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Chapter 6: Problem 38

A ball of mass \(m\) moving at a speed \(v\) makes a head on collision with an identical ball at rest. 'The kinetic energy of balls after the collision is \(\frac{3}{4} t h\) of the original. If \(e\) is the coefficient of restitution. Then: (a) \(v_{1}-\frac{1 \times \frac{1}{\sqrt{2}}}{2} v\) (b) \(v_{2}-\frac{1-\frac{1}{\sqrt{2}}}{2} v\) (c) \(e-\frac{1}{\sqrt{2}}\) (d) \(e-\frac{1}{2}\)

Short Answer

Expert verified
The answer is (c) \(e - \frac{1}{\sqrt{2}} = 0\).

Step by step solution

01

Initial Conditions and Equations

We start by setting up the initial and final conditions of the collision. Initially, one ball with mass \( m \) moves with velocity \( v \) and the other identical ball is at rest. The initial kinetic energy \( K_i \) is given by \[ K_i = \frac{1}{2}mv^2 \].
02

Condition after Collision

According to the problem, the kinetic energy of the balls after the collision is \( \frac{3}{4} \) of the original kinetic energy. Therefore, the final kinetic energy \( K_f \) is \[ K_f = \frac{3}{4} \cdot \frac{1}{2}mv^2 = \frac{3}{8}mv^2 \].
03

Applying the Coefficient of Restitution

The coefficient of restitution \( e \) relates the relative velocity of separation to the relative velocity of approach. Mathematically, it's given by \[ e = \frac{v_2 - v_1}{v - 0} \] where \( v_1 \) and \( v_2 \) are the velocities of the first and second balls post-collision, respectively.
04

Conservation of Momentum

Using the law of conservation of momentum, we get: \[ mv = mv_1 + mv_2 \]. Simplifying, \[ v = v_1 + v_2 \].
05

Solving for Velocities

From the conservation of energy and momentum equations, we can set up the following system of equations: \( mv_1 + mv_2 = mv \) and \( \frac{1}{2}m(v_1^2 + v_2^2) = \frac{3}{8}mv^2 \). Use these to solve for \( v_1 \) and \( v_2 \).
06

Analysis and Conclusion

Solving the above equations, we find the velocities \( v_1 = \left( \frac{1}{\sqrt{2}} \right)v \) and \( v_2 = \left( 1 - \frac{1}{\sqrt{2}} \right)v \). Plugging these values back into the equation for \( e \), we find \( e = \frac{1}{\sqrt{2}} \).
07

Final Answer Selection

Based on the calculated values, the final answer is (c) \( e - \frac{1}{\sqrt{2}} = 0 \).

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Key Concepts

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

Coefficient of Restitution
The coefficient of restitution, often denoted by \( e \), is a measure used in physics to describe how "bouncy" a collision is. It's a crucial concept in understanding the energy behavior between two colliding objects. In essence, it compares the speed at which two objects separate to the speed at which they came together before the collision.
  • If \( e = 1 \), the collision is perfectly elastic, meaning no kinetic energy is lost, and the objects bounce off each other without losing speed.
  • If \( e = 0 \), the collision is perfectly inelastic, where the objects stick together, and there is maximum kinetic energy loss.
  • For our specific scenario with two identical balls, we use the equation: \[ e = \frac{v_2 - v_1}{v} \]
Here, \( v_2 - v_1 \) is the relative velocity of the balls post-collision. Understanding the coefficient of restitution helps predict the velocities of the balls after they collide. For instance, in this exercise, we eventually find \( e = \frac{1}{\sqrt{2}} \), indicating a partially elastic collision with some energy conservation.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In collisions, kinetic energy is a key factor because it can be conserved or converted into other forms of energy. The initial kinetic energy in our exercise is given by \( K_i = \frac{1}{2}mv^2 \).
In a real collision, not all kinetic energy is necessarily conserved. Instead, some may transform into sound, heat, or deformation energy. In the problem we're solving:
  • The total initial kinetic energy is \( K_i = \frac{1}{2}mv^2 \).
  • Post-collision, the kinetic energy is \( K_f = \frac{3}{8}mv^2 \), which indicates a reduction by \( \frac{1}{4} \) of the initial energy.
  • This transformation matches the characteristics of an inelastic collision where kinetic energy decreases but the total energy in closed systems remains constant.
Recognizing how kinetic energy changes can offer insights into the nature of the collision and the energy transformations that took place during impact.
Conservation of Momentum
Momentum, a fundamental principle in physics, must be conserved in any collision scenario. This principle is encapsulated by the formula \[ mv = mv_1 + mv_2 \]. This equation states that the momentum before the collision (one moving ball) must equal the momentum after (two moving balls).
In this particular exercise:
  • The initial momentum, \( mv \), equals the sum of the momentum of the two balls post-collision \( (mv_1 + mv_2) \).
  • By conserving momentum, we can establish a relationship between initial and final velocities.
Momentum conservation allows us to determine how speeds will change post-collision. Here, it enables us to set up equations that ultimately guide us to solving for the post-collision velocities, \( v_1 \) and \( v_2 \), which are crucial for calculating both kinetic energy and the coefficient of restitution.
Head-on Collision
A head-on collision occurs when two objects collide directly along the line of their motion. In our scenario, one ball is initially stationary while the other moves directly towards it. This straightforward alignment simplifies the collision dynamics.
In such collisions:
  • The relative speed of the balls' approach and separation can be directly analyzed.
  • Using the conservation laws (momentum and restitution), we can predict the outcomes with precision.
  • Special attention is paid to these collisions because they clearly demonstrate the interplay of kinetic energy, momentum, and restitution.
Understanding head-on collisions helps make sense of more complex collision scenarios by offering a simplified model where the effects of collisions are aligned along a single axis.

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Most popular questions from this chapter

A uniform metal rod of length \(1 \mathrm{~m}\) is bent at \(90^{\circ}\) so as to form two arms of equal length. The centre of mass of this bent rod is (a) on the bisector of the angle, \(\left(\frac{1}{\sqrt{2}}\right) \mathrm{m}\) from vertex (b) on the bisector of the angle, \(\left(\frac{1}{2 \sqrt{2}}\right) \mathrm{m}\) from vertex (c) on the bisector of the angle, \(\left(\frac{1}{2}\right) \mathrm{m}\) from vertex (d) on the bisector of the angle, \(\left(\frac{1}{4 \sqrt{2}}\right)\) m from vertex

A loaded spring gun of mass \(M\) fires a shot of mass \(m\) with a velocity \(v\) at an angle of elevation \(\theta\). The gun is initially at rest on a horizontal frictionless surface. After firing the centre of mass of the gun- shot system (a) Moves with a velocity \(\frac{\mathrm{v} m}{M}\) (b) Moves with a velocity \(\frac{v m}{M \cos 0}\) in the horizontal direction (c) Moves with a velocity \(\frac{m v \sin \theta}{M+m}\) along vertical (d) Moves with velocity \(v(M-m) /(M+m)\) in horizontal direction

Two stars of masses \(M\) and \(m, M\) being greater than \(m\) are separated by a distance \(D .\) At a point between the two stars, their gravitational fields are equal in magnitude but opposite in direction. At that point a test object will feel no force. This point is: (a) At the c,m, of the two-star system (b) Between the c.m. and the mid point between the two stars (c) Between the c.m. and the star of Mass \(M\) (d) Between the mid-point of the two stars and the star of mass \(m\)

A ball of mass \(M\) strikes another ball of mass \(m\) at rest. If they separate in mutually perpendicular directions, then the coefficient of impact (e) is (a) \(\frac{M}{m}\) (b) \(\frac{m}{M}\) (c) \(\frac{m}{2 M}\) (d) Zero

Locate the centre of mass of a uniform straight rod of mass \(m\) and length \(L\).
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