Question

Three bodies A, B and C have equal kinetic energies and their masses are 400 g, 1.2 kg and 1.6 kg respectively. The ratio of their linear momenta is :

• A. \(1 : \sqrt{3} : 2\)
• B. \(1 : \sqrt{3} : \sqrt{2}\)
• C. \(\sqrt{2} : \sqrt{3} : 1\)
• D. \(\sqrt{3} : \sqrt{2} : 1\)

Answer 1

The Correct Option is A

To solve the problem of finding the ratio of linear momenta for three bodies with equal kinetic energies, we start by considering the relationship between kinetic energy and momentum.

The kinetic energy (\(KE\)) of a body is given by the formula:

\(KE = \frac{1}{2}mv^2\)

where \(m\) is the mass and \(v\) is the velocity of the body.

The linear momentum (\(p\)) of a body is given by:

\(p = mv\)

We are told that the kinetic energies of bodies A, B, and C are equal. Therefore, for each body:

\(\frac{1}{2} m_A v_A^2 = \frac{1}{2} m_B v_B^2 = \frac{1}{2} m_C v_C^2\)

Since the kinetic energies are equal, we can write:

\(m_A v_A^2 = m_B v_B^2 = m_C v_C^2\)

Now, we can express the momentum for each body in terms of its velocity:

\(v_A = \sqrt{\frac{2 \times KE}{m_A}}\), \(v_B = \sqrt{\frac{2 \times KE}{m_B}}\), \(v_C = \sqrt{\frac{2 \times KE}{m_C}}\)

Then, the momentum for each body is:

\(p_A = m_A \times v_A = m_A \times \sqrt{\frac{2 \times KE}{m_A}}\), \(p_B = m_B \times v_B = m_B \times \sqrt{\frac{2 \times KE}{m_B}}\), \(p_C = m_C \times v_C = m_C \times \sqrt{\frac{2 \times KE}{m_C}}\)

Simplifying further, we find the momentum expressions:

\(p_A = \sqrt{2 \times KE \times m_A}\), \(p_B = \sqrt{2 \times KE \times m_B}\), \(p_C = \sqrt{2 \times KE \times m_C}\)

Given masses are \(m_A = 400\) g, \(m_B = 1.2\) kg, and \(m_C = 1.6\) kg. Converting \(m_A\) to kg, we have \(m_A = 0.4\) kg.

Using these masses to find the ratio of their momentum:

Momentum ratio:

\(\frac{p_A}{p_B} = \frac{\sqrt{0.4}}{\sqrt{1.2}} = \frac{\sqrt{1}}{\sqrt{3}} = \frac{1}{\sqrt{3}}\)

\(\frac{p_B}{p_C} = \frac{\sqrt{1.2}}{\sqrt{1.6}} = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2}\)

Solving for the combined ratio gives:

\(p_A : p_B : p_C = 1 : \sqrt{3} : 2\)

Thus, the correct option is \(1 : \sqrt{3} : 2\).

Answer 2

The kinetic energy is given by:
\[ KE = \frac{P^2}{2m}, \quad \text{where } P \propto \sqrt{m}. \]
For masses \(m_A = 400 \, \text{g}, \, m_B = 1.2 \, \text{kg}, \, m_C = 1.6 \, \text{kg}\):
\[ P_A : P_B : P_C = \sqrt{400} : \sqrt{1200} : \sqrt{1600}. \]
Simplify:
\[ P_A : P_B : P_C = 1 : \sqrt{3} : 2. \]
Final Answer: 1 : \(\sqrt{3}\) : 2.