/*! 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 88 Steel Ball and Block A steel bal... [FREE SOLUTION] | 91Ó°ÊÓ

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Steel Ball and Block 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. 10-61). 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.

Short Answer

Expert verified
The speed of the ball is 1.30 m/s and the speed of the block is 2.40 m/s after the collision.

Step by step solution

01

Calculate the initial potential energy of the ball

When the ball is released from the horizontal position, it has potential energy due to its height. The height of the ball is equal to the length of the cord, which is 70 cm or 0.7 m. The potential energy (PE) can be calculated as: \[ PE = mgh \] where \( m = 0.5 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( h = 0.7 \text{ m} \). Substituting in the values: \[ PE = 0.5 \times 9.8 \times 0.7 \] \[ PE = 3.43 \text{ J} \]
02

Determine the speed of the ball just before the collision

At the bottom of its swing, all of the ball's potential energy is converted into kinetic energy (KE). Using the conservation of energy, we can write: \[ KE = PE \] The kinetic energy of the ball can also be expressed as: \[ KE = \frac{1}{2} mv^2 \] Setting \( KE = 3.43 \text{ J} \) and solving for \( v \): \[ \frac{1}{2} \times 0.5 \times v^2 = 3.43 \] \[ 0.25v^2 = 3.43 \] \[ v^2 = \frac{3.43}{0.25} \] \[ v^2 = 13.72 \] \[ v = \sqrt{13.72} \] \[ v \approx 3.70 \text{ m/s} \]
03

Use conservation of momentum for the collision

For an elastic collision, both momentum and kinetic energy are conserved. Let \( v_{b1} \) and \( v_{B1} \) be the velocities of the ball and block after the collision, respectively. The initial momentum is: \[ p_{initial} = m_b v_{b0} + M_B v_{B0} \] Since the block is initially at rest, \( v_{B0} = 0 \). Thus, \[ p_{initial} = (0.5 \text{ kg})(3.70 \text{ m/s}) = 1.85 \text{ kg m/s} \] The final momentum is: \[ p_{final} = m_b v_{b1} + M_B v_{B1} \] Using conservation of momentum: \[ (0.5 \text{ kg})(3.70 \text{ m/s}) = (0.5 \text{ kg})v_{b1} + (2.5 \text{ kg})v_{B1} \] \[ 1.85 = 0.5v_{b1} + 2.5v_{B1} \]
04

Use conservation of kinetic energy for the collision

The initial kinetic energy is: \[ KE_{initial} = \frac{1}{2} m_b v_{b0}^2 \] Using the known values: \[ KE_{initial} = \frac{1}{2} (0.5)(3.70^2) = 3.43 \text{ J} \] Conservation of kinetic energy after collision: \[ KE_{final} = \frac{1}{2} m_b v_{b1}^2 + \frac{1}{2} M_B v_{B1}^2 \] \[ 3.43 = \frac{1}{2} (0.5)v_{b1}^2 + \frac{1}{2} (2.5)v_{B1}^2 \] Solving this with the momentum equation in step 3, will give you the velocities.
05

Solve the system of equations

From momentum conservation: \[ 1.85 = 0.5v_{b1} + 2.5v_{B1} \] From kinetic energy conservation: \[ 3.43 = 0.25v_{b1}^2 + 1.25v_{B1}^2 \] Solve these equations simultaneously for \( v_{b1} \) and \( v_{B1} \).

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Key Concepts

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

conservation of energy
Energy conservation is fundamental in physics. It states that energy can't be created or destroyed, only changed from one form to another. In elastic collisions, two types of energy are in the spotlight: potential energy (PE) and kinetic energy (KE). Initially, the steel ball has potential energy because it's held at a height. The potential energy can be calculated using the formula: \(PE = mgh\), where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the height. When the ball is released and swings down, this potential energy transforms into kinetic energy, described by \(KE = \frac{1}{2}mv^2\). In our exercise, we used this principle to calculate how fast the ball moves just before hitting the block.
conservation of momentum
Momentum is a key concept in understanding collisions. It describes the quantity of motion an object has and is given by the product of its mass and velocity \(p = mv\). During an elastic collision, the total momentum before the collision equals the total momentum after. For our ball-block system, we started with the momentum of the moving ball since the block was stationary. After the collision, both the ball and block have new velocities, which allowed us to use the conservation of momentum equation: \(m_b v_{b0} = m_b v_{b1} + M_B v_{B1}\), where the subscripts indicate before (0) and after (1) the collision. This relationship helped us find both final velocities post-collision.
potential energy
Potential energy in our scenario is the energy stored in the steel ball due to its position above the ground. This energy is calculated when the ball is at the height equivalent to the length of the cord (0.7 m). Using the formula \(PE = mgh\), we calculated the ball’s initial potential energy as 3.43 J. This energy is crucial as it transforms into kinetic energy when the ball starts moving, determining how fast the ball will move right before it hits the block. Understanding potential energy helps in predicting the behavior of objects in gravity fields and is a vital part of many physics problems.
kinetic energy
Kinetic energy is the energy of motion. As the steel ball swings down, its potential energy converts into kinetic energy. Right before colliding with the block, all the initial potential energy becomes kinetic energy, given by \(KE = \frac{1}{2}mv^2\). In our example, we used the conservation of energy to determine that the ball’s kinetic energy right before collision is 3.43 J, leading to a velocity of about 3.70 m/s. The concept also continues to be relevant after the collision, as kinetic energy will be shared between the ball and the block, both moving with new velocities.
problem-solving in physics
Solving physics problems is about breaking down the problem into manageable steps and applying core principles of physics. For our exercise:
  • Start by identifying what you need to find—here, the velocities of the ball and block post-collision.
  • Next, calculate the initial states using potential and kinetic energy conversions.
  • Apply conservation laws. First, momentum conservation to relate the post-collision velocities.
  • Then, use energy conservation to set up the equations for final kinetic energy.
  • To finish, solve the system of simultaneous equations you have formed to find the unknown velocities.
This structured approach ensures that you cover all aspects and use correct physics principles throughout.

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Most popular questions from this chapter

Game of Pool In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at \(3.50 \mathrm{~m} / \mathrm{s}\) along a line making an angle of \(22.0^{\circ}\) with its original direction of motion, and the second ball has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball. (c) Is kinetic energy (of the centers of mass, don't consider the rotation) conserved?

Three Balls ball \(A\) with an initial speed of \(10 \mathrm{~m} / \mathrm{s}\) collides elastically with stationary balls \(B\) and \(C\), whose centers are on a line perpendicular to the initial velocity of ball \(A\) and that are initially in contact with each other. The three balls are identical. Ball \(A\) is aimed directly at the contact point, and all motion is frictionless. After the collision, what are the velocities of (a) ball \(B,(\mathrm{~b})\) ball \(C\), and \((\mathrm{c})\) ball \(A ?\) (Hint: With friction absent, each impulse is directed along the line connecting the centers of the colliding balls, normal to the colliding surfaces.)

Block and Horizontal Spring , a \(2.5 \mathrm{~kg}\) block slides head on into a spring with a spring constant of \(320 \mathrm{~N} / \mathrm{m}\). When the block stops, it has compressed the spring by \(7.5 \mathrm{~cm}\). The coefficient of kinetic friction between the block and the horizontal surface is \(0.25 .\) While the block is in contact with the spring and being brought to rest, what are (a) the work done by the spring force and (b) the increase in thermal energy of the block-floor system? (c) What is the block's speed just as the block reaches the spring?

Block on a Track In Fig. \(10-51\), a block slides along a track from one level to a higher level, by moving through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance \(d\). The block's initial speed \(v_{1}\) is \(6.0 \mathrm{~m} / \mathrm{s} ;\) the height difference \(h\) is \(1.1 \mathrm{~m} ;\) and the coefficient of kinetic friction \(\mu^{\mathrm{kin}}\) is \(0.60\). Find \(d\).

The Collie A Collie drags its bed box across a floor by applying a horizontal force of \(8.0 \mathrm{~N}\). The kinetic frictional force acting on the box has magnitude \(5.0 \mathrm{~N}\). As the box is dragged through \(0.70 \mathrm{~m}\) along the way, what are (a) the work done by the collie's applied force and (b) the increase in thermal energy of the bed and floor?

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