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A particle moving horizontally with velocity u strikes a fixed frictionless sohere at height R/2 above the centre of sphere. After striking the sphere velocity of particle changes to vertically upward. If R be the radius of sphere the maximum height attained by particle with respect to centre of sphere is?
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A particle moving horizontally with velocity u strikes a fixed frictio...
Given information:
- A particle is moving horizontally with velocity u.
- The particle strikes a fixed frictionless sphere at a height R/2 above the center of the sphere.
- After striking the sphere, the velocity of the particle changes to vertically upward.
To find:
The maximum height attained by the particle with respect to the center of the sphere.
Explanation:
Step 1: Analyzing the initial velocity:
- The particle is moving horizontally with a velocity u before striking the sphere.
- Since there is no friction, the horizontal component of the velocity remains constant throughout the motion.
Step 2: Collision with the sphere:
- When the particle strikes the sphere at a height R/2 above the center, it experiences an upward normal force from the sphere.
- This upward normal force changes the direction of the particle's velocity from horizontal to vertical.
- The horizontal component of the velocity becomes zero after the collision, and only the vertical component remains.
Step 3: Conservation of energy:
- At the highest point of the particle's trajectory, its vertical velocity becomes zero.
- Using the principle of conservation of mechanical energy, we can equate the initial kinetic energy to the potential energy at the highest point.
- The initial kinetic energy is given by (1/2)mu^2, where m is the mass of the particle.
- The potential energy at the highest point is given by mgh, where g is the acceleration due to gravity and h is the maximum height attained by the particle.
- Equating these two expressions, we get (1/2)mu^2 = mgh.
Step 4: Solving for maximum height:
- Canceling the mass of the particle from both sides of the equation, we get (1/2)u^2 = gh.
- Since the velocity u is given and the acceleration due to gravity g is known, we can solve for h.
- Rearranging the equation, we get h = (1/2)u^2/g.
Step 5: Maximum height with respect to the center of the sphere:
- The maximum height attained by the particle is R/2 above the center of the sphere.
- Adding R/2 to the equation obtained in step 4, we get the maximum height with respect to the center of the sphere as h_max = R/2 + (1/2)u^2/g.
Final answer:
The maximum height attained by the particle with respect to the center of the sphere is R/2 + (1/2)u^2/g.
- A particle is moving horizontally with velocity u.
- The particle strikes a fixed frictionless sphere at a height R/2 above the center of the sphere.
- After striking the sphere, the velocity of the particle changes to vertically upward.
To find:
The maximum height attained by the particle with respect to the center of the sphere.
Explanation:
Step 1: Analyzing the initial velocity:
- The particle is moving horizontally with a velocity u before striking the sphere.
- Since there is no friction, the horizontal component of the velocity remains constant throughout the motion.
Step 2: Collision with the sphere:
- When the particle strikes the sphere at a height R/2 above the center, it experiences an upward normal force from the sphere.
- This upward normal force changes the direction of the particle's velocity from horizontal to vertical.
- The horizontal component of the velocity becomes zero after the collision, and only the vertical component remains.
Step 3: Conservation of energy:
- At the highest point of the particle's trajectory, its vertical velocity becomes zero.
- Using the principle of conservation of mechanical energy, we can equate the initial kinetic energy to the potential energy at the highest point.
- The initial kinetic energy is given by (1/2)mu^2, where m is the mass of the particle.
- The potential energy at the highest point is given by mgh, where g is the acceleration due to gravity and h is the maximum height attained by the particle.
- Equating these two expressions, we get (1/2)mu^2 = mgh.
Step 4: Solving for maximum height:
- Canceling the mass of the particle from both sides of the equation, we get (1/2)u^2 = gh.
- Since the velocity u is given and the acceleration due to gravity g is known, we can solve for h.
- Rearranging the equation, we get h = (1/2)u^2/g.
Step 5: Maximum height with respect to the center of the sphere:
- The maximum height attained by the particle is R/2 above the center of the sphere.
- Adding R/2 to the equation obtained in step 4, we get the maximum height with respect to the center of the sphere as h_max = R/2 + (1/2)u^2/g.
Final answer:
The maximum height attained by the particle with respect to the center of the sphere is R/2 + (1/2)u^2/g.
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A particle moving horizontally with velocity u strikes a fixed frictionless sohere at height R/2 above the centre of sphere. After striking the sphere velocity of particle changes to vertically upward. If R be the radius of sphere the maximum height attained by particle with respect to centre of sphere is?
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A particle moving horizontally with velocity u strikes a fixed frictionless sohere at height R/2 above the centre of sphere. After striking the sphere velocity of particle changes to vertically upward. If R be the radius of sphere the maximum height attained by particle with respect to centre of sphere is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle moving horizontally with velocity u strikes a fixed frictionless sohere at height R/2 above the centre of sphere. After striking the sphere velocity of particle changes to vertically upward. If R be the radius of sphere the maximum height attained by particle with respect to centre of sphere is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moving horizontally with velocity u strikes a fixed frictionless sohere at height R/2 above the centre of sphere. After striking the sphere velocity of particle changes to vertically upward. If R be the radius of sphere the maximum height attained by particle with respect to centre of sphere is?.
A particle moving horizontally with velocity u strikes a fixed frictionless sohere at height R/2 above the centre of sphere. After striking the sphere velocity of particle changes to vertically upward. If R be the radius of sphere the maximum height attained by particle with respect to centre of sphere is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle moving horizontally with velocity u strikes a fixed frictionless sohere at height R/2 above the centre of sphere. After striking the sphere velocity of particle changes to vertically upward. If R be the radius of sphere the maximum height attained by particle with respect to centre of sphere is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moving horizontally with velocity u strikes a fixed frictionless sohere at height R/2 above the centre of sphere. After striking the sphere velocity of particle changes to vertically upward. If R be the radius of sphere the maximum height attained by particle with respect to centre of sphere is?.
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