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Chapter 2: Problem 1

A ball impinges directly on another ball at rest. The first ball is brought to rest by the impact. If half of the kinetic energy is lost by the impact, the value of coefficient of restitution is (1) \(\frac{1}{2}\) (2) \(\frac{1}{\sqrt{3}}\) (3) \(\frac{1}{\sqrt{2}}\) (4) \(\frac{\sqrt{3}}{2}\)

Short Answer

Expert verified
The coefficient of restitution is \( \frac{1}{\sqrt{2}} \).

Step by step solution

01

Identify Given Information and Formulae

The problem given describes an elastic collision where one ball, with mass \( m_1 \), is moving with initial velocity \( u_1 \) and impinges on another stationary ball with mass \( m_2 \). The first ball comes to rest after impact, so its final velocity \( v_1 = 0 \). It is stated that half of the kinetic energy is lost due to the impact.
02

Initial and Final Kinetic Energy

The initial kinetic energy of the system is \( KE_{initial} = \frac{1}{2}m_1 u_1^2 \) because the second ball is initially at rest. After the impact, the kinetic energy of the first ball is zero (since \( v_1 = 0 \)). Since half of the initial kinetic energy is lost, the kinetic energy of the system after impact is \( KE_{final} = \frac{1}{4}m_1 u_1^2 \). So, the kinetic energy of the second ball \( \frac{1}{2}m_2 v_2^2 = \frac{1}{4}m_1 u_1^2 \).
03

Apply Conservation of Momentum

According to the conservation of momentum, the total momentum before and after the collision is the same. Thus, \( m_1 u_1 = m_2 v_2 \). Solving this gives \( v_2 = \frac{m_1 u_1}{m_2} \).
04

Relate Kinetic Energy and Momentum

From Step 3, we have \( v_2 = \frac{m_1 u_1}{m_2} \). Substitute this into the kinetic energy equation: \( \frac{1}{2}m_2 v_2^2 = \frac{1}{4}m_1 u_1^2 \). Substituting \( v_2 \), we have \( \frac{1}{2} m_2 (\frac{m_1 u_1}{m_2})^2 = \frac{1}{4} m_1 u_1^2 \). Simplifying, we find that \( \frac{m_1^2}{m_2} \times \frac{u_1^2}{m_2} = \frac{1}{4} \times m_1 u_1^2 \).
05

Solve for Coefficient of Restitution

Use the relationship: coefficient of restitution \( e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{v_2}{u_1} \) since \( v_1 = 0 \) and \( u_2 = 0 \). From our equation, \( e = \frac{m_1}{m_2} \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \). This matches the given option.

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Key Concepts

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

Elastic Collision
An elastic collision is a type of collision where the total kinetic energy of the system is conserved before and after the impact. This is an ideal scenario since, in real-life collisions, some energy is usually lost to sound, heat, or deformation. Nevertheless, elastic collisions are important in physics as they allow us to apply specific conservation principles to solve problems.

In the given problem, a ball makes a direct impact with another stationary ball. The first ball is brought to rest after the collision. Even though the kinetic energy changes, the nature of the collision鈥攖he complete halting of one ball鈥攍eads us to classify it under elastic collisions. It's essential to remember that in an ideal elastic collision, both momentum and kinetic energy are considered conserved. However, this exercise introduces loss to illustrate practical considerations.
  • The first ball stops moving post-collision, which is a typical outcome in many elastic collision exercises.
  • The energy retained by the second ball is half of the initial energy, highlighting the energy dissipation considered for this situation.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion, defined by the equation \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity. In the collision described in the exercise, two main points about kinetic energy stand out:

Initially, only the first ball has kinetic energy because it is the only one moving. At the end of the collision, the energy distribution changes. The kinetic energy of the stationary ball becomes important because it's this energy that demonstrates the transfer from the first ball.
  • The initial kinetic energy of the moving ball is \( \frac{1}{2}m_1 u_1^2 \).
  • After the collision, the energy drops such that the moving ball now carries the remaining kinetic energy. Half of the initial energy is lost, leading to the equation for the second ball: \( \frac{1}{2}m_2 v_2^2 = \frac{1}{4}m_1 u_1^2 \).
Conservation of Momentum
The principle of conservation of momentum states that in a closed system, the sum of momenta before and after a collision remains constant, provided no external force acts on the system. It can be expressed mathematically for the given collision as:

\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Given that the second ball is initially at rest \((u_2 = 0)\) and the first ball comes to a rest afterwards \((v_1 = 0)\), the equation simplifies to:
\[ m_1 u_1 = m_2 v_2 \]
This straightforward relation allows us to solve for velocity after the collision, showing how momentum transfers from one object to another.
  • This transfer is essential in determining many other aspects of the collision, like the coefficient of restitution.
  • The simplified equation proves that momentum, unlike kinetic energy, perfectly stays in the system, offering a powerful way to predict outcomes in closed systems.

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Most popular questions from this chapter

A block of mass ' \(m\) ' is hanging from a massless spring of spring constant \(k\). It is in equilibrium under the influence of gravitational force. Another particle of same mass ' \(m\) ' moving upwards with velocity \(u_{0}\) hits the block and sticks to it. For the subsequent motion, choose the incorrect statements: (1) Velocity of the combined mass must be maximum at natural length of the spring. (2) Velocity of the combined mass must be maximum at the new equilibrium position. (3) Velocity of the combined mass must be maximum at the instant particle hits the block. (4) Velocity of the combined mass must be maximum at a point lying between old equilibrium position and natural length.

Two identical billiard balls undergo an oblique elastic collision. Initially, one of the balls is stationary. If the initially stationary ball after collision moves in a direction which makes an angle of \(37^{\circ}\) with direction of initial motion of the moving ball, then the angle through which initially moving ball will be deflected is (1) \(37^{\circ}\) (2) \(60^{\circ}\) (3) \(53^{\circ}\) (4) \(>53^{\circ}\)

In column I, nature of collision between two bodies is give while in column II some physical quantity that may remain conserved during the collision is given. Match the entries of column I with the entries of column II. $$ \begin{array}{|l|l|l|} \hline {\text { Column I }} & {\text { Column II }} \\ \hline \text { i. } \quad \text { Elastic collision } & \text { a. } \begin{array}{l} \text { kinetic energy is } \\ \text { conserved } \end{array} \\ \hline \text { ii. } \text { Inelastic collision } & \text { b. } \begin{array}{l} \text { kinetic energy of the } \\ \text { system may increase } \end{array} \\ \hline \begin{array}{l} \text { iii. } \text { Perfectly inelastic } \\ \text { collision } \end{array} & \begin{array}{l} \text { c. } \text { kinetic energy is not } \\ \text { conserved } \end{array} \\ \hline \text { iv. } \begin{array}{l} \text { Collision between two } \\ \text { cars moving at high } \\ \text { speeds } \end{array} & \text { d. } \begin{array}{r} \text { total energy is } \\ \text { conserved } \end{array} \\ \hline \end{array} $$

A particle at rest is constrained to move on a smooth horizontal surface, Another identical particle hits the stationary particle with a velocity \(v\) at an angle \(\theta=60^{\circ}\) with horizontal. If the particles move together, the velocity of the combination just after the impact is equal to (I) \(v\) (2) \(w / 2\) (3) \(\sqrt{3} v / 4\) (4) \(v / 4\)

A small ball thrown at an initial velocity \(u=25 \mathrm{~m} / \mathrm{s}\) directed at an angle \(\theta=37^{\circ}\) above the horizontal collides elastically with a vertical massive smooth wall moving with a uniform horizontal velocity \(u / 5\) towards the ball. After collision with the wall the ball returns to the point from where it was thrown. Determine the time \(t\) (in \(\mathrm{s}\) ) from the beginning of motion of the ball to the moment of its impact with the wall. (Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) )
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