danielesther442
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8. A stationary bomb of mass [tex]5 \, \text{kg}[/tex] explodes into two parts. One part, [tex]A[/tex], with mass [tex]2 \, \text{kg}[/tex] flies off with a velocity of [tex]6 \, \text{m/s}[/tex], and another part, [tex]B[/tex], with mass [tex]3 \, \text{kg}[/tex] flies off with a certain velocity in the opposite direction.

Calculate:
i. The velocity of part [tex]B[/tex].
ii. The total kinetic energy produced by the explosion.



Answer :

GinnyAnswer
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].

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