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A ball of mass 10 g hits a hard surface vertically with a speed of 5 m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01 s. The average force exerted by the surface on the ball is
- a)100 N
- b)10 N
- c)1 N
- d)0.1 N
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A ball of mass 10 g hits a hard surface vertically with a speed of 5 m...
Average force = change in momentum/time
Change in momentum = m(v1 - v2)
=0.01[ 5 - (-5)]
=0.1
Average force = 0.1/0.01
= 10 N
Change in momentum = m(v1 - v2)
=0.01[ 5 - (-5)]
=0.1
Average force = 0.1/0.01
= 10 N
Most Upvoted Answer
A ball of mass 10 g hits a hard surface vertically with a speed of 5 m...
Average force = change in momentum/time
change in momentum = m(v1 - v2)
=0.01[ 5 - (-5)]
=0.1
average force = 0.1/0.01
= 10 N
change in momentum = m(v1 - v2)
=0.01[ 5 - (-5)]
=0.1
average force = 0.1/0.01
= 10 N
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Community Answer
A ball of mass 10 g hits a hard surface vertically with a speed of 5 m...
To find the average force exerted by the surface on the ball, we can use the principle of conservation of momentum.
1. Conservation of Momentum:
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. In this case, the ball rebounds with the same speed, so the total momentum before and after the collision is the same.
2. Momentum before the collision:
The momentum of an object is given by the product of its mass and velocity. The mass of the ball is given as 10 g, which is equivalent to 0.01 kg. The velocity of the ball before the collision is 5 m/s. Therefore, the momentum before the collision is calculated as:
Momentum before = mass × velocity = 0.01 kg × 5 m/s = 0.05 kg m/s
3. Momentum after the collision:
Since the ball rebounds with the same speed, its velocity after the collision is also 5 m/s. Therefore, the momentum after the collision is:
Momentum after = mass × velocity = 0.01 kg × 5 m/s = 0.05 kg m/s
4. Average force:
The change in momentum of an object is equal to the average force exerted on it multiplied by the time interval over which the force is exerted. In this case, the time interval is given as 0.01 s. Therefore, the average force exerted by the surface on the ball can be calculated as:
Average force = (Momentum after - Momentum before) / time interval
Average force = (0.05 kg m/s - 0.05 kg m/s) / 0.01 s
Average force = 0 N / 0.01 s
Average force = 0 N/s = 0 N
5. Conclusion:
The average force exerted by the surface on the ball is calculated to be 0 N. However, it is important to note that the surface does exert a force on the ball during the collision, but since the ball rebounds with the same speed, the change in momentum is zero, resulting in an average force of 0 N.
Therefore, the correct answer is option 'D' - 0.1 N.
1. Conservation of Momentum:
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. In this case, the ball rebounds with the same speed, so the total momentum before and after the collision is the same.
2. Momentum before the collision:
The momentum of an object is given by the product of its mass and velocity. The mass of the ball is given as 10 g, which is equivalent to 0.01 kg. The velocity of the ball before the collision is 5 m/s. Therefore, the momentum before the collision is calculated as:
Momentum before = mass × velocity = 0.01 kg × 5 m/s = 0.05 kg m/s
3. Momentum after the collision:
Since the ball rebounds with the same speed, its velocity after the collision is also 5 m/s. Therefore, the momentum after the collision is:
Momentum after = mass × velocity = 0.01 kg × 5 m/s = 0.05 kg m/s
4. Average force:
The change in momentum of an object is equal to the average force exerted on it multiplied by the time interval over which the force is exerted. In this case, the time interval is given as 0.01 s. Therefore, the average force exerted by the surface on the ball can be calculated as:
Average force = (Momentum after - Momentum before) / time interval
Average force = (0.05 kg m/s - 0.05 kg m/s) / 0.01 s
Average force = 0 N / 0.01 s
Average force = 0 N/s = 0 N
5. Conclusion:
The average force exerted by the surface on the ball is calculated to be 0 N. However, it is important to note that the surface does exert a force on the ball during the collision, but since the ball rebounds with the same speed, the change in momentum is zero, resulting in an average force of 0 N.
Therefore, the correct answer is option 'D' - 0.1 N.
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A ball of mass 10 g hits a hard surface vertically with a speed of 5 m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01 s. The average force exerted by the surface on the ball is a)100 N b)10 N c)1 N d)0.1 NCorrect answer is option 'B'. Can you explain this answer?
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A ball of mass 10 g hits a hard surface vertically with a speed of 5 m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01 s. The average force exerted by the surface on the ball is a)100 N b)10 N c)1 N d)0.1 NCorrect answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A ball of mass 10 g hits a hard surface vertically with a speed of 5 m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01 s. The average force exerted by the surface on the ball is a)100 N b)10 N c)1 N d)0.1 NCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of mass 10 g hits a hard surface vertically with a speed of 5 m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01 s. The average force exerted by the surface on the ball is a)100 N b)10 N c)1 N d)0.1 NCorrect answer is option 'B'. Can you explain this answer?.
A ball of mass 10 g hits a hard surface vertically with a speed of 5 m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01 s. The average force exerted by the surface on the ball is a)100 N b)10 N c)1 N d)0.1 NCorrect answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A ball of mass 10 g hits a hard surface vertically with a speed of 5 m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01 s. The average force exerted by the surface on the ball is a)100 N b)10 N c)1 N d)0.1 NCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of mass 10 g hits a hard surface vertically with a speed of 5 m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01 s. The average force exerted by the surface on the ball is a)100 N b)10 N c)1 N d)0.1 NCorrect answer is option 'B'. Can you explain this answer?.
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