Two balls of masses `m_(1)` and `m_(2)` are separated from each other by a powder charge placed between them. The whole system is at rest on the ground. Suddenly the powder charge explodes and masses are pushed apart. The mass `m_(1)` travels a distance `s_(1)` and stops. If the coefficients of friction between the balls and ground are same, the mass `m_(2)` stops after travelling the distance

Options

A. `s_(2)=(m_(1))/(m_(2))s_(1)`
B. `s_(2)=(m_(2))/(m_(1))s_(1)`
C. `s_(2)=(m_(1)^(2))/(m_(2)^(2))s_(1)`
D. `s_(2)=(m_(2)^(2))/(m_(1)^(2))s_(1)`

Answer

To solve the problem step by step, we will analyze the motion of the two masses after the explosion of the powder charge and apply the principles of Newton's laws of motion and kinematics.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Two masses \( m_1 \) and \( m_2 \) are at rest on the ground before the explosion.
- After the explosion, both masses are pushed apart.

2. **Understanding the Motion of Mass \( m_1 \):**
- Mass \( m_1 \) travels a distance \( s_1 \) before coming to a stop.
- The force acting on \( m_1 \) is the frictional force, which opposes its motion.
- The frictional force \( F_f \) can be expressed as:
\[ F_f = \mu \cdot m_1 \cdot g \]
- Here, \( \mu \) is the coefficient of friction, and \( g \) is the acceleration due to gravity.

3. **Calculate the Acceleration of Mass \( m_1 \):**
- The acceleration \( a \) due to friction is given by:
\[ a = -\frac{F_f}{m_1} = -\frac{\mu \cdot m_1 \cdot g}{m_1} = -\mu g \]
- The negative sign indicates that the acceleration is in the opposite direction of the motion.

4. **Apply Kinematic Equation for Mass \( m_1 \):**
- Using the kinematic equation:
\[ v^2 = u^2 + 2as \]
- For mass \( m_1 \), the final velocity \( v = 0 \) (it stops), initial velocity \( u \) is unknown, and \( s = s_1 \):
\[ 0 = u^2 + 2(-\mu g)s_1 \]
- Rearranging gives:
\[ u^2 = 2\mu g s_1 \]

5. **Momentum Conservation Between Masses:**
- When the powder charge explodes, momentum is conserved:
\[ m_1 u = m_2 v \]
- Rearranging gives:
\[ v = \frac{m_1}{m_2} u \]

6. **Calculate the Distance \( s_2 \) for Mass \( m_2 \):**
- Substitute \( v \) into the kinematic equation for mass \( m_2 \):
\[ 0 = v^2 + 2(-\mu g)s_2 \]
- Substituting \( v = \frac{m_1}{m_2} u \):
\[ 0 = \left(\frac{m_1}{m_2} u\right)^2 + 2(-\mu g)s_2 \]
- Rearranging gives:
\[ s_2 = \frac{(m_1^2 u^2)}{(m_2^2)(2\mu g)} \]

7. **Substituting \( u^2 \) from Step 4:**
- Substitute \( u^2 = 2\mu g s_1 \) into the equation for \( s_2 \):
\[ s_2 = \frac{(m_1^2)(2\mu g s_1)}{(m_2^2)(2\mu g)} = \frac{m_1^2}{m_2^2} s_1 \]

8. **Final Result:**
- The distance \( s_2 \) that mass \( m_2 \) travels before stopping is:
\[ s_2 = \frac{m_1^2}{m_2^2} s_1 \]

### Conclusion:
The distance \( s_2 \) that mass \( m_2 \) travels is proportional to the square of the ratio of the masses and the distance \( s_1 \) that mass \( m_1 \) travels.