A shell of mass 5 m, acted upon by no external force and initially at rest , burts into three fragments of masses m, 2 m, 2 m respectively . The first two fragments move in opposite directions with velocites magnitude 2 v and v respectively . The third fragment will

Options

A. Move with a velocity v in direction perpendicular to the other two
B. Move with a velocity 2 v in the direction of velocity of the first fragment
C. Move with a velocity v in the direction of velocity of the second fragment
D. Be at rest

Answer

To solve the problem, we will use the principle of conservation of momentum. Here's the step-by-step solution:

### Step 1: Understand the Initial Conditions
The shell has a total mass of \(5m\) and is initially at rest. Therefore, the initial momentum \(P_{\text{initial}}\) is:
\[ P_{\text{initial}} = 0 \]

### Step 2: Identify the Masses and Velocities of the Fragments
The shell bursts into three fragments:
- Fragment 1: Mass = \(m\), Velocity = \(2v\) (moving in the positive direction)
- Fragment 2: Mass = \(2m\), Velocity = \(-v\) (moving in the negative direction)
- Fragment 3: Mass = \(2m\), Velocity = \(v_3\) (unknown direction and magnitude)

### Step 3: Write the Final Momentum Equation
According to the conservation of momentum:
\[ P_{\text{initial}} = P_{\text{final}} \]
Thus, we have:
\[ 0 = (m \cdot 2v) + (2m \cdot -v) + (2m \cdot v_3) \]

### Step 4: Simplify the Equation
Substituting the values into the momentum equation:
\[ 0 = 2mv - 2mv + 2mv_3 \]
This simplifies to:
\[ 0 = 2mv_3 \]

### Step 5: Solve for \(v_3\)
From the equation \(0 = 2mv_3\), we can conclude that:
\[ v_3 = 0 \]

### Step 6: Conclusion
The third fragment, which has a mass of \(2m\), does not move. Therefore, it is at rest.

### Final Answer
The third fragment will be at rest.

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