Question

Two masses \(m_1\) and \(m_2\) moving with velocities \(V_1\) and \(V_2\) in opposite directions collide elastically and after collision \(m_1\) and \(m_2\) move with velocities \(V_2\) and \(V_1\) respectively. The ratio \(\frac{m_2}{m_1}\) is

• A. \(0.25\)
• B. \(0.50\)
• C. \(0.75\)
• D. \(1.0\)

Hint:

Exchange of velocities needs equal masses.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In an elastic collision, momentum and kinetic energy are both conserved. The velocities after collision are given by standard formulas.

Step 2: Write the final velocity of m1:
With \(V_2\) taken as opposite sign, the general result in one dimension is
\[ v_1'=\frac{(m_1-m_2)v_1+2m_2v_2}{m_1+m_2} \]
We are told \(v_1'=v_2\) (the velocity values are exchanged).

Step 3: Set up:
\[ (m_1-m_2)v_1+2m_2v_2=(m_1+m_2)v_2 \]
\[ (m_1-m_2)v_1=(m_1-m_2)v_2 \]

Step 4: Solve:
Because \(v_1\ne v_2\) (they move in opposite directions), \(m_1-m_2=0\), so \(m_1=m_2\).

Step 5: Choose:
\(\dfrac{m_2}{m_1}=1\), option (D).

Final Answer:
The masses must be equal, so the ratio is 1. \[ \boxed{1.0} \]