Question

Two unequal masses possess the same K.E. Then, the heavier mass has

• A. greater momentum
• B. smaller momentum
• C. the same momentum as the lighter mass
• D. greater speed

Answer 1

The correct option is A

greater momentum

Step 1: Given data

Consider two unequal masses $m_1$and $m_2$, such that

$m_1>m_2$

The two unequal masses possess the same kinetic energy,

$K.E{.}_1=K.E{.}_2$

Step 2: Finding the ratio of their velocities

The kinetic energy of a body is given by:

$K.E.=\frac{1}{2}m{v}^2$, where $m$ is the mass of the body and $v$ is its velocity.

If the two masses possess the same kinetic energy then,

$$\begin{gathered} K.E{.}_1=K.E{.}_2 \\ \frac{1}{2}{m}_1{v}_1^2=\frac{1}{2}{m}_2{v}_2^2 \\ \frac{v_1^2}{v_2^2}=\frac{m_2}{m_1} \\ \frac{v_1}{v_2}=\sqrt{\frac{m_2}{m_1}} \end{gathered}$$

Step 3: Find the ratio of their momentum

The linear momentum of a body is given by

$p=mv$ , where $m$ is the mass of the body and $v$ is its velocity.

For $m_1>m_2$, the ratio of linear momentum is

$$\begin{gathered} \frac{p_1}{p_2}=\frac{m_1{v}_1}{m_2{v}_2} \\ =\frac{m_1}{m_2}\times \sqrt{\frac{m_2}{m_1}} \\ =\sqrt{\frac{m_1}{m_2}} \end{gathered}$$

As $m_1>m_2$ , $p_1>p_2$

Thus, momentum is greater for heavier mass $m_1$.

Hence, the correct option is A.