To solve this problem, we need to apply the principle of conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces act on it. In this case, we have an isolated particle that explodes into two fragments, and we can analyze their motion based on the information provided.
Understanding the Situation
Initially, the particle of mass \( m \) is moving along the x-axis. When it explodes, it breaks into two fragments: one with mass \( \frac{m}{4} \) and the other with mass \( \frac{3m}{4} \). Since the explosion occurs in a horizontal plane, we can assume that the initial momentum of the system is entirely in the x-direction.
Applying Conservation of Momentum
Before the explosion, the momentum of the particle can be expressed as:
- Initial momentum, \( p_{\text{initial}} = mv \) (where \( v \) is the initial velocity along the x-axis).
After the explosion, the momentum of the two fragments can be expressed as:
- Momentum of the smaller fragment (mass \( \frac{m}{4} \)): \( p_1 = \frac{m}{4} v_1 \)
- Momentum of the larger fragment (mass \( \frac{3m}{4} \)): \( p_2 = \frac{3m}{4} v_2 \)
According to the conservation of momentum:
Initial momentum = Final momentum
Thus, we have:
mv = \( \frac{m}{4} v_1 + \frac{3m}{4} v_2 \)
Analyzing the Fragments' Positions
We know that the smaller fragment is at \( y = +15 \) cm after the explosion. Since the explosion occurs in a horizontal plane, the vertical position of the larger fragment must be determined based on the conservation of momentum in the x-direction. Since there are no external forces acting in the y-direction, the total vertical momentum before and after the explosion must also be zero.
Finding the Larger Fragment's Position
Let’s denote the vertical position of the larger fragment as \( y_2 \). Since the total vertical momentum before the explosion was zero (the original particle was moving horizontally), the sum of the vertical components of the momentum after the explosion must also equal zero:
\( \frac{m}{4} v_{1y} + \frac{3m}{4} v_{2y} = 0 \)
Given that the smaller fragment is at \( y = +15 \) cm, we can assume it has a vertical velocity component \( v_{1y} \) that is positive. For the momentum to balance, the larger fragment must have a vertical velocity component \( v_{2y} \) that is negative. Thus, we can express the relationship as:
\( \frac{m}{4} v_{1y} = -\frac{3m}{4} v_{2y} \)
Conclusion on the Larger Fragment's Position
Since the smaller fragment is at \( y = +15 \) cm, and the larger fragment must have a negative vertical position to balance the momentum, we can conclude that the larger fragment is at:
\( y_2 = -\frac{3}{4} \cdot 15 \text{ cm} = -11.25 \text{ cm} \)
Therefore, the larger fragment is at \( y = -11.25 \) cm. This negative value indicates that it is below the original height of the particle before the explosion.
