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Step-by-Step Solution
Step 1: Identify the Initial Scenario
A single block of mass 3M is moving with velocity V0 = 40 m/s. It splits into two parts:
A smaller part of mass M (ratio 1 part).
A larger part of mass 2M (ratio 2 parts).
The smaller part moves in the same direction with a velocity V1 = 60 m/s. We need to find the final velocity of the larger part (V2) and then determine the fractional change in kinetic energy due to this splitting.
Step 2: Apply Conservation of Linear Momentum
Before splitting, the total momentum of the system is
$ p_{\text{initial}} = (3M)(V_0). $
After splitting, the two parts have masses M and 2M and velocities V1 and V2 respectively. Therefore,
$ p_{\text{final}} = M \,V_1 + 2M \,V_2. $
By conservation of momentum:
$ 3M \,V_0 = M \,V_1 \;+\; 2M \,V_2. $
Dividing throughout by M:
$ 3 V_0 = V_1 + 2 V_2. $
Step 3: Substitute Known Values and Solve for V2
We know V0 = 40 m/s and V1 = 60 m/s. Hence,
$ 3 \times 40 = 60 + 2 V_2 \quad \Rightarrow \quad 120 = 60 + 2V_2 \quad \Rightarrow \quad 2V_2 = 60 \quad \Rightarrow \quad V_2 = 30 \text{ m/s}. $
Step 4: Compute the Initial and Final Kinetic Energies
• Initial Kinetic Energy (before splitting):
$ K_{\text{initial}} = \tfrac{1}{2} (3M) {V_0}^2 = \tfrac{3}{2} M (40)^2 = 3 \times 20 \times 40 M = \tfrac{3}{2}M \times 1600. $
(We will keep it symbolic as (1/2)(3M)V02 in the final fraction.)
• Final Kinetic Energy (after splitting):
$ K_{\text{final}} = \tfrac{1}{2} \,M\, {V_1}^2 + \tfrac{1}{2}\, (2M)\, {V_2}^2. $
Substitute V1=60 m/s and V2=30 m/s:
$ K_{\text{final}} = \tfrac{1}{2} M \times (60)^2 \;+\; \tfrac{1}{2} \,(2M) \times (30)^2. $
$ = \tfrac{1}{2}M \times 3600 + (M) \times 900. $
$ = 1800 M + 900 M = 2700 M. $
Step 5: Calculate the Fractional Change in Kinetic Energy
The fractional change is given by
$ \tfrac{\Delta K.E.}{K.E._{\text{initial}}} = \tfrac{K_{\text{final}} - K_{\text{initial}}}{K_{\text{initial}}}. $
For clarity, keep the expression in terms of V0, V1, and V2:
$ \tfrac{\Delta K.E.}{K_{\text{initial}}} = \tfrac{\tfrac{1}{2}M{V_1}^2 + \tfrac{1}{2} (2M){V_2}^2 - \tfrac{1}{2} (3M){V_0}^2}{\tfrac{1}{2} (3M){V_0}^2}
= \tfrac{V_1^2 + 2{V_2}^2 - 3{V_0}^2}{3{V_0}^2}. $
Substituting V0 = 40 m/s, V1 = 60 m/s, V2 = 30 m/s:
$ \tfrac{3600 + 2 \times 900 - 3 \times 1600}{3 \times 1600} = \tfrac{3600 + 1800 - 4800}{4800} = \tfrac{600}{4800} = \tfrac{1}{8}. $
Hence, the fractional change in kinetic energy is $ \tfrac{1}{8} $.
Answer: $ \tfrac{1}{8} $
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