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A particle of mass m moving with velocity u1 collides elastically with another particle of same mass moving with velocity u2 in the same direction. After collision their speeds are v1 and v2 respectively then
(A) u1 + v1 = v2 + u2
(B) u1 – v1 = v2 + u2
(A) u1 + v1 = v2 + u2
(B) u1 – v1 = v2 + u2
- a)Equation B is correct but not A
- b)Equation A is correct but not B
- c)Both the equations A and B are incorrect
- d)Both the equations A and B are correct
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A particle of massmmoving with velocityu1collides elastically with ano...
Equation A is correct but not B
(For elastic collision e = 1)
⇒ 1(u2 – u1) = –(v2 – v1)
⇒ u2 + v2 = u1 + v1
The correct answer is: Equation A is correct but not B
(For elastic collision e = 1)⇒ 1(u2 – u1) = –(v2 – v1)
⇒ u2 + v2 = u1 + v1
The correct answer is: Equation A is correct but not B
Most Upvoted Answer
A particle of massmmoving with velocityu1collides elastically with ano...
Equation A is correct but not B
(For elastic collision e = 1)
⇒ 1(u2 – u1) = –(v2 – v1)
⇒ u2 + v2 = u1 + v1
The correct answer is: Equation A is correct but not B
(For elastic collision e = 1)⇒ 1(u2 – u1) = –(v2 – v1)
⇒ u2 + v2 = u1 + v1
The correct answer is: Equation A is correct but not B
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Community Answer
A particle of massmmoving with velocityu1collides elastically with ano...
V1 = u2
(C)u1(v1-v2) = u2(v2-v1)
(D)u1(v1+v2) = u2(v1+v2)
Let the masses of both particles be m, and their velocities before the collision be u1 and u2 respectively. After the collision, their velocities become v1 and v2 respectively.
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
Initial momentum = m * u1 + m * u2
Final momentum = m * v1 + m * v2
Since the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Initial kinetic energy = (1/2) * m * (u1)^2 + (1/2) * m * (u2)^2
Final kinetic energy = (1/2) * m * (v1)^2 + (1/2) * m * (v2)^2
Using the conservation of momentum and kinetic energy, we can solve for v1 and v2.
By conservation of momentum:
m * u1 + m * u2 = m * v1 + m * v2
By conservation of kinetic energy:
(1/2) * m * (u1)^2 + (1/2) * m * (u2)^2 = (1/2) * m * (v1)^2 + (1/2) * m * (v2)^2
Simplifying the equations, we get:
u1 + u2 = v1 + v2 (1)
u1^2 + u2^2 = v1^2 + v2^2 (2)
Multiplying equation (1) by u1 and equation (2) by u2, we get:
u1 * (u1 + u2) = u1 * (v1 + v2)
u1^2 + u1 * u2 = u1 * v1 + u1 * v2 (3)
u2 * (u1 + u2) = u2 * (v1 + v2)
u1 * u2 + u2^2 = u2 * v1 + u2 * v2 (4)
Subtracting equation (4) from equation (3), we get:
u1^2 - u2^2 = u1 * v1 - u2 * v1 + u1 * v2 - u2 * v2
(u1 + u2)(u1 - u2) = v1(u1 - u2) + v2(u1 - u2)
(u1 - u2)(u1 + u2 - v1 - v2) = 0
Since u1 and u2 are not equal, we can divide both sides of the equation by (u1 - u2) to get:
u1 + u2 = v1 + v2
Therefore, the correct option is (A) u1 + u2 = v1 + v2.
(C)u1(v1-v2) = u2(v2-v1)
(D)u1(v1+v2) = u2(v1+v2)
Let the masses of both particles be m, and their velocities before the collision be u1 and u2 respectively. After the collision, their velocities become v1 and v2 respectively.
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
Initial momentum = m * u1 + m * u2
Final momentum = m * v1 + m * v2
Since the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Initial kinetic energy = (1/2) * m * (u1)^2 + (1/2) * m * (u2)^2
Final kinetic energy = (1/2) * m * (v1)^2 + (1/2) * m * (v2)^2
Using the conservation of momentum and kinetic energy, we can solve for v1 and v2.
By conservation of momentum:
m * u1 + m * u2 = m * v1 + m * v2
By conservation of kinetic energy:
(1/2) * m * (u1)^2 + (1/2) * m * (u2)^2 = (1/2) * m * (v1)^2 + (1/2) * m * (v2)^2
Simplifying the equations, we get:
u1 + u2 = v1 + v2 (1)
u1^2 + u2^2 = v1^2 + v2^2 (2)
Multiplying equation (1) by u1 and equation (2) by u2, we get:
u1 * (u1 + u2) = u1 * (v1 + v2)
u1^2 + u1 * u2 = u1 * v1 + u1 * v2 (3)
u2 * (u1 + u2) = u2 * (v1 + v2)
u1 * u2 + u2^2 = u2 * v1 + u2 * v2 (4)
Subtracting equation (4) from equation (3), we get:
u1^2 - u2^2 = u1 * v1 - u2 * v1 + u1 * v2 - u2 * v2
(u1 + u2)(u1 - u2) = v1(u1 - u2) + v2(u1 - u2)
(u1 - u2)(u1 + u2 - v1 - v2) = 0
Since u1 and u2 are not equal, we can divide both sides of the equation by (u1 - u2) to get:
u1 + u2 = v1 + v2
Therefore, the correct option is (A) u1 + u2 = v1 + v2.
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A particle of massmmoving with velocityu1collides elastically with another particle of same mass moving with velocityu2in the same direction. After collision their speeds arev1andv2respectively then(A)u1+v1=v2+u2(B)u1–v1=v2+u2a)EquationBis correct but notAb)EquationAis correct but notBc)Both the equationsAandBare incorrectd)Both the equationsAandBare correctCorrect answer is option 'B'. Can you explain this answer?
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A particle of massmmoving with velocityu1collides elastically with another particle of same mass moving with velocityu2in the same direction. After collision their speeds arev1andv2respectively then(A)u1+v1=v2+u2(B)u1–v1=v2+u2a)EquationBis correct but notAb)EquationAis correct but notBc)Both the equationsAandBare incorrectd)Both the equationsAandBare correctCorrect answer is option 'B'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A particle of massmmoving with velocityu1collides elastically with another particle of same mass moving with velocityu2in the same direction. After collision their speeds arev1andv2respectively then(A)u1+v1=v2+u2(B)u1–v1=v2+u2a)EquationBis correct but notAb)EquationAis correct but notBc)Both the equationsAandBare incorrectd)Both the equationsAandBare correctCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of massmmoving with velocityu1collides elastically with another particle of same mass moving with velocityu2in the same direction. After collision their speeds arev1andv2respectively then(A)u1+v1=v2+u2(B)u1–v1=v2+u2a)EquationBis correct but notAb)EquationAis correct but notBc)Both the equationsAandBare incorrectd)Both the equationsAandBare correctCorrect answer is option 'B'. Can you explain this answer?.
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