Question

A block moving horizontally on a smooth surface with a speed of 40 m/s splits into two parts with masses in the ratio of \(1 : 2\). If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is :

• A. \(\frac{1}{8}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{1}{3}\)
• D. \(\frac{2}{3}\)

Hint:

In explosion/splitting problems, the kinetic energy of the system always increases (\(\Delta K>0\)) because chemical or internal potential energy is converted into kinetic energy. If you get a negative value, recheck your momentum conservation calculation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When a block splits, the total linear momentum is conserved because no external horizontal force acts on the system. The kinetic energy changes due to the internal energy released during the splitting process.
Step 2: Key Formula or Approach:
1. Conservation of Momentum: \(M v = m_1 v_1 + m_2 v_2\)
2. Initial Kinetic Energy: \(K_i = \frac{1}{2} M v^2\)
3. Final Kinetic Energy: \(K_f = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2\)
4. Fractional change: \(\frac{\Delta K}{K_i} = \frac{K_f - K_i}{K_i}\)
Step 3: Detailed Explanation:
Let total mass be \(M\). Masses are \(m_1 = \frac{1}{3}M\) (smaller) and \(m_2 = \frac{2}{3}M\) (larger).
Initial velocity \(v = 40\) m/s.
Smaller part velocity \(v_1 = 60\) m/s.
Using momentum conservation:
\[ M(40) = \left( \frac{M}{3} \right)(60) + \left( \frac{2M}{3} \right) v_2 \]
\[ 40 = 20 + \frac{2}{3} v_2 \implies 20 = \frac{2}{3} v_2 \implies v_2 = 30 \text{ m/s} \]
Initial Kinetic Energy:
\[ K_i = \frac{1}{2} M (40)^2 = 800M \]
Final Kinetic Energy:
\[ K_f = \frac{1}{2} \left( \frac{M}{3} \right) (60)^2 + \frac{1}{2} \left( \frac{2M}{3} \right) (30)^2 \]
\[ K_f = \frac{M}{6} (3600) + \frac{M}{3} (900) = 600M + 300M = 900M \]
Change in Kinetic Energy:
\[ \Delta K = K_f - K_i = 900M - 800M = 100M \]
Fractional change:
\[ \frac{\Delta K}{K_i} = \frac{100M}{800M} = \frac{1}{8} \]
Step 4: Final Answer:
The fractional change in kinetic energy is \(\frac{1}{8}\).