An object of mass 3m, initially at rest, explodes breaking into two fragments of mass m and 2m, respectively. Which one of the following statements concerning the fragments after the explosion is true?

Options

A. They will fly off at right angles.
B. They will fly off in the same direction,
C. The smaller fragment will have twice the speed of the larger fragment
D. The larger fragment will have twice the speed of the smaller fragment.

Answer

To solve the problem, we will apply the principle of conservation of momentum. Here are the steps to arrive at the solution:

### Step 1: Understand the Initial Conditions
The total mass of the object before the explosion is \(3m\), and it is initially at rest. Therefore, the initial momentum \(P_{initial}\) is:
\[ P_{initial} = 0 \]

### Step 2: Define the Masses of the Fragments
After the explosion, the object breaks into two fragments:
- Fragment 1 has mass \(m\)
- Fragment 2 has mass \(2m\)

### Step 3: Apply Conservation of Momentum
According to the law of conservation of momentum, the total momentum before the explosion must equal the total momentum after the explosion. Thus:
\[ P_{initial} = P_{final} \]
Since \(P_{initial} = 0\), we have:
\[ 0 = m v_1 + 2m v_2 \]
where \(v_1\) is the velocity of the fragment with mass \(m\) and \(v_2\) is the velocity of the fragment with mass \(2m\).

### Step 4: Rearranging the Equation
We can simplify the equation by dividing through by \(m\) (assuming \(m \neq 0\)):
\[ 0 = v_1 + 2v_2 \]
This implies:
\[ v_1 = -2v_2 \]

### Step 5: Interpret the Result
The negative sign indicates that the direction of \(v_1\) is opposite to that of \(v_2\). This means that the smaller fragment (mass \(m\)) moves in the opposite direction to the larger fragment (mass \(2m\)).

### Step 6: Determine the Relationship Between Speeds
From the equation \(v_1 = -2v_2\), we see that the speed of the smaller fragment (mass \(m\)) is twice that of the larger fragment (mass \(2m\)). Therefore, if we denote the speed of the larger fragment as \(v\), then:
\[ |v_1| = 2|v_2| \]

### Conclusion
The smaller fragment (mass \(m\)) moves with a speed that is twice the speed of the larger fragment (mass \(2m\)). Thus, the correct statement concerning the fragments after the explosion is that the smaller fragment moves faster than the larger fragment.

### Final Answer
The smaller fragment will have twice the speed of the larger fragment.

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