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Chapter 4: Problem 503

The coefficient of restitution e for a perfectly elastic collision is (A) 1 (B) 0 (C) \(\infty\) (D) \(-1\)

Short Answer

Expert verified
For a perfectly elastic collision, the coefficient of restitution is 1. This is because both momentum and kinetic energy are conserved in such a collision. The correct answer is (A) 1.

Step by step solution

01

Understanding the coefficient of restitution

The coefficient of restitution is a measure of the "bounciness" of a collision between two objects. It is defined as the relative speed of separation divided by the relative speed of approach before the collision. Mathematically, it can be represented as follows: \( e = \frac{relative\,speed\,following\,the\,collision}{relative\,speed\,before\,the\,collision} \) For a perfectly elastic collision, we have the conservation of both momentum and kinetic energy.
02

Conservation of Momentum

Momentum is always conserved in a collision, regardless of whether it is elastic or inelastic. The total momentum before the collision must equal the total momentum after the collision. \( m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \) Here, \(m_1\) and \(m_2\) are the masses of the two objects, and \(v_{1i}\), \(v_{2i}\), \(v_{1f}\), and \(v_{2f}\) are their initial and final velocities, respectively.
03

Conservation of Kinetic Energy

In a perfectly elastic collision, kinetic energy is conserved as well. This means that the total kinetic energy before the collision must equal the total kinetic energy after the collision. \( \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 \)
04

Finding the Coefficient of Restitution for a Perfectly Elastic Collision

Using both conservation laws, we can find the value of \(e\) for a perfectly elastic collision. Solving for the relative speeds of separation and approach, we can express \(e\) as follows: \( e=\frac{(v_{1f}-v_{2f})}{(v_{2i}-v_{1i})} \) After solving the equations for conservation of momentum and conservation of kinetic energy simultaneously and simplifying, we obtain: \( e = 1 \) Thus, for a perfectly elastic collision, the coefficient of restitution is 1.
05

Answer

The correct answer is (A) 1, which is the coefficient of restitution for a perfectly elastic collision.

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Most popular questions from this chapter

An ice-cream has a marked value of \(700 \mathrm{kcal}\). How many kilo-watt- hour of energy will it deliver to the body as it is digested $(\mathrm{J}=4.2 \mathrm{~J} / \mathrm{cal})$ (A) \(0.81 \mathrm{kwh}\) (B) \(0.90 \mathrm{kwh}\) (C) \(1.11 \mathrm{kwh}\) (D) \(0.71 \mathrm{kwh}\)

A spring gun of spring constant \(90 \times 10^{2} \mathrm{~N} / \mathrm{M}\) is compressed \(4 \mathrm{~cm}\) by a ball of mass \(16 \mathrm{~g}\). If the trigger is pulled, calculate the velocity of the ball. (A) \(60 \mathrm{~m} / \mathrm{s}\) (B) \(3 \mathrm{~m} / \mathrm{s}\) (C) \(90 \mathrm{~m} / \mathrm{s}\) (D) \(30 \mathrm{~m} / \mathrm{s}\)

A bullet of mass \(\mathrm{m}\) moving with velocity \(\mathrm{v}\) strikes a block of mass \(\mathrm{M}\) at rest and gets embedded into it. The kinetic energy of the composite block will be (A) \((1 / 2) \mathrm{mv}^{2} \times[\mathrm{M} /(\mathrm{m}+\mathrm{m})]\) (B) \((1 / 2) \mathrm{mv}^{2} \times[(\mathrm{m}+\mathrm{m}) / \mathrm{M}]\) (C) \((1 / 2) \mathrm{MV}^{2} \times[\mathrm{m} /(\mathrm{m}+\mathrm{M})]\) (D) \((1 / 2) \mathrm{mv}^{2} \times[\mathrm{m} /(\mathrm{m}+\mathrm{M})]\)

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