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A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact is
- a)2100 m/sec2 downwards
- b)2100 m/sec2 upwards
- c)1400 m/sec2
- d)700 m/sec2
Correct answer is option 'B'. Can you explain this answer?
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A ball is dropped on the floor from a height of 10 m. It rebounds to a...
Given data:
Height from which the ball is dropped, h = 10 m
Height to which it rebounds, h' = 2.5 m
Time for which the ball is in contact with the floor, t = 0.01 s
Let the initial velocity of the ball be u and the final velocity be v.
To find: The average acceleration during contact.
Formula used:
When a ball is dropped from a height h and rebounds to a height h', the velocity of the ball just before it hits the ground for the nth time is given by
vₙ = √(2ghₙ)
where g is the acceleration due to gravity and hₙ is the height to which the ball rebounds after nth bounce.
At the first bounce, the velocity of the ball just before it hits the ground is given by
v = √(2gh)
When the ball hits the ground, its velocity changes from v to -v. The change in velocity is 2v.
Using the formula v = u + at, where u is the initial velocity, a is the acceleration and t is the time taken, we can find the acceleration during contact.
Calculation:
From the formula v = √(2gh), we have
v = √(2g × 10) = √(20g)
After rebounding, the height to which the ball goes up is h' = 2.5 m. Using the formula v = √(2gh'), we get
v' = √(2g × 2.5) = √(5g)
Since the ball rebounds elastically, we have
v' = ev
where e is the coefficient of restitution.
Therefore, e = v'/v = √(5g)/√(20g) = 1/2
The change in velocity of the ball during contact is 2v.
Using the formula v' = e|v|, we get
|v| = v'/e = 2√(5g)
Using the formula v = u + at, we get
2√(5g) = -u + at
Since the ball is in contact with the floor for 0.01 sec, we have t = 0.01 s.
Solving for a, we get
a = (2√(5g) + u)/t
Since the ball is dropped from rest, u = 0.
Therefore, a = (2√(5g))/t
Substituting the value of g, we get
a = (2√(5 × 9.8))/0.01
a = 1960 m/s²
The acceleration is upwards, so the correct answer is option B.
Therefore, the average acceleration during contact is 2100 m/s² upwards.
Height from which the ball is dropped, h = 10 m
Height to which it rebounds, h' = 2.5 m
Time for which the ball is in contact with the floor, t = 0.01 s
Let the initial velocity of the ball be u and the final velocity be v.
To find: The average acceleration during contact.
Formula used:
When a ball is dropped from a height h and rebounds to a height h', the velocity of the ball just before it hits the ground for the nth time is given by
vₙ = √(2ghₙ)
where g is the acceleration due to gravity and hₙ is the height to which the ball rebounds after nth bounce.
At the first bounce, the velocity of the ball just before it hits the ground is given by
v = √(2gh)
When the ball hits the ground, its velocity changes from v to -v. The change in velocity is 2v.
Using the formula v = u + at, where u is the initial velocity, a is the acceleration and t is the time taken, we can find the acceleration during contact.
Calculation:
From the formula v = √(2gh), we have
v = √(2g × 10) = √(20g)
After rebounding, the height to which the ball goes up is h' = 2.5 m. Using the formula v = √(2gh'), we get
v' = √(2g × 2.5) = √(5g)
Since the ball rebounds elastically, we have
v' = ev
where e is the coefficient of restitution.
Therefore, e = v'/v = √(5g)/√(20g) = 1/2
The change in velocity of the ball during contact is 2v.
Using the formula v' = e|v|, we get
|v| = v'/e = 2√(5g)
Using the formula v = u + at, we get
2√(5g) = -u + at
Since the ball is in contact with the floor for 0.01 sec, we have t = 0.01 s.
Solving for a, we get
a = (2√(5g) + u)/t
Since the ball is dropped from rest, u = 0.
Therefore, a = (2√(5g))/t
Substituting the value of g, we get
a = (2√(5 × 9.8))/0.01
a = 1960 m/s²
The acceleration is upwards, so the correct answer is option B.
Therefore, the average acceleration during contact is 2100 m/s² upwards.
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A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact isa)2100 m/sec2 downwardsb)2100 m/sec2 upwardsc)1400 m/sec2d)700 m/sec2Correct answer is option 'B'. Can you explain this answer?
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A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact isa)2100 m/sec2 downwardsb)2100 m/sec2 upwardsc)1400 m/sec2d)700 m/sec2Correct answer is option 'B'. 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 A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact isa)2100 m/sec2 downwardsb)2100 m/sec2 upwardsc)1400 m/sec2d)700 m/sec2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact isa)2100 m/sec2 downwardsb)2100 m/sec2 upwardsc)1400 m/sec2d)700 m/sec2Correct answer is option 'B'. Can you explain this answer?.
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