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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

\(1 \mathrm{~kg}\) apple gives \(25 \mathrm{KJ}\) energy to a monkey. How much height he can climb by using this energy if his efficiency is \(40 \%\). (mass of monkey \(=25 \mathrm{~kg}\) and \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(20 \mathrm{~m}\) (B) \(4 \mathrm{~m}\) (C) \(30 \mathrm{~m}\) (D) \(40 \mathrm{~m}\)

Bansi does a given amount of work in \(30 \mathrm{sec}\). Jaimeen does the same amount of work .. in \(15 \mathrm{sec}\). The ratio of the output power of Bansi to the Jaimeen is.... (A) \(1: 1\) (B) \(1: 2\) (C) \(2: 1\) (D) \(5: 3\)

A uniform chain of length \(2 \mathrm{~m}\) is kept on a table such that a length of \(50 \mathrm{~cm}\) hangs freely from the edge of the table. The total mass of the chain is \(5 \mathrm{~kg}\). What is the work done in pulling the entire chain on the table. $\left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{2}\right)$ (A) \(7.2 \mathrm{~J}\) (B) \(3 \mathrm{~J}\) (C) \(4.6 \mathrm{~J}\) (D) \(120 \mathrm{~J}\)

What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of \(18 \mathrm{~cm}\) (Take \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(0.4 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(1.8 \mathrm{~m} / \mathrm{s}\) (D) \(0.6 \mathrm{~m} / \mathrm{s}\)

With what velocity should a student of mass \(40 \mathrm{~kg}\) run so that his kinetic energy becomes \(160 \mathrm{~J}\) ? (A) \(4 \mathrm{~m} / \mathrm{s}\) (B) \(\sqrt{8} \mathrm{~m} / \mathrm{s}\) (C) \(16 \mathrm{~m} / \mathrm{s}\) (D) \(8 \mathrm{~m} / \mathrm{s}\)
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