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A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with a speed v each in mutually perpendicular directions. The minimum energy released in the process of explosion is :
- a)(3/2) mv2
- b)(2/3) mv2
- c)(4/3) mv2
- d)(3/4) mv2
Correct answer is option 'A'. Can you explain this answer?
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A particle of mass4mwhich is at rest explodes into three fragments. Tw...
The correct answer is: (3/2) mv2
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A particle of mass4mwhich is at rest explodes into three fragments. Tw...
Explanation:
Given:
Mass of the particle before explosion = 4m
Mass of each fragment after explosion = m
Speed of each fragment after explosion = v
To find:
Minimum energy released in the process of explosion.
- To solve this problem, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy.
Conservation of momentum:
Before the explosion, the particle is at rest, so the initial momentum is zero.
After the explosion, the two fragments are moving in mutually perpendicular directions, so the momentum of each fragment is mv.
According to the principle of conservation of momentum:
Initial momentum = Final momentum
0 = 2mv + 2mv
0 = 4mv
From this equation, we can conclude that the magnitude of velocity (v) of each fragment is zero. However, since we are looking for the minimum energy released, we need to consider the case where the fragments have non-zero velocities.
Conservation of kinetic energy:
The initial kinetic energy of the particle is zero since it is at rest.
After the explosion, the fragments have kinetic energy due to their non-zero velocities.
The total initial kinetic energy = 0
The total final kinetic energy = KE1 + KE2 + KE3
Since the fragments have equal masses and velocities, the kinetic energy of each fragment is given by:
KE = (1/2)mv^2
The total final kinetic energy = 2 * (1/2)mv^2 + (1/2)mv^2 + (1/2)mv^2
= mv^2 + mv^2 + mv^2
= 3mv^2
Therefore, the minimum energy released in the process of explosion is 3mv^2, which corresponds to option A: (3/2)mv^2.
Given:
Mass of the particle before explosion = 4m
Mass of each fragment after explosion = m
Speed of each fragment after explosion = v
To find:
Minimum energy released in the process of explosion.
- To solve this problem, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy.
Conservation of momentum:
Before the explosion, the particle is at rest, so the initial momentum is zero.
After the explosion, the two fragments are moving in mutually perpendicular directions, so the momentum of each fragment is mv.
According to the principle of conservation of momentum:
Initial momentum = Final momentum
0 = 2mv + 2mv
0 = 4mv
From this equation, we can conclude that the magnitude of velocity (v) of each fragment is zero. However, since we are looking for the minimum energy released, we need to consider the case where the fragments have non-zero velocities.
Conservation of kinetic energy:
The initial kinetic energy of the particle is zero since it is at rest.
After the explosion, the fragments have kinetic energy due to their non-zero velocities.
The total initial kinetic energy = 0
The total final kinetic energy = KE1 + KE2 + KE3
Since the fragments have equal masses and velocities, the kinetic energy of each fragment is given by:
KE = (1/2)mv^2
The total final kinetic energy = 2 * (1/2)mv^2 + (1/2)mv^2 + (1/2)mv^2
= mv^2 + mv^2 + mv^2
= 3mv^2
Therefore, the minimum energy released in the process of explosion is 3mv^2, which corresponds to option A: (3/2)mv^2.
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