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A shell of mass 5M, acted upon by no external force and initially at rest, bursts into three fragments of masses M, 2M and 2M respectively. The first two fragments move in opposite directions with velocities of magnitudes 2V and V respectively. The third fragment will
- a)move with a velocity V in a direction perpendicular to the other two
- b)move with a velocity 2V in the direction of velocity of the first fragment
- c)be at rest
- d)move with a velocity V in the direction of velocity of the second fragment
Correct answer is option 'C'. Can you explain this answer?
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A shell of mass 5M, acted upon by no external force and initially at r...
By conservation of momentum
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A shell of mass 5M, acted upon by no external force and initially at r...
Solution:
Given:
Mass of shell (initial) = 5M
Mass of first fragment = M
Mass of second fragment = 2M
Mass of third fragment = 2M
Velocities:
Magnitude of velocity of first fragment = 2V
Magnitude of velocity of second fragment = V
The problem states that the shell is acted upon by no external force, which means that the total momentum before and after the explosion should remain conserved.
1. Conservation of Momentum:
The initial momentum of the system is zero since the shell is initially at rest. Therefore, according to the law of conservation of momentum, the total momentum after the explosion should also be zero.
2. Finding the Velocities of the Fragments:
Let's assume the velocity of the third fragment after the explosion is V3.
The momentum of the first fragment (M1) is given by:
M1 = M * 2V (since mass = M and velocity = 2V)
The momentum of the second fragment (M2) is given by:
M2 = 2M * V (since mass = 2M and velocity = V)
The momentum of the third fragment (M3) is given by:
M3 = 2M * V3 (since mass = 2M and velocity = V3)
3. Applying Conservation of Momentum:
According to the law of conservation of momentum, the sum of the momenta of the individual fragments should be zero.
Therefore, M1 + M2 + M3 = 0
Substituting the values of M1, M2, and M3:
2M * 2V + 2M * V + 2M * V3 = 0
4. Simplifying the Equation:
4MV + 2MV + 2MV3 = 0
8MV + 2MV3 = 0
2MV + MV3 = 0
Dividing both sides of the equation by M:
2V + V3 = 0
Simplifying further:
V3 = -2V
5. Conclusion:
The negative sign indicates that the velocity of the third fragment is in the opposite direction to the first two fragments. Since the velocity is non-zero, it means that the third fragment is moving. Therefore, the correct answer is option 'C' - the third fragment will be at rest.
Given:
Mass of shell (initial) = 5M
Mass of first fragment = M
Mass of second fragment = 2M
Mass of third fragment = 2M
Velocities:
Magnitude of velocity of first fragment = 2V
Magnitude of velocity of second fragment = V
The problem states that the shell is acted upon by no external force, which means that the total momentum before and after the explosion should remain conserved.
1. Conservation of Momentum:
The initial momentum of the system is zero since the shell is initially at rest. Therefore, according to the law of conservation of momentum, the total momentum after the explosion should also be zero.
2. Finding the Velocities of the Fragments:
Let's assume the velocity of the third fragment after the explosion is V3.
The momentum of the first fragment (M1) is given by:
M1 = M * 2V (since mass = M and velocity = 2V)
The momentum of the second fragment (M2) is given by:
M2 = 2M * V (since mass = 2M and velocity = V)
The momentum of the third fragment (M3) is given by:
M3 = 2M * V3 (since mass = 2M and velocity = V3)
3. Applying Conservation of Momentum:
According to the law of conservation of momentum, the sum of the momenta of the individual fragments should be zero.
Therefore, M1 + M2 + M3 = 0
Substituting the values of M1, M2, and M3:
2M * 2V + 2M * V + 2M * V3 = 0
4. Simplifying the Equation:
4MV + 2MV + 2MV3 = 0
8MV + 2MV3 = 0
2MV + MV3 = 0
Dividing both sides of the equation by M:
2V + V3 = 0
Simplifying further:
V3 = -2V
5. Conclusion:
The negative sign indicates that the velocity of the third fragment is in the opposite direction to the first two fragments. Since the velocity is non-zero, it means that the third fragment is moving. Therefore, the correct answer is option 'C' - the third fragment will be at rest.
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A shell of mass 5M, acted upon by no external force and initially at rest, bursts into three fragments of masses M, 2M and 2M respectively. The first two fragments move in opposite directions with velocities of magnitudes 2V and V respectively. The third fragment willa)move with a velocity V in a direction perpendicular to the other twob)move with a velocity 2V in the direction of velocity of the first fragmentc)be at restd)move with a velocity V in the direction of velocity of the second fragmentCorrect answer is option 'C'. Can you explain this answer?
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A shell of mass 5M, acted upon by no external force and initially at rest, bursts into three fragments of masses M, 2M and 2M respectively. The first two fragments move in opposite directions with velocities of magnitudes 2V and V respectively. The third fragment willa)move with a velocity V in a direction perpendicular to the other twob)move with a velocity 2V in the direction of velocity of the first fragmentc)be at restd)move with a velocity V in the direction of velocity of the second fragmentCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A shell of mass 5M, acted upon by no external force and initially at rest, bursts into three fragments of masses M, 2M and 2M respectively. The first two fragments move in opposite directions with velocities of magnitudes 2V and V respectively. The third fragment willa)move with a velocity V in a direction perpendicular to the other twob)move with a velocity 2V in the direction of velocity of the first fragmentc)be at restd)move with a velocity V in the direction of velocity of the second fragmentCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A shell of mass 5M, acted upon by no external force and initially at rest, bursts into three fragments of masses M, 2M and 2M respectively. The first two fragments move in opposite directions with velocities of magnitudes 2V and V respectively. The third fragment willa)move with a velocity V in a direction perpendicular to the other twob)move with a velocity 2V in the direction of velocity of the first fragmentc)be at restd)move with a velocity V in the direction of velocity of the second fragmentCorrect answer is option 'C'. Can you explain this answer?.
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