A body of mass 8 kg and another of mass 2 kg are moving with equal kinetic energy. The ratio of their respective momenta will be

Question

Three masses $200\,\text{kg}$, $300\,\text{kg}$ and $400\,\text{kg}$ are placed at the vertices of an equilateral triangle of side $20\,\text{m}$. They are rearranged on the vertices of a bigger triangle of side $25\,\text{m}$ with the same centre. The work done in this process is ___ J.
(Gravitational constant $G = 6.7 \times 10^{-11}\,\text{N m}^2\text{kg}^{-2}$)

Answer 1

The problem asks for the ratio of the momenta of two bodies with different masses but equal kinetic energy. Let's solve this step-by-step:

Given:

Mass of the first body, $m_1 = 8 \, \text{kg}$.
Mass of the second body, $m_2 = 2 \, \text{kg}$.
Both bodies have equal kinetic energy.

Kinetic Energy: The kinetic energy (KE) is defined by the formula:

$\text{KE} = \frac{1}{2} mv^2$

For the two bodies, since their kinetic energies are equal:

$\frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_2 v_2^2$

We can omit $\frac{1}{2}$ from both sides:

$m_1 v_1^2 = m_2 v_2^2$

Plug in the values of $m_1$ and $m_2$:

$8 v_1^2 = 2 v_2^2$

Solving for the ratio of the speeds:

$v_1^2 = \frac{2}{8} v_2^2$

$v_1^2 = \frac{1}{4} v_2^2$

$v_1 = \frac{1}{2} v_2$

Momentum Formula: The momentum ($p$) is given by:

$p = mv$

The momentum for both bodies is:

$p_1 = m_1 v_1$

$p_2 = m_2 v_2$

Plug in the values and expressions:

$p_1 = 8 v_1 = 8 \left(\frac{1}{2} v_2 \right) = 4 v_2$

$p_2 = 2 v_2$

Calculate the ratio of momenta:

$\frac{p_1}{p_2} = \frac{4 v_2}{2 v_2} = \frac{4}{2} = 2:1$

Conclusion: The ratio of their respective momenta is $2:1$. Hence, the correct answer is

2:1

.

A body of mass 8 kg and another of mass 2 kg are moving with equal kinetic energy. The ratio of their respective momenta will be

The Correct Option is B

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