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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 and 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
- 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 'D'. Can you explain this answer?
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Most Upvoted Answer
Two spheres A and B of masses m1 and m2 respectively collide. A is at...
According to law of conservation of linear momentum along x-axis, we get
m1 × 0 + m2 × v = m1v′cosθ
m2v = m1v′cosθ ......(i)
According to law of conservation of linear momentum long y-axis, we get
Divide (ii) by (i), we get
tanθ = −1/2 or θ = tan−1(−1/2) to the x-axis
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Two spheres A and B of masses m1 and m2 respectively collide. A is at...
Given:
- Sphere A is at rest initially.
- Sphere B has a mass of m2 and is moving with velocity v along the x-axis.
- After the collision, sphere B has a velocity of v/2 in a direction perpendicular to the original direction.
To find:
The direction in which sphere A moves after the collision.
Solution:
Step 1: Analyzing the collision
- In this collision, both momentum and kinetic energy are conserved.
- Momentum conservation: The total momentum before the collision is equal to the total momentum after the collision.
- Kinetic energy conservation: The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Step 2: Using momentum conservation
- Since sphere A is at rest initially, its initial momentum is zero.
- The momentum of sphere B before the collision is given by: momentum_B_initial = m2 * v (along the x-axis).
- The momentum of sphere B after the collision is given by: momentum_B_final = m2 * (v/2) (perpendicular to the x-axis).
- The momentum of sphere A after the collision is given by: momentum_A_final = m1 * v_A_final (unknown direction).
- Using momentum conservation, we can write the equation:
momentum_B_initial + momentum_A_initial = momentum_B_final + momentum_A_final
m2 * v = m2 * (v/2) + m1 * v_A_final
Step 3: Using kinetic energy conservation
- The kinetic energy of sphere B before the collision is given by: KE_B_initial = (1/2) * m2 * v^2
- The kinetic energy of sphere B after the collision is given by: KE_B_final = (1/2) * m2 * (v/2)^2 = (1/8) * m2 * v^2 (perpendicular to the x-axis).
- The kinetic energy of sphere A after the collision is given by: KE_A_final = (1/2) * m1 * v_A_final^2 (unknown value).
- Using kinetic energy conservation, we can write the equation:
KE_B_initial + KE_A_initial = KE_B_final + KE_A_final
(1/2) * m2 * v^2 = (1/8) * m2 * v^2 + (1/2) * m1 * v_A_final^2
Step 4: Solving the equations
- From the momentum conservation equation, we can simplify it as:
m2 * v = (m2/2) * v + m1 * v_A_final
(v_A_final) = (v/2) * (1 - m2/m1)
- Substituting this value of v_A_final in the kinetic energy conservation equation, we get:
(1/2) * m2 * v^2 = (1/8) * m2 * v^2 + (1/2) * m1 * [(v/2) * (1 - m2/m1)]^2
(1/2) * m2 * v^2 = (1/8) * m2 * v^2 + (1/2) * m1 * [(v^2/4) * (1
- Sphere A is at rest initially.
- Sphere B has a mass of m2 and is moving with velocity v along the x-axis.
- After the collision, sphere B has a velocity of v/2 in a direction perpendicular to the original direction.
To find:
The direction in which sphere A moves after the collision.
Solution:
Step 1: Analyzing the collision
- In this collision, both momentum and kinetic energy are conserved.
- Momentum conservation: The total momentum before the collision is equal to the total momentum after the collision.
- Kinetic energy conservation: The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Step 2: Using momentum conservation
- Since sphere A is at rest initially, its initial momentum is zero.
- The momentum of sphere B before the collision is given by: momentum_B_initial = m2 * v (along the x-axis).
- The momentum of sphere B after the collision is given by: momentum_B_final = m2 * (v/2) (perpendicular to the x-axis).
- The momentum of sphere A after the collision is given by: momentum_A_final = m1 * v_A_final (unknown direction).
- Using momentum conservation, we can write the equation:
momentum_B_initial + momentum_A_initial = momentum_B_final + momentum_A_final
m2 * v = m2 * (v/2) + m1 * v_A_final
Step 3: Using kinetic energy conservation
- The kinetic energy of sphere B before the collision is given by: KE_B_initial = (1/2) * m2 * v^2
- The kinetic energy of sphere B after the collision is given by: KE_B_final = (1/2) * m2 * (v/2)^2 = (1/8) * m2 * v^2 (perpendicular to the x-axis).
- The kinetic energy of sphere A after the collision is given by: KE_A_final = (1/2) * m1 * v_A_final^2 (unknown value).
- Using kinetic energy conservation, we can write the equation:
KE_B_initial + KE_A_initial = KE_B_final + KE_A_final
(1/2) * m2 * v^2 = (1/8) * m2 * v^2 + (1/2) * m1 * v_A_final^2
Step 4: Solving the equations
- From the momentum conservation equation, we can simplify it as:
m2 * v = (m2/2) * v + m1 * v_A_final
(v_A_final) = (v/2) * (1 - m2/m1)
- Substituting this value of v_A_final in the kinetic energy conservation equation, we get:
(1/2) * m2 * v^2 = (1/8) * m2 * v^2 + (1/2) * m1 * [(v/2) * (1 - m2/m1)]^2
(1/2) * m2 * v^2 = (1/8) * m2 * v^2 + (1/2) * m1 * [(v^2/4) * (1
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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 and 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 directiona)same as that of Bb)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 'D'. 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 and 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 directiona)same as that of Bb)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 'D'. 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 and 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 directiona)same as that of Bb)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 'D'. 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 and 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 directiona)same as that of Bb)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 'D'. Can you explain this answer?.
Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v and 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 directiona)same as that of Bb)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 'D'. 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 and 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 directiona)same as that of Bb)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 'D'. 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 and 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 directiona)same as that of Bb)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 'D'. Can you explain this answer?.
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