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A radioactive nucleus of mass M moving along the positive x-direction with speed v emits an alpha particle of mass m. If the alpha particle proceeds along the positive y-direction, the centre of mass of the system (made of the daughter nucleus and the alpha particle) will:
- a)move along the positive x-direction with speed equal to v
- b)remain at rest
- c)move along the positive x-direction with speed less than v
- d)move along the positive x-direction with speed greater than v
Correct answer is option 'A'. Can you explain this answer?
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A radioactive nucleus of mass M moving along the positive x-direction ...
The momentum of the initial particle remains conserved. Hence the emitted particles’ COM will remain along +ve x-axis.
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A radioactive nucleus of mass M moving along the positive x-direction ...
**Explanation:**
When a radioactive nucleus emits an alpha particle, the momentum of the system should be conserved. According to the law of conservation of momentum, the total momentum before and after the emission should be the same.
Let's consider the initial momentum of the system before the emission:
Initial momentum = Momentum of the nucleus + Momentum of the alpha particle
The momentum of the nucleus can be calculated as the product of its mass (M) and velocity (v):
Momentum of the nucleus = M * v
Since the alpha particle is emitted in the positive y-direction, its momentum can be calculated as the product of its mass (m) and velocity (v_alpha):
Momentum of the alpha particle = m * v_alpha
To conserve momentum, the total momentum after the emission should be equal to the initial momentum:
Total momentum after emission = Momentum of the daughter nucleus + Momentum of the alpha particle
Since the alpha particle proceeds along the positive y-direction, the momentum of the daughter nucleus should be in the negative x-direction to balance the momentum. Therefore, the momentum of the daughter nucleus can be calculated as the product of its mass (M - m) and velocity (v_daughter):
Momentum of the daughter nucleus = (M - m) * v_daughter
Now, applying the law of conservation of momentum:
Momentum of the nucleus + Momentum of the alpha particle = Momentum of the daughter nucleus + Momentum of the alpha particle
M * v + m * v_alpha = (M - m) * v_daughter + m * v_alpha
Since the question states that the alpha particle proceeds along the positive y-direction, the velocity of the alpha particle in the x-direction (v_alpha) is zero. Hence, the equation becomes:
M * v = (M - m) * v_daughter
Now, solving for the velocity of the daughter nucleus (v_daughter):
v_daughter = (M * v) / (M - m)
Since the mass of the alpha particle (m) is much smaller than the mass of the nucleus (M), the denominator (M - m) is approximately equal to M. Hence, the equation becomes:
v_daughter = (M * v) / M = v
Therefore, the velocity of the daughter nucleus is equal to the velocity of the original nucleus (v), and it moves along the positive x-direction. Hence, the correct answer is option 'A' - the center of mass of the system moves along the positive x-direction with a speed equal to v.
When a radioactive nucleus emits an alpha particle, the momentum of the system should be conserved. According to the law of conservation of momentum, the total momentum before and after the emission should be the same.
Let's consider the initial momentum of the system before the emission:
Initial momentum = Momentum of the nucleus + Momentum of the alpha particle
The momentum of the nucleus can be calculated as the product of its mass (M) and velocity (v):
Momentum of the nucleus = M * v
Since the alpha particle is emitted in the positive y-direction, its momentum can be calculated as the product of its mass (m) and velocity (v_alpha):
Momentum of the alpha particle = m * v_alpha
To conserve momentum, the total momentum after the emission should be equal to the initial momentum:
Total momentum after emission = Momentum of the daughter nucleus + Momentum of the alpha particle
Since the alpha particle proceeds along the positive y-direction, the momentum of the daughter nucleus should be in the negative x-direction to balance the momentum. Therefore, the momentum of the daughter nucleus can be calculated as the product of its mass (M - m) and velocity (v_daughter):
Momentum of the daughter nucleus = (M - m) * v_daughter
Now, applying the law of conservation of momentum:
Momentum of the nucleus + Momentum of the alpha particle = Momentum of the daughter nucleus + Momentum of the alpha particle
M * v + m * v_alpha = (M - m) * v_daughter + m * v_alpha
Since the question states that the alpha particle proceeds along the positive y-direction, the velocity of the alpha particle in the x-direction (v_alpha) is zero. Hence, the equation becomes:
M * v = (M - m) * v_daughter
Now, solving for the velocity of the daughter nucleus (v_daughter):
v_daughter = (M * v) / (M - m)
Since the mass of the alpha particle (m) is much smaller than the mass of the nucleus (M), the denominator (M - m) is approximately equal to M. Hence, the equation becomes:
v_daughter = (M * v) / M = v
Therefore, the velocity of the daughter nucleus is equal to the velocity of the original nucleus (v), and it moves along the positive x-direction. Hence, the correct answer is option 'A' - the center of mass of the system moves along the positive x-direction with a speed equal to v.
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A radioactive nucleus of mass M moving along the positive x-direction with speed v emits an alpha particle of mass m. If the alpha particle proceeds along the positive y-direction, the centre of mass of the system (made of the daughter nucleus and the alpha particle) will:a)move along the positive x-direction with speed equal to vb)remain at restc)move along the positive x-direction with speed less than vd)move along the positive x-direction with speed greater than vCorrect answer is option 'A'. Can you explain this answer?
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A radioactive nucleus of mass M moving along the positive x-direction with speed v emits an alpha particle of mass m. If the alpha particle proceeds along the positive y-direction, the centre of mass of the system (made of the daughter nucleus and the alpha particle) will:a)move along the positive x-direction with speed equal to vb)remain at restc)move along the positive x-direction with speed less than vd)move along the positive x-direction with speed greater than vCorrect answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A radioactive nucleus of mass M moving along the positive x-direction with speed v emits an alpha particle of mass m. If the alpha particle proceeds along the positive y-direction, the centre of mass of the system (made of the daughter nucleus and the alpha particle) will:a)move along the positive x-direction with speed equal to vb)remain at restc)move along the positive x-direction with speed less than vd)move along the positive x-direction with speed greater than vCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A radioactive nucleus of mass M moving along the positive x-direction with speed v emits an alpha particle of mass m. If the alpha particle proceeds along the positive y-direction, the centre of mass of the system (made of the daughter nucleus and the alpha particle) will:a)move along the positive x-direction with speed equal to vb)remain at restc)move along the positive x-direction with speed less than vd)move along the positive x-direction with speed greater than vCorrect answer is option 'A'. Can you explain this answer?.
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