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Chapter 9: Problem 55

A \(5.0 \mathrm{~kg}\) block with a speed of \(3.0 \mathrm{~m} / \mathrm{s}\) collides with a 10 \(\mathrm{kg}\) block that has a speed of \(2.0 \mathrm{~m} / \mathrm{s}\) in the same direction. After the collision, the \(10 \mathrm{~kg}\) block travels in the original direction with a speed of \(2.5 \mathrm{~m} / \mathrm{s}\). (a) What is the velocity of the \(5.0 \mathrm{~kg}\) block immediately after the collision? (b) By how much does the total kinetic energy of the system of two blocks change because of the collision? (c) Suppose, instead, that the \(10 \mathrm{~kg}\) block ends up with a speed of \(4.0 \mathrm{~m} / \mathrm{s}\). What then is the change in the total kinetic energy? (d) Account for the result you obtained in (c).

Short Answer

Expert verified
(a) 2.0 m/s. (b) Decrease by 1.25 J. (c) Increase by 40 J. (d) Requires energy input explanation.

Step by step solution

01

Identify Known Values

We have two blocks with masses and velocities:- Block 1: mass, \( m_1 = 5.0 \) kg, initial velocity, \( u_1 = 3.0 \) m/s.- Block 2: mass, \( m_2 = 10 \) kg, initial velocity, \( u_2 = 2.0 \) m/s. After the collision, the velocity of Block 2 is \( v_2 = 2.5 \) m/s.
02

Apply Conservation of Momentum (a)

Momentum before collision:\[ p_{initial} = m_1 \cdot u_1 + m_2 \cdot u_2 = 5.0 \times 3.0 + 10 \times 2.0 = 15 + 20 = 35 \text{ kg m/s}. \]Momentum after collision should be equal to momentum before collision:\[ p_{final} = m_1 \cdot v_1 + m_2 \cdot v_2, \]\[ 35 = 5.0 \cdot v_1 + 10 \cdot 2.5. \]Solve for \( v_1 \):\[ 35 = 5.0 \cdot v_1 + 25. \]\[ 10 = 5.0 \cdot v_1. \]\[ v_1 = 2.0 \text{ m/s}. \]The velocity of the 5.0 kg block after the collision is 2.0 m/s.
03

Calculate Change in Kinetic Energy (b)

Initial kinetic energy:\[ KE_{initial} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} (5.0)(3.0)^2 + \frac{1}{2} (10)(2.0)^2 \]\[ KE_{initial} = 22.5 + 20 = 42.5 \text{ J}. \]Final kinetic energy:\[ KE_{final} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (5.0)(2.0)^2 + \frac{1}{2} (10)(2.5)^2 \]\[ KE_{final} = 10 + 31.25 = 41.25 \text{ J}. \]Change in kinetic energy:\[ \Delta KE = KE_{final} - KE_{initial} = 41.25 - 42.5 = -1.25 \text{ J}. \]
04

Analyze Change for Alternate Speed (c)

Suppose the \(10 \text{ kg}\) block ends with a speed of \(4.0 \text{ m/s}\):**Find \(v_1\):**\[ 35 = 5.0 \cdot v_1 + 10 \cdot 4.0, \]\[ 35 = 5.0 \cdot v_1 + 40, \]\[ -5 = 5.0 \cdot v_1, \]\[ v_1 = -1.0 \text{ m/s}. \]Initial kinetic energy remains \( 42.5 \text{ J} \).New final kinetic energy:\[ KE_{final} = \frac{1}{2} (5.0)(-1.0)^2 + \frac{1}{2} (10)(4.0)^2, \]\[ KE_{final} = 2.5 + 80 = 82.5 \text{ J}. \]Change in kinetic energy:\[ \Delta KE = 82.5 - 42.5 = 40 \text{ J}. \]
05

Explain the Result in (d)

The final scenario in (c) depicts an inelastic collision where the system gains kinetic energy by storing potential energy or using other energy sources. The change suggests an energy input or conversion not accounted for solely by kinetic factors. It is physically improbable without external energy contributing to the system.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It can be calculated using the formula: \[ KE = \frac{1}{2}mv^2 \] where:
  • \( KE \) is the kinetic energy,
  • \( m \) is the mass of the object, and
  • \( v \) is the velocity of the object.
The faster an object moves, the greater its kinetic energy. Similarly, the heavier the object, the more kinetic energy it will have at a given speed. In collision problems, we often calculate the kinetic energy before and after the event to understand energy changes. These changes can indicate if some energy was lost as heat or sound in the collision, as often happens in inelastic collisions.
Inelastic Collision
In an inelastic collision, the objects involved do not retain their complete kinetic energy. Instead, some energy is usually lost to heat, sound, or deformation. However, momentum is always conserved in such collisions.
In the exercise, two blocks collide, and after the collision, their velocities change, but the total momentum of the system remains constant. This principle helps us calculate unknown velocities post-collision.
The characteristic of inelastic collisions is that the total kinetic energy after the collision is generally less than the total kinetic energy before the collision due to these energy transformations. For learners, it's crucial to understand that while kinetic energy is not conserved in inelastic collisions, momentum always is.
Potential Energy
Potential energy is stored energy that depends on the position or configuration of a system. For example, gravitational potential energy depends on an object's height and mass, using the formula: \[ PE = mgh \] where:
  • \( PE \) is the potential energy,
  • \( m \) is the mass,
  • \( g \) is the acceleration due to gravity, and
  • \( h \) is the height.
While potential energy wasn't directly calculated in this specific exercise, understanding energy conservation requires recognizing how potential energy can convert to kinetic energy and vice versa. In certain scenarios, potential energy stored in a system might transform, affecting the kinetic energy calculations.
Mass and Velocity Calculations
Calculating mass and velocity accurately is vital for analyzing collisions. These calculations stem from the conservation of momentum principle, expressed as: \[ p = mv \] where:
  • \( p \) is momentum,
  • \( m \) is mass, and
  • \( v \) is velocity.
By calculating the initial momentum of the system and knowing the final momentum should remain equal, we can solve for unknown velocities post-collision. Applying these principles in step-by-step calculations allows us to find how velocity changes during collisions.
This skill is indispensable in physics to understand how interacting objects influence each other's motion, especially in complex systems.

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Most popular questions from this chapter

A \(2100 \mathrm{~kg}\) truck traveling north at \(41 \mathrm{~km} / \mathrm{h}\) turns east and accelerates to \(51 \mathrm{~km} / \mathrm{h} .\) (a) What is the change in the truck's kinetic energy? What are the (b) magnitude and (c) direction of the change in its momentum?

A \(2140 \mathrm{~kg}\) railroad flatcar, which can move with negligible friction, is motionless next to a platform. A \(242 \mathrm{~kg}\) sumo wrestler runs at \(5.3 \mathrm{~m} / \mathrm{s}\) along the platform (parallel to the track) and then jumps onto the flatcar. What is the speed of the flatcar if he then (a) stands on it, \((\mathrm{b})\) runs at \(5.3 \mathrm{~m} / \mathrm{s}\) relative to it in his original direction, and (c) turns and runs at \(5.3 \mathrm{~m} / \mathrm{s}\) relative to the flatcar opposite his original direction?

An old Chrysler with mass \(2400 \mathrm{~kg}\) is moving along a straight stretch of road at \(80 \mathrm{~km} / \mathrm{h}\). It is followed by a Ford with mass 1600 kg moving at \(60 \mathrm{~km} / \mathrm{h}\). How fast is the center of mass of the two cars moving?

A steel ball of mass \(0.500 \mathrm{~kg}\) is fastened to a cord that is \(70.0 \mathrm{~cm}\) long and fixed at the far end. The ball is then released when the cord is horizontal (Fig. 9-65). At the bottom of its path, the ball strikes a \(2.50 \mathrm{~kg}\) steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.

A projectile proton with a speed of \(500 \mathrm{~m} / \mathrm{s}\) collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at \(60^{\circ}\) from the original direction. After the collision, what are the speeds of (a) the target proton and (b) the projectile proton?
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