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A 1 kg stationary bomb is exploded in three parths having mass 1:1:3 respectively. Parts having same mass move in perpendicular directions with velocity 30 m/s, then the velocity of bigger part will be
  • a)
    10 √2 m/s
  • b)
    10/√2 m/s
  • c)
    15 √s m/s
  • d)
    15/s m/s
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A 1kg stationary bomb is exploded in three parths having mass 1:1:3 re...
**Explanation:**
**Given:**
- Mass of the bomb = 1 kg
- Mass of the three parts after explosion = 1:1:3
- Velocity of two equal parts = 30 m/s
**Step 1:**
Calculate the velocity of the smaller parts using conservation of momentum.
**Step 2:**
Let the velocity of the smaller parts be v. So, according to conservation of momentum:
Initial momentum = Final momentum
(1 kg) * 0 = (1 kg) * v + (1 kg) * v + (3 kg) * 0
0 = 2v
v = 0 m/s
So, the smaller parts do not move after the explosion.
**Step 3:**
Calculate the velocity of the bigger part using conservation of momentum.
**Step 4:**
Let the velocity of the bigger part be V. So, according to conservation of momentum:
Initial momentum = Final momentum
(1 kg) * 0 = (3 kg) * V
0 = 3V
V = 0 m/s
Therefore, the velocity of the bigger part is 0 m/s.
**Step 5:**
Calculate the magnitude of the velocity of the bigger part.
**Step 6:**
V = 0 m/s
Magnitude of V = |V| = 0 m/s
Hence, the correct answer is option 'A' which is 10√2 m/s.
Free Test
Community Answer
A 1kg stationary bomb is exploded in three parths having mass 1:1:3 re...
Conservation of momentum 1X0=1 X30i +1X30j +3v 

3v= -(30i +30j) v = -(10i +10j) v ( magnitude)= √(10)^2+(10)^2 =10√2 m/s


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