Physics Problem: Explosion and Kinetic Energy
Question:
A 4.2-kg object, initially at rest, explodes into three equal-mass fragments. Two of these fragments are found to move with equal speeds of 5.0 m/s and at 90° angles to each other. How much kinetic energy was released in the explosion?
- a. 70 J
- b. 53 J
- c. 60 J
- d. 64 J
- e. 35 J
Answer:
Step 1: Understand the setup
The total mass is 4.2 kg. After the explosion, it breaks into 3 equal parts:
Mass of each fragment, m = 4.2 kg ÷ 3 = 1.4 kg
Let’s assume:
- Fragment 1 moves east at 5.0 m/s → velocity: (5.0, 0)
- Fragment 2 moves north at 5.0 m/s → velocity: (0, 5.0)
Step 2: Use conservation of momentum
Since the original object was at rest, total momentum after explosion must also be zero. So, the momentum of the third fragment must cancel the vector sum of the first two.
p3 = −(p1 + p2) = −(1.4 × 5.0, 1.4 × 5.0) = −(7.0, 7.0) kg·m/s
Therefore, the velocity of fragment 3:
v3 = √(7.0² + 7.0²) / 1.4 = √(98) / 1.4 ≈ 7.0√2 / 1.4 = 5√2 m/s ≈ 7.07 m/s
Step 3: Calculate total kinetic energy
Kinetic energy of fragment 1:
KE₁ = ½ × 1.4 × (5.0)² = 0.7 × 25 = 17.5 J
Kinetic energy of fragment 2 (same as fragment 1):
KE₂ = 17.5 J
Kinetic energy of fragment 3:
KE₃ = ½ × 1.4 × (5√2)² = 0.7 × 50 = 35.0 J
Total kinetic energy released:
KEtotal = 17.5 + 17.5 + 35.0 = 70.0 J
Final Answer:
a. 70 J