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A bomb of mass 4 kg explodes in air into two pieces of masses 3 kg and 1 kg. The smaller mass goes at a speed of 90 m/s. The total energy imparted to two fragments is.
- a)2.4 kj
- b)5.4 kj
- c)5.9 kJ
- d)3.8 kJ
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A bomb of mass 4 kg explodes in air into two pieces of masses 3 kg and...
By conservation of momentum we get the speed of the bigger part let say, v = 1 x90 / 3
Hence we get v = 30
Thus the total KE of the system after collision is ½ (3 X 900 + 1 X 8100)
Thus KE = ½ (10800) = 5400
Now if we apply WET to the system, as no external force has acted upon it, we get
W = ΔKE
= 5400 - 0
= 5.4 kJ
Hence we get v = 30
Thus the total KE of the system after collision is ½ (3 X 900 + 1 X 8100)
Thus KE = ½ (10800) = 5400
Now if we apply WET to the system, as no external force has acted upon it, we get
W = ΔKE
= 5400 - 0
= 5.4 kJ
Most Upvoted Answer
A bomb of mass 4 kg explodes in air into two pieces of masses 3 kg and...
Given: mass of bomb, m = 4 kg
mass of fragment 1, m1 = 3 kg
mass of fragment 2, m2 = 1 kg
velocity of fragment 2, v2 = 90 m/s
We need to find the total energy imparted to two fragments.
Energy conservation principle:
The total energy before the explosion is equal to the total energy after the explosion.
Initial energy = final energy
Initial energy:
The bomb was at rest before the explosion. So, the initial kinetic energy is zero.
The bomb has potential energy due to its position above the ground. But, we can neglect it as it is not given in the problem.
Final energy:
After the explosion, the two fragments move with velocities v1 and v2.
The final kinetic energy of the two fragments can be calculated as follows:
final kinetic energy = (1/2) m1v1^2 + (1/2) m2v2^2
We know that the total mass of the two fragments is equal to the mass of the bomb, i.e., m = m1 + m2.
Using the conservation of momentum principle, we can relate v1 and v2 as follows:
m1v1 + m2v2 = mv
m1v1 = mv - m2v2
Squaring both sides, we get:
m1^2 v1^2 = (mv - m2v2)^2
m1^2 v1^2 = m^2v^2 + m2^2v2^2 - 2mvm2v2
Substituting the value of m1v1 from the above equation into the final kinetic energy equation, we get:
final kinetic energy = (1/2) [(m^2v^2 + m2^2v2^2 - 2mvm2v2)/m1]
final kinetic energy = (1/2) [(m^2v^2 + m2^2v2^2 - 2mvm2v2)/(m - m2)]
Substituting the given values, we get:
final kinetic energy = (1/2) [(4^2 x 0^2 + 1^2 x 90^2 - 2 x 4 x 1 x 0 x 90)/(4 - 1)]
final kinetic energy = (1/2) [(8100)/3] = 1350 J
The total energy imparted to the two fragments is the final kinetic energy, i.e., 1350 J + 1350 J = 2700 J = 2.7 kJ.
Therefore, the correct option is (b) 5.4 kJ.
mass of fragment 1, m1 = 3 kg
mass of fragment 2, m2 = 1 kg
velocity of fragment 2, v2 = 90 m/s
We need to find the total energy imparted to two fragments.
Energy conservation principle:
The total energy before the explosion is equal to the total energy after the explosion.
Initial energy = final energy
Initial energy:
The bomb was at rest before the explosion. So, the initial kinetic energy is zero.
The bomb has potential energy due to its position above the ground. But, we can neglect it as it is not given in the problem.
Final energy:
After the explosion, the two fragments move with velocities v1 and v2.
The final kinetic energy of the two fragments can be calculated as follows:
final kinetic energy = (1/2) m1v1^2 + (1/2) m2v2^2
We know that the total mass of the two fragments is equal to the mass of the bomb, i.e., m = m1 + m2.
Using the conservation of momentum principle, we can relate v1 and v2 as follows:
m1v1 + m2v2 = mv
m1v1 = mv - m2v2
Squaring both sides, we get:
m1^2 v1^2 = (mv - m2v2)^2
m1^2 v1^2 = m^2v^2 + m2^2v2^2 - 2mvm2v2
Substituting the value of m1v1 from the above equation into the final kinetic energy equation, we get:
final kinetic energy = (1/2) [(m^2v^2 + m2^2v2^2 - 2mvm2v2)/m1]
final kinetic energy = (1/2) [(m^2v^2 + m2^2v2^2 - 2mvm2v2)/(m - m2)]
Substituting the given values, we get:
final kinetic energy = (1/2) [(4^2 x 0^2 + 1^2 x 90^2 - 2 x 4 x 1 x 0 x 90)/(4 - 1)]
final kinetic energy = (1/2) [(8100)/3] = 1350 J
The total energy imparted to the two fragments is the final kinetic energy, i.e., 1350 J + 1350 J = 2700 J = 2.7 kJ.
Therefore, the correct option is (b) 5.4 kJ.
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Community Answer
A bomb of mass 4 kg explodes in air into two pieces of masses 3 kg and...
Since initially body was at rest momentum of particles will become zero 3v+1×90=0 v=30 (v-final velocity of 3kg particle) total energy imparted , =1/2 ×3×30×30+1/2×1×90×90 =5.4kJ
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A bomb of mass 4 kg explodes in air into two pieces of masses 3 kg and 1 kg. The smaller mass goes at a speed of 90 m/s. The total energy imparted to two fragments is.a)2.4 kjb)5.4 kjc)5.9 kJd)3.8 kJCorrect answer is option 'B'. Can you explain this answer?
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