stick together or deform, leading to a loss of kinetic energy in the system. Unlike elastic collisions, in which kinetic energy is conserved, inelastic collisions usually involve energy transformation into internal energy, such as heat or sound.

For instance, when two cars collide and crumple, the kinetic energy is not the same before and after the crash because part of it is used to deform the cars. Similarly, in the given exercise, when two wooden blocks collide and come to a stop or move together after the collision, some kinetic energy is lost. The degree of inelasticity can range from partial to perfectly inelastic, but the momentum is always conserved in such collisions, which leads us to the vital principle of 'conservation of momentum'.
Momentum and Collisions
Momentum is a measure of the quantity of motion an object has and is a vector quantity, meaning it has both magnitude and direction. It is expressed as the product of an object's mass (\( m \) and its velocity (\( v \): \( p = mv \). When discussing collisions, the Principle of Conservation of Momentum states that the total momentum of a system remains constant if no external forces act on it.

In the exercise at hand, two blocks collide on a frictionless surface, which implies no external forces are at play, so their total momentum before and after the collision must be the same. By calculating and comparing the momentum of each block before and after the event, we apply this principle to predict their velocities post-collision. Understanding momentum's role in collisions helps us solve problems involving multiple interacting objects and their subsequent motions.

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Most popular questions from this chapter

A \(68.5 \mathrm{~kg}\) astronaut is doing a repair in space on the orbiting space station. She throws a 2.25 kg tool away from her at \(3.20 \mathrm{~m} / \mathrm{s}\) relative to the space station. What will be the change in her speed as a result of this throw?

\(A 12.0 \mathrm{~kg}\) shell is launched at an angle of \(55.0^{\circ}\) above the horizontal with an initial speed of \(150 \mathrm{~m} / \mathrm{s}\). At its highest point, the shell explodes into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Ignore air resistance. If the heavier fragment lands back at the point from which the shell was launched, where will the lighter fragment land, and how much energy was relcased in the explosion?

Two identical \(0.900 \mathrm{~kg}\) masses are pressed against opposite ends of a light spring of force constant \(1.75 \mathrm{~N} / \mathrm{cm}\), compressing the spring by \(20.0 \mathrm{~cm}\) from its normal length. Find the speed of each mass when it has moved free of the spring on a frictionless, horizontal table.

On a very muddy football field, a \(110 \mathrm{~kg}\) linebucker tackles an \(85 \mathrm{~kg}\) halthnck. Immediately hefore the collision, the linchacker is slipping with a velocity of \(8.8 \mathrm{~m} / \mathrm{s}\) north and the halthack is sliding with a velocity of \(7.2 \mathrm{~m} / \mathrm{s}\) east. What is the velocity (magnitude and direction) at which the two players move together immediately after the cullision?

A10.0g marble slides to the left at a speed of \(0.400 \mathrm{~m} / \mathrm{s}\) on the frictionless, horizontal surface of an icy New York sidewalk and has a head-on, clastic collision with a larger \(30.0 \mathrm{~g}\) marble sliding to the right at speed of \(0.200 \mathrm{~m} / \mathrm{s}\) (Fig. \(\mathrm{E} 8.48\) ). (a) Find the velocity of each marble (magnitude and dircction) after the collision. (since the collision is head-on, all motion is along a line. (b) Calculate the change in momentum (the momentum after the collision minus the momentum before the collision) for each marble. Compare your values for each marble. (c) Calculate the change in kinetic energy (the kinctic energy after the collision minus the kinetic energy before the collision) for each marble. Compare your values for each marble.

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