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Two particles A and B of mass m and one particle C of mass are kept on the x axis in thr order ABC, Particle A is given a velocity vi . Consequently there are two collisions, both of which are completely inelastic. If the net energy loss because of these collisions is 7/8 of the initial energy,, the value of M is (ignore frictional losses)
- a)8 m
- b)6 m
- c)4 m
- d)2 m
Correct answer is option 'B'. Can you explain this answer?
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Two particles A and B of mass m and one particle C of mass are kept on...
Given:
- Particles A and B have mass m
- Particle C has mass M
- Initial velocity of particle A is vi
- There are two completely inelastic collisions
- Net energy loss due to collisions is 7/8 of the initial energy
To Find:
The value of M
Solution:
Step 1: Conservation of Momentum
In an inelastic collision, the total momentum before and after the collision remains conserved. Therefore, we can write the equation for conservation of momentum for the first collision as:
m * vi = (m + m + M) * vf1
where vf1 is the final velocity of the three particles after the first collision.
Similarly, for the second collision, the equation for conservation of momentum can be written as:
(m + m + M) * vf1 = (2m + M) * vf2
where vf2 is the final velocity of the two particles A and B after the second collision.
Step 2: Initial Energy
The initial energy of the system is given by the kinetic energy of particle A, which can be calculated as:
Initial Energy = (1/2) * m * vi^2
Step 3: Final Energy
The final energy of the system after the two collisions can be calculated as the sum of the kinetic energies of particles A and B. Since the collisions are completely inelastic, the masses of A and B combine after the collisions.
Final Energy = (1/2) * (2m + M) * vf2^2
Step 4: Energy Loss
The energy loss due to the collisions can be calculated as the difference between the initial energy and the final energy:
Energy Loss = Initial Energy - Final Energy
= (1/2) * m * vi^2 - (1/2) * (2m + M) * vf2^2
Given that the net energy loss is 7/8 of the initial energy, we have:
(7/8) * Initial Energy = Energy Loss
Substituting the values from the above equations, we get:
(7/8) * (1/2) * m * vi^2 = (1/2) * m * vi^2 - (1/2) * (2m + M) * vf2^2
After simplifying, we have:
(7/8) = 1 - (1/2) * (2m + M) * vf2^2 / (m * vi^2)
Step 5: Solving for M
Now, let's substitute the value of vf2 from the conservation of momentum equation for the second collision. After simplifying, we get:
vf2 = vf1 * (m + m + M) / (2m + M)
Substituting this value in the equation obtained in Step 4, we can solve for M:
(7/8) = 1 - (1/2) * (2m + M) * [vf1 * (m + m + M) / (2m + M)]^2 / (m * vi^2)
Simplifying further, we get:
(7/8) = 1 - (1/2) * vf1^2
- Particles A and B have mass m
- Particle C has mass M
- Initial velocity of particle A is vi
- There are two completely inelastic collisions
- Net energy loss due to collisions is 7/8 of the initial energy
To Find:
The value of M
Solution:
Step 1: Conservation of Momentum
In an inelastic collision, the total momentum before and after the collision remains conserved. Therefore, we can write the equation for conservation of momentum for the first collision as:
m * vi = (m + m + M) * vf1
where vf1 is the final velocity of the three particles after the first collision.
Similarly, for the second collision, the equation for conservation of momentum can be written as:
(m + m + M) * vf1 = (2m + M) * vf2
where vf2 is the final velocity of the two particles A and B after the second collision.
Step 2: Initial Energy
The initial energy of the system is given by the kinetic energy of particle A, which can be calculated as:
Initial Energy = (1/2) * m * vi^2
Step 3: Final Energy
The final energy of the system after the two collisions can be calculated as the sum of the kinetic energies of particles A and B. Since the collisions are completely inelastic, the masses of A and B combine after the collisions.
Final Energy = (1/2) * (2m + M) * vf2^2
Step 4: Energy Loss
The energy loss due to the collisions can be calculated as the difference between the initial energy and the final energy:
Energy Loss = Initial Energy - Final Energy
= (1/2) * m * vi^2 - (1/2) * (2m + M) * vf2^2
Given that the net energy loss is 7/8 of the initial energy, we have:
(7/8) * Initial Energy = Energy Loss
Substituting the values from the above equations, we get:
(7/8) * (1/2) * m * vi^2 = (1/2) * m * vi^2 - (1/2) * (2m + M) * vf2^2
After simplifying, we have:
(7/8) = 1 - (1/2) * (2m + M) * vf2^2 / (m * vi^2)
Step 5: Solving for M
Now, let's substitute the value of vf2 from the conservation of momentum equation for the second collision. After simplifying, we get:
vf2 = vf1 * (m + m + M) / (2m + M)
Substituting this value in the equation obtained in Step 4, we can solve for M:
(7/8) = 1 - (1/2) * (2m + M) * [vf1 * (m + m + M) / (2m + M)]^2 / (m * vi^2)
Simplifying further, we get:
(7/8) = 1 - (1/2) * vf1^2
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Two particles A and B of mass m and one particle C of mass are kept on the x axis in thr order ABC, Particle A is given a velocity vi. Consequently there are two collisions, both of which are completely inelastic. If the net energy loss because of these collisions is 7/8 of the initial energy,, the value of M is (ignore frictional losses)a)8 mb)6 mc)4 md)2 mCorrect answer is option 'B'. Can you explain this answer?
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