A 10 g bullet travelling at 200 m/s strikes and remains embedded in a 2 kg target which is originally at rest but free to move. At what speed does the target move off ?

Answer

To solve the problem of a bullet embedding itself in a target, we can use the principle of conservation of momentum. Here’s a step-by-step solution:

### Step 1: Convert the mass of the bullet to kilograms
The mass of the bullet is given as 10 grams. To convert grams to kilograms:
\[ \text{Mass of bullet} = 10 \, \text{g} = \frac{10}{1000} \, \text{kg} = 0.01 \, \text{kg} \]

### Step 2: Identify the mass of the target
The mass of the target is given as 2 kg.

### Step 3: Calculate the initial momentum before the collision
The bullet is moving at a velocity of 200 m/s, and the target is at rest (velocity = 0 m/s). The initial momentum (p_initial) can be calculated as:
\[ p_{\text{initial}} = \text{mass of bullet} \times \text{velocity of bullet} + \text{mass of target} \times \text{velocity of target} \]
\[ p_{\text{initial}} = (0.01 \, \text{kg} \times 200 \, \text{m/s}) + (2 \, \text{kg} \times 0 \, \text{m/s}) = 2 \, \text{kg m/s} \]

### Step 4: Calculate the total mass after the collision
After the bullet embeds itself in the target, the total mass (m_total) of the system is:
\[ m_{\text{total}} = \text{mass of bullet} + \text{mass of target} = 0.01 \, \text{kg} + 2 \, \text{kg} = 2.01 \, \text{kg} \]

### Step 5: Set up the equation for momentum after the collision
Let \( v \) be the common velocity of the bullet and the target after the collision. The momentum after the collision (p_final) is:
\[ p_{\text{final}} = m_{\text{total}} \times v = 2.01 \, \text{kg} \times v \]

### Step 6: Apply the law of conservation of momentum
According to the law of conservation of momentum:
\[ p_{\text{initial}} = p_{\text{final}} \]
Substituting the values we have:
\[ 2 \, \text{kg m/s} = 2.01 \, \text{kg} \times v \]

### Step 7: Solve for \( v \)
Rearranging the equation to solve for \( v \):
\[ v = \frac{2 \, \text{kg m/s}}{2.01 \, \text{kg}} \approx 0.99 \, \text{m/s} \]

### Final Answer
The speed at which the target moves off after the bullet embeds itself is approximately **0.99 m/s**.

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