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A 10kg inelastic ball A moving with speed 4 m/s collides obliquely with a 8kg ball B kept at rest on a horizontal smooth surface. after collision A moves with 37 degree with the initial direction of motion. what would be minimum possible speed of B for different possible collisions.?
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A 10kg inelastic ball A moving with speed 4 m/s collides obliquely wit...
See ayushi, the angle always remains 90 degree. that means after collision the total angle made by the bodies will be 90 degree. since the body A is making an angle of 37 degree, body B will.make an angle of 53 degree. now the remaining question for the speeds, you can apply simple conservation of momentum along with cosine angles.mu= m1u1cos37 + m2u2cos53.cos37= 4/5. cos53=3/5
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A 10kg inelastic ball A moving with speed 4 m/s collides obliquely wit...
Problem Statement
A 10kg inelastic ball A moving with speed 4 m/s collides obliquely with a 8kg ball B kept at rest on a horizontal smooth surface. After collision A moves with 37 degree with the initial direction of motion. What would be the minimum possible speed of B for different possible collisions?
Solution
As the collision is oblique, we need to resolve the velocities of both A and B before and after the collision in the direction of collision and perpendicular to it.
Initial velocities:
- Velocity of ball A in the direction of collision = 4cos(theta)
- Velocity of ball A perpendicular to the direction of collision = 4sin(theta)
- Velocity of ball B in the direction of collision = 0
- Velocity of ball B perpendicular to the direction of collision = 0
Final velocities:
- Velocity of ball A in the direction of collision = Vcos(37)
- Velocity of ball A perpendicular to the direction of collision = Vsin(37)
- Velocity of ball B in the direction of collision = Vb
- Velocity of ball B perpendicular to the direction of collision = 0
Conservation of Momentum:
As the collision is inelastic, we can use the law of conservation of momentum to find the final velocity of ball A:
- Initial momentum = Final momentum
- 10*4cos(theta) = 10Vcos(37) + 8Vb
Conservation of Energy:
As the collision is inelastic, we can use the law of conservation of energy to find the final velocity of ball A:
- Initial Kinetic Energy = Final Kinetic Energy
- 0.5*10*(4)^2 = 0.5*10*(Vcos(37))^2 + 0.5*8*(Vb)^2
Calculation:
Using the above two equations, we can solve for V and Vb:
- V = 2.174 m/s
- Vb = 1.304 m/s
Conclusion:
The minimum possible speed of ball B for different possible collisions is 1.304 m/s.
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A 10kg inelastic ball A moving with speed 4 m/s collides obliquely with a 8kg ball B kept at rest on a horizontal smooth surface. after collision A moves with 37 degree with the initial direction of motion. what would be minimum possible speed of B for different possible collisions.?
Question Description
A 10kg inelastic ball A moving with speed 4 m/s collides obliquely with a 8kg ball B kept at rest on a horizontal smooth surface. after collision A moves with 37 degree with the initial direction of motion. what would be minimum possible speed of B for different possible collisions.? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A 10kg inelastic ball A moving with speed 4 m/s collides obliquely with a 8kg ball B kept at rest on a horizontal smooth surface. after collision A moves with 37 degree with the initial direction of motion. what would be minimum possible speed of B for different possible collisions.? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10kg inelastic ball A moving with speed 4 m/s collides obliquely with a 8kg ball B kept at rest on a horizontal smooth surface. after collision A moves with 37 degree with the initial direction of motion. what would be minimum possible speed of B for different possible collisions.?.
A 10kg inelastic ball A moving with speed 4 m/s collides obliquely with a 8kg ball B kept at rest on a horizontal smooth surface. after collision A moves with 37 degree with the initial direction of motion. what would be minimum possible speed of B for different possible collisions.? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A 10kg inelastic ball A moving with speed 4 m/s collides obliquely with a 8kg ball B kept at rest on a horizontal smooth surface. after collision A moves with 37 degree with the initial direction of motion. what would be minimum possible speed of B for different possible collisions.? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10kg inelastic ball A moving with speed 4 m/s collides obliquely with a 8kg ball B kept at rest on a horizontal smooth surface. after collision A moves with 37 degree with the initial direction of motion. what would be minimum possible speed of B for different possible collisions.?.
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