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A small ball thrown with an initial velocity μ directed at an angle θ = 37° above the horizontal
collides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontal
velocity u/5 towards ball. After collision with the wall, the ball returns to the point from where it
was thrown. Neglect friction between ball and wall. The time t from beginning of motion of the ball
till the moment of its impact with the wall is (tan37° = 3/4)
collides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontal
velocity u/5 towards ball. After collision with the wall, the ball returns to the point from where it
was thrown. Neglect friction between ball and wall. The time t from beginning of motion of the ball
till the moment of its impact with the wall is (tan37° = 3/4)
- a)3μ/5g
- b)18μ/25g
- c)54μ/125g
- d)54μ/25g
Correct answer is option 'C'. Can you explain this answer?
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A small ball thrown with an initial velocity μ directed at an ang...
Given information:
- Initial velocity of the ball = u
- Angle of projection = 37 degrees
- Coefficient of restitution = 1/4
- Velocity of the wall = u/5
To find: Time taken by the ball to collide with the wall and return to the initial point.
Analysis:
1. Motion of the ball before collision:
- The horizontal component of velocity remains constant throughout the motion.
- The vertical component of velocity changes due to gravity and the angle of projection.
2. Collision with the wall:
- The ball collides with the wall inelastically, which means it loses some of its kinetic energy.
- The velocity of the wall changes the horizontal component of the ball's velocity.
- The vertical component of the ball's velocity remains unchanged.
3. Motion of the ball after collision:
- The ball moves in a parabolic path due to gravity and the angle of projection.
- The horizontal component of velocity remains constant throughout the motion.
- The vertical component of velocity changes due to gravity.
Solution:
1. Time taken to reach the wall:
- The horizontal distance traveled by the ball before collision = u*t*cos(37)
- The horizontal distance traveled by the wall = (u/5)*t
- Equating the above two distances, we get:
u*t*cos(37) = (u/5)*t
t = 5*cos(37)
2. Velocity of the ball after collision:
- The horizontal component of velocity remains constant: u*cos(37)
- The vertical component of velocity changes due to gravity: u*sin(37)/4 (due to coefficient of restitution)
3. Time taken to return to the initial point:
- The vertical displacement of the ball = -2*h (where h is the maximum height reached by the ball before collision)
- The time taken to reach the maximum height = u*sin(37)/(2*g)
- The time taken to return to the initial point = u*sin(37)/g
4. Total time taken:
- Total time taken = Time taken to reach the wall + Time taken to return to the initial point
- Substituting the values, we get:
Total time taken = 5*cos(37) + (4*u*sin(37)/g)
Total time taken = 54/125*u/g
Therefore, the correct option is (c) 54/125g.
- Initial velocity of the ball = u
- Angle of projection = 37 degrees
- Coefficient of restitution = 1/4
- Velocity of the wall = u/5
To find: Time taken by the ball to collide with the wall and return to the initial point.
Analysis:
1. Motion of the ball before collision:
- The horizontal component of velocity remains constant throughout the motion.
- The vertical component of velocity changes due to gravity and the angle of projection.
2. Collision with the wall:
- The ball collides with the wall inelastically, which means it loses some of its kinetic energy.
- The velocity of the wall changes the horizontal component of the ball's velocity.
- The vertical component of the ball's velocity remains unchanged.
3. Motion of the ball after collision:
- The ball moves in a parabolic path due to gravity and the angle of projection.
- The horizontal component of velocity remains constant throughout the motion.
- The vertical component of velocity changes due to gravity.
Solution:
1. Time taken to reach the wall:
- The horizontal distance traveled by the ball before collision = u*t*cos(37)
- The horizontal distance traveled by the wall = (u/5)*t
- Equating the above two distances, we get:
u*t*cos(37) = (u/5)*t
t = 5*cos(37)
2. Velocity of the ball after collision:
- The horizontal component of velocity remains constant: u*cos(37)
- The vertical component of velocity changes due to gravity: u*sin(37)/4 (due to coefficient of restitution)
3. Time taken to return to the initial point:
- The vertical displacement of the ball = -2*h (where h is the maximum height reached by the ball before collision)
- The time taken to reach the maximum height = u*sin(37)/(2*g)
- The time taken to return to the initial point = u*sin(37)/g
4. Total time taken:
- Total time taken = Time taken to reach the wall + Time taken to return to the initial point
- Substituting the values, we get:
Total time taken = 5*cos(37) + (4*u*sin(37)/g)
Total time taken = 54/125*u/g
Therefore, the correct option is (c) 54/125g.
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A small ball thrown with an initial velocity μ directed at an angle θ = 37° above the horizontalcollides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontalvelocity u/5 towards ball. After collision with the wall, the ball returns to the point from where itwas thrown. Neglect friction between ball and wall. The time t from beginning of motion of the balltill the moment of its impact with the wall is (tan37° = 3/4)a)3μ/5gb)18μ/25gc)54μ/125gd)54μ/25gCorrect answer is option 'C'. Can you explain this answer?
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A small ball thrown with an initial velocity μ directed at an angle θ = 37° above the horizontalcollides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontalvelocity u/5 towards ball. After collision with the wall, the ball returns to the point from where itwas thrown. Neglect friction between ball and wall. The time t from beginning of motion of the balltill the moment of its impact with the wall is (tan37° = 3/4)a)3μ/5gb)18μ/25gc)54μ/125gd)54μ/25gCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A small ball thrown with an initial velocity μ directed at an angle θ = 37° above the horizontalcollides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontalvelocity u/5 towards ball. After collision with the wall, the ball returns to the point from where itwas thrown. Neglect friction between ball and wall. The time t from beginning of motion of the balltill the moment of its impact with the wall is (tan37° = 3/4)a)3μ/5gb)18μ/25gc)54μ/125gd)54μ/25gCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A small ball thrown with an initial velocity μ directed at an angle θ = 37° above the horizontalcollides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontalvelocity u/5 towards ball. After collision with the wall, the ball returns to the point from where itwas thrown. Neglect friction between ball and wall. The time t from beginning of motion of the balltill the moment of its impact with the wall is (tan37° = 3/4)a)3μ/5gb)18μ/25gc)54μ/125gd)54μ/25gCorrect answer is option 'C'. Can you explain this answer?.
A small ball thrown with an initial velocity μ directed at an angle θ = 37° above the horizontalcollides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontalvelocity u/5 towards ball. After collision with the wall, the ball returns to the point from where itwas thrown. Neglect friction between ball and wall. The time t from beginning of motion of the balltill the moment of its impact with the wall is (tan37° = 3/4)a)3μ/5gb)18μ/25gc)54μ/125gd)54μ/25gCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A small ball thrown with an initial velocity μ directed at an angle θ = 37° above the horizontalcollides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontalvelocity u/5 towards ball. After collision with the wall, the ball returns to the point from where itwas thrown. Neglect friction between ball and wall. The time t from beginning of motion of the balltill the moment of its impact with the wall is (tan37° = 3/4)a)3μ/5gb)18μ/25gc)54μ/125gd)54μ/25gCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A small ball thrown with an initial velocity μ directed at an angle θ = 37° above the horizontalcollides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontalvelocity u/5 towards ball. After collision with the wall, the ball returns to the point from where itwas thrown. Neglect friction between ball and wall. The time t from beginning of motion of the balltill the moment of its impact with the wall is (tan37° = 3/4)a)3μ/5gb)18μ/25gc)54μ/125gd)54μ/25gCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A small ball thrown with an initial velocity μ directed at an angle θ = 37° above the horizontalcollides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontalvelocity u/5 towards ball. After collision with the wall, the ball returns to the point from where itwas thrown. Neglect friction between ball and wall. The time t from beginning of motion of the balltill the moment of its impact with the wall is (tan37° = 3/4)a)3μ/5gb)18μ/25gc)54μ/125gd)54μ/25gCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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A small ball thrown with an initial velocity μ directed at an angle θ = 37° above the horizontalcollides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontalvelocity u/5 towards ball. After collision with the wall, the ball returns to the point from where itwas thrown. Neglect friction between ball and wall. The time t from beginning of motion of the balltill the moment of its impact with the wall is (tan37° = 3/4)a)3μ/5gb)18μ/25gc)54μ/125gd)54μ/25gCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A small ball thrown with an initial velocity μ directed at an angle θ = 37° above the horizontalcollides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontalvelocity u/5 towards ball. After collision with the wall, the ball returns to the point from where itwas thrown. Neglect friction between ball and wall. The time t from beginning of motion of the balltill the moment of its impact with the wall is (tan37° = 3/4)a)3μ/5gb)18μ/25gc)54μ/125gd)54μ/25gCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A small ball thrown with an initial velocity μ directed at an angle θ = 37° above the horizontalcollides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontalvelocity u/5 towards ball. After collision with the wall, the ball returns to the point from where itwas thrown. Neglect friction between ball and wall. The time t from beginning of motion of the balltill the moment of its impact with the wall is (tan37° = 3/4)a)3μ/5gb)18μ/25gc)54μ/125gd)54μ/25gCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A small ball thrown with an initial velocity μ directed at an angle θ = 37° above the horizontalcollides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontalvelocity u/5 towards ball. After collision with the wall, the ball returns to the point from where itwas thrown. Neglect friction between ball and wall. The time t from beginning of motion of the balltill the moment of its impact with the wall is (tan37° = 3/4)a)3μ/5gb)18μ/25gc)54μ/125gd)54μ/25gCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
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