A cannon shell is fired to hit a target at a horizontal distance R. However, it breaks into two equal parts at its highest point. One part (A) returns to the cannon. The other part:

Options

A. (a)Will fall at a distance of R beyond the target
B. (b)Will fall at a distance of 3R beyond the target
C. (c)Will hit the target
D. (d)Have nine times the kinetic energy of A just after explosion

Explanation

To solve the problem step-by-step, we will analyze the situation of the cannon shell being fired, breaking into two parts, and determining the distances and velocities of each part after the explosion.

### Step 1: Understand the Initial Conditions
- A cannon shell is fired to hit a target at a horizontal distance \( R \).
- At its highest point, the shell breaks into two equal parts (let's denote them as Part A and Part B).

### Step 2: Analyze the Center of Mass (COM)
- The center of mass of the system (the cannon shell) can be calculated using the formula:
\[ R_{COM} = \frac{m_1 R_1 + m_2 R_2}{m_1 + m_2} \]
- Since both parts are equal in mass (let's denote each mass as \( m \)), and at the highest point, the horizontal distance to the target is \( R \), we can denote:
- \( m_1 = m \) (mass of Part A)
- \( R_1 = 0 \) (Part A returns to the cannon)
- \( m_2 = m \) (mass of Part B)
- \( R_2 = R \) (Part B moves forward)

### Step 3: Calculate the Center of Mass
- Substituting the values into the COM equation:
\[ R_{COM} = \frac{m \cdot 0 + m \cdot R}{m + m} = \frac{mR}{2m} = \frac{R}{2} \]
- This indicates that the center of mass of the two parts is at a distance \( \frac{R}{2} \) from the cannon.

### Step 4: Determine the Distance Traveled by Each Part
- Part A returns to the cannon, so its distance is \( 0 \).
- Part B, which moves forward, will travel an additional distance beyond the target. Since the center of mass is at \( \frac{R}{2} \) and the total distance to the target is \( R \), Part B will travel:
\[ R + \frac{R}{2} = \frac{3R}{2} \]
- Therefore, Part B falls at a distance \( R \) beyond the target.

### Step 5: Calculate the Velocities of Each Part
- Before the explosion, the velocity of the shell at its highest point is \( v \).
- After the explosion, Part A returns with a velocity \( -v \) (negative indicates direction towards the cannon).
- For Part B, we can use the conservation of momentum. The total momentum before the explosion is equal to the total momentum after the explosion:
\[ mv = m(-v) + mv_B \]
where \( v_B \) is the velocity of Part B.
- Rearranging gives:
\[ mv + mv = mv_B \implies 2mv = mv_B \implies v_B = 2v \]

### Step 6: Calculate the Kinetic Energy of Each Part
- The kinetic energy of Part A (returning to the cannon):
\[ KE_A = \frac{1}{2} m (-v)^2 = \frac{1}{2} mv^2 \]
- The kinetic energy of Part B:
\[ KE_B = \frac{1}{2} m (2v)^2 = \frac{1}{2} m \cdot 4v^2 = 2mv^2 \]
- The ratio of kinetic energies:
\[ \frac{KE_B}{KE_A} = \frac{2mv^2}{\frac{1}{2} mv^2} = 4 \]

### Conclusion
- Part A returns to the cannon, and Part B travels \( R \) beyond the target.
- The kinetic energy of Part B is \( 4 \) times that of Part A just after the explosion.