Question

A body of mass \(5\,kg\) makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to \(\dfrac{1}{10}\)th of its original velocity. Then the mass of the second body is

• A. \(4.09\,kg\)
• B. \(0.5\,kg\)
• C. \(5\,kg\)
• D. \(5.09\,kg\)

Hint:

For elastic collision with one body at rest, use: \(v_1 = \frac{m_1-m_2}{m_1+m_2}u_1\).

Answer 1

The Correct Option is A

Step 1: Use elastic collision formula for 1D.
For elastic collision with second body at rest:
\[ v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1 \]
Step 2: Substitute given values.
Here \(m_1 = 5\), \(u_1 = u\), and after collision:
\[ v_1 = \frac{u}{10} \]
So:
\[ \frac{u}{10} = \frac{5 - m_2}{5 + m_2}u \]
Step 3: Solve for \(m_2\).
Cancel \(u\):
\[ \frac{1}{10} = \frac{5 - m_2}{5 + m_2} \]
Cross multiply:
\[ 5 + m_2 = 10(5 - m_2) \]
\[ 5 + m_2 = 50 - 10m_2 \]
\[ 11m_2 = 45 \Rightarrow m_2 = \frac{45}{11} = 4.09\,kg \]
Final Answer: \[ \boxed{4.09\,kg} \]