Question

A block moving horizontally on a smooth surface with a speed of \( 40 \, ms^{-1} \) splits into two equal parts. If one of the parts moves at \( 60 \, ms^{-1} \) in the same direction, then the fractional change in the kinetic energy will be \( x : 4 \) where \( x = \dots\dots\dots \).

Hint:

In explosions or splitting where internal energy is released, the final kinetic energy is always greater than the initial kinetic energy, even though momentum remains conserved.

Answer 1

Correct Answer: 1

Step 1: Understanding the Concept:
This problem involves the conservation of linear momentum and the calculation of kinetic energy before and after the explosion (splitting) of the block.
Step 2: Key Formula or Approach:
1. Conservation of Momentum: \( m_1 v_1 + m_2 v_2 = M V \).
2. Kinetic Energy: \( K = \frac{1}{2} m v^2 \).
3. Fractional Change: \( \frac{\Delta K}{K_i} = \frac{K_f - K_i}{K_i} \).
Step 3: Detailed Explanation:
Let the total mass be \( M \). The block splits into two equal parts, so each mass is \( M/2 \).
1. Initial State:
- Initial speed \( V = 40 \, ms^{-1} \).
- Initial Momentum: \( P_i = M(40) \).
- Initial Kinetic Energy: \( K_i = \frac{1}{2} M (40)^2 = 800 M \).
2. Final State:
- Mass 1 (\( M/2 \)) moves at \( v_1 = 60 \, ms^{-1} \).
- Mass 2 (\( M/2 \)) moves at speed \( v_2 \).
- By Momentum Conservation: \( M(40) = \frac{M}{2}(60) + \frac{M}{2} v_2 \).
\[ 40 = 30 + \frac{v_2}{2} \implies \frac{v_2}{2} = 10 \implies v_2 = 20 \, ms^{-1} \]
- Final Kinetic Energy: \( K_f = \frac{1}{2} (\frac{M}{2}) (60)^2 + \frac{1}{2} (\frac{M}{2}) (20)^2 \)
\[ K_f = \frac{M}{4} (3600 + 400) = \frac{4000 M}{4} = 1000 M \]
3. Fractional Change:
- Change in KE: \( \Delta K = K_f - K_i = 1000 M - 800 M = 200 M \).
- Fractional Change: \( \frac{\Delta K}{K_i} = \frac{200 M}{800 M} = \frac{1}{4} \).
Comparing with \( x : 4 \), we get \( x = 1 \).
Step 4: Final Answer:
The value of x is 1.