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Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity v/2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction.
- a)Same as that of B [2012]
- b)Opposite to that of B
- c)θ = tan–1 (1/2) to the x-axis
- d)θ = tan–1 (–1/2) to the x-axis
Correct answer is option 'C'. Can you explain this answer?
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Two spheres A and B of masses m1 and m2 respectively collide. A is at ...
Perpendicular to the original direction of B
d)Cannot be determined
Answer:
c) Perpendicular to the original direction of B
Explanation:
We can solve this problem using conservation of momentum and conservation of kinetic energy.
Before the collision, the total momentum of the system is:
p = m2v
Since sphere A is at rest, its momentum is zero.
The total kinetic energy of the system before the collision is:
K = (1/2)m2v^2
After the collision, the spheres move in different directions. Let the velocity of sphere A be u and the velocity of sphere B be w. Then, the total momentum of the system after the collision is:
p' = m1u + m2w
Since sphere B moves perpendicular to its original direction, we can write:
w = 2v
Using conservation of momentum, we have:
m2v = m1u + m2(2v)
Simplifying, we get:
u = (m2 - 2m1)v / m1
Now, using conservation of kinetic energy, we have:
(1/2)m1u^2 + (1/2)m2w^2 = (1/2)m2v^2
Substituting the values of u and w, we get:
(m2 - 2m1)v^2 = m1u^2
Simplifying, we get:
u = sqrt((m2 - 2m1)/m1) v
Since m1 and m2 are positive, (m2 - 2m1)/m1 is negative. Therefore, u is imaginary, which means that sphere A moves in a direction perpendicular to the original direction of sphere B. Hence, the answer is option c) Perpendicular to the original direction of B.
d)Cannot be determined
Answer:
c) Perpendicular to the original direction of B
Explanation:
We can solve this problem using conservation of momentum and conservation of kinetic energy.
Before the collision, the total momentum of the system is:
p = m2v
Since sphere A is at rest, its momentum is zero.
The total kinetic energy of the system before the collision is:
K = (1/2)m2v^2
After the collision, the spheres move in different directions. Let the velocity of sphere A be u and the velocity of sphere B be w. Then, the total momentum of the system after the collision is:
p' = m1u + m2w
Since sphere B moves perpendicular to its original direction, we can write:
w = 2v
Using conservation of momentum, we have:
m2v = m1u + m2(2v)
Simplifying, we get:
u = (m2 - 2m1)v / m1
Now, using conservation of kinetic energy, we have:
(1/2)m1u^2 + (1/2)m2w^2 = (1/2)m2v^2
Substituting the values of u and w, we get:
(m2 - 2m1)v^2 = m1u^2
Simplifying, we get:
u = sqrt((m2 - 2m1)/m1) v
Since m1 and m2 are positive, (m2 - 2m1)/m1 is negative. Therefore, u is imaginary, which means that sphere A moves in a direction perpendicular to the original direction of sphere B. Hence, the answer is option c) Perpendicular to the original direction of B.
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Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity v/2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction.a)Same as that of B [2012]b)Opposite to that of Bc)θ = tan–1 (1/2) to the x-axisd)θ = tan–1 (–1/2) to the x-axisCorrect answer is option 'C'. Can you explain this answer?
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Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity v/2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction.a)Same as that of B [2012]b)Opposite to that of Bc)θ = tan–1 (1/2) to the x-axisd)θ = tan–1 (–1/2) to the x-axisCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity v/2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction.a)Same as that of B [2012]b)Opposite to that of Bc)θ = tan–1 (1/2) to the x-axisd)θ = tan–1 (–1/2) to the x-axisCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity v/2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction.a)Same as that of B [2012]b)Opposite to that of Bc)θ = tan–1 (1/2) to the x-axisd)θ = tan–1 (–1/2) to the x-axisCorrect answer is option 'C'. Can you explain this answer?.
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