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Chapter 4: Problem 454

Two bodies of masses \(\mathrm{m}\) and \(3 \mathrm{~m}\) have same momentum. their respective kinetic energies \(E_{1}\) and \(E_{2}\) are in the ratio..... (A) \(1: 3\) (B) \(3: 1\) (C) \(1: 3\) (D) \(1: 6\)

Short Answer

Expert verified
The ratio of the kinetic energies \(E_1\) and \(E_2\) is \(3:1\).

Step by step solution

01

Momenta of the objects

We are given that the two objects have the same momentum. So let's write the momenta for both objects: Object 1 (mass = m): Momenta = m * v1 Object 2 (mass = 3m): Momenta = 3m * v2 According to the problem, the momenta are equal, so: \(m * v1 = 3m * v2\)
02

Solving for one of the speeds

Dividing both sides of the equation by m gives: \(v1 = 3 * v2\) Now let's solve for either v1 or v2. We choose v2 for simplicity: \(v2 = \frac{v1}{3}\)
03

Writing expressions for kinetic energies

Now let's write the kinetic energy expressions for each object: Object 1 (mass = m): \(E_1 = \frac{1}{2} * m * v1^2\) Object 2 (mass = 3m): \(E_2 = \frac{1}{2} * 3m * v2^2\)
04

Substituting expression for v2 and simplifying

Now, let's substitute the expression for v2 from Step 2 into the expression for E2: \(E_2 = \frac{1}{2} * 3m * \left(\frac{v1}{3}\right)^2\) And simplifying E2: \(E_2 = \frac{1}{2} * 3m * \frac{v1^2}{9}\) \(E_2 = \frac{1}{6} * m * v1^2\)
05

Finding the ratio of kinetic energies

To find the ratio of the kinetic energies, we will divide \(E_1\) by \(E_2\): \(\frac{E_1}{E_2} = \frac{\frac{1}{2} * m * v1^2}{\frac{1}{6} * m * v1^2}\) Cancelling the terms \(m * v1^2\): \(\frac{E_1}{E_2} = \frac{\frac{1}{2}}{\frac{1}{6}}\) Now, dividing the fractions: \(\frac{E_1}{E_2} = \frac{1}{2} * \frac{6}{1}\) \(\frac{E_1}{E_2} = 3\) Thus, \(E_1\) and \(E_2\) are in the ratio \(3:1\), which corresponds to option (B).

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Most popular questions from this chapter

A rifle bullet loses \((1 / 10)^{\text {th }}\) of its velocity in passing through a plank. The least number of such planks required just to stop the bullet is (A) 5 (B) 10 (C) 11 (D) 20

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