Sure, let's solve this problem step-by-step.
Given:
- Total mass of the bomb before explosion: \(5 \text{ kg}\)
- Mass of part A (\(m_A\)): \(2 \text{ kg}\)
- Mass of part B (\(m_B\)): \(3 \text{ kg}\)
- Velocity of part A (\(v_A\)): \(6 \text{ m/s}\) (in a certain direction)
To Find:
i. Velocity of part B (\(v_B\))
ii. Total kinetic energy produced by the explosion
Steps to Solution:
### i. Velocity of Part B
1. Law of Conservation of Momentum:
The total momentum before the explosion should equal the total momentum after the explosion. Since the bomb was initially stationary, the initial momentum is zero.
[tex]\[
\text{Initial Momentum} = 0
\][/tex]
The momentum of part A after the explosion:
[tex]\[
P_A = m_A \cdot v_A = 2 \text{ kg} \cdot 6 \text{ m/s} = 12 \text{ kg}\cdot \text{m/s}
\][/tex]
The momentum of part B after the explosion must balance this to make the total final momentum zero (since the initial was zero):
[tex]\[
P_B = m_B \cdot v_B
\][/tex]
Setting the total momentum to zero after the explosion:
[tex]\[
P_A + P_B = 0
\][/tex]
[tex]\[
12 \text{ kg}\cdot \text{m/s} + (3 \text{ kg} \cdot v_B) = 0
\][/tex]
Solving for \(v_B\):
[tex]\[
3 \text{ kg} \cdot v_B = -12 \text{ kg}\cdot \text{m/s}
\][/tex]
[tex]\[
v_B = -4 \text{ m/s}
\][/tex]
Thus, the velocity of part B is \( -4 \text{ m/s} \). The negative sign indicates that it is moving in the opposite direction to part A.
### ii. Total Kinetic Energy Produced by the Explosion
2. Kinetic Energy Calculation:
The kinetic energy (KE) of an object is given by the formula:
[tex]\[
KE = \frac{1}{2} m v^2
\][/tex]
For part A:
[tex]\[
KE_A = \frac{1}{2} \cdot m_A \cdot v_A^2 = \frac{1}{2} \cdot 2 \text{ kg} \cdot (6 \text{ m/s})^2 = \frac{1}{2} \cdot 2 \cdot 36 = 36 \text{ J}
\][/tex]
For part B:
[tex]\[
KE_B = \frac{1}{2} \cdot m_B \cdot v_B^2 = \frac{1}{2} \cdot 3 \text{ kg} \cdot (-4 \text{ m/s})^2 = \frac{1}{2} \cdot 3 \cdot 16 = 24 \text{ J}
\][/tex]
Total kinetic energy produced by the explosion:
[tex]\[
KE_{total} = KE_A + KE_B = 36 \text{ J} + 24 \text{ J} = 60 \text{ J}
\][/tex]
Summary:
i. The velocity of part B is \( -4 \text{ m/s} \) (opposite direction).
ii. The total kinetic energy produced by the explosion is [tex]\( 60 \text{ J} \)[/tex].
