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A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 m. Ball hits the ground and bounces off the ground. Upon impact with the ground, the velocity reduces by 20%. The height (in m) to which the ball will rise is ______
(Important - Enter only the numerical value in the answer)
Correct answer is '0.64'. Can you explain this answer?
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A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 ...
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Given data:
- Mass of the ball (m) = 0.1 kg
- Initial height (h) = 1 m
- Velocity reduction upon impact = 20%
Key concept:
When the ball hits the ground, it undergoes an elastic collision, which means that the total mechanical energy of the system (ball and Earth) is conserved. Therefore, we can use the principle of conservation of mechanical energy to solve this problem.
Principle of conservation of mechanical energy:
The total mechanical energy of a system is the sum of its potential energy and kinetic energy. In this case, the mechanical energy of the ball is conserved, so we can write:
Initial mechanical energy = Final mechanical energy
Calculation:
1. Initial mechanical energy:
- The initial mechanical energy of the ball is given by the potential energy at height h, which is equal to mgh, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
- Therefore, the initial mechanical energy is: E1 = mgh = 0.1 kg * 9.8 m/s^2 * 1 m = 0.98 Joules.
2. Final mechanical energy:
- The final mechanical energy of the ball is the sum of its potential energy at the maximum height reached and its kinetic energy at that point.
- Let the maximum height reached be h'.
- The potential energy at height h' is given by mgh', and the kinetic energy at that point is given by (1/2)mv^2, where v is the velocity of the ball at that point.
- Since the velocity reduces by 20% upon impact with the ground, the velocity after the impact is 0.8 times the velocity before the impact.
- Therefore, the final mechanical energy is: E2 = mgh' + (1/2)mv^2 = 0.1 kg * 9.8 m/s^2 * h' + (1/2) * 0.1 kg * (0.8v)^2.
3. Applying the principle of conservation of mechanical energy:
- From the principle of conservation of mechanical energy, we have E1 = E2.
- Substituting the values of E1 and E2, we get: 0.98 Joules = 0.1 kg * 9.8 m/s^2 * h' + (1/2) * 0.1 kg * (0.8v)^2.
4. Solving for h':
- Rearranging the equation, we have: 0.98 Joules - 0.1 kg * 9.8 m/s^2 * h' = (1/2) * 0.1 kg * (0.8v)^2.
- Simplifying further, we get: 0.98 - 0.98h' = 0.04 * v^2.
- Dividing both sides by 0.04, we have: 24.5 - 24.5h' = v^2.
- Since the velocity reduces by 20%, we can write: v = 0.8v.
- Substituting this value in the equation, we get: 24.5 - 24.5h' = (0.8v)^
- Mass of the ball (m) = 0.1 kg
- Initial height (h) = 1 m
- Velocity reduction upon impact = 20%
Key concept:
When the ball hits the ground, it undergoes an elastic collision, which means that the total mechanical energy of the system (ball and Earth) is conserved. Therefore, we can use the principle of conservation of mechanical energy to solve this problem.
Principle of conservation of mechanical energy:
The total mechanical energy of a system is the sum of its potential energy and kinetic energy. In this case, the mechanical energy of the ball is conserved, so we can write:
Initial mechanical energy = Final mechanical energy
Calculation:
1. Initial mechanical energy:
- The initial mechanical energy of the ball is given by the potential energy at height h, which is equal to mgh, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
- Therefore, the initial mechanical energy is: E1 = mgh = 0.1 kg * 9.8 m/s^2 * 1 m = 0.98 Joules.
2. Final mechanical energy:
- The final mechanical energy of the ball is the sum of its potential energy at the maximum height reached and its kinetic energy at that point.
- Let the maximum height reached be h'.
- The potential energy at height h' is given by mgh', and the kinetic energy at that point is given by (1/2)mv^2, where v is the velocity of the ball at that point.
- Since the velocity reduces by 20% upon impact with the ground, the velocity after the impact is 0.8 times the velocity before the impact.
- Therefore, the final mechanical energy is: E2 = mgh' + (1/2)mv^2 = 0.1 kg * 9.8 m/s^2 * h' + (1/2) * 0.1 kg * (0.8v)^2.
3. Applying the principle of conservation of mechanical energy:
- From the principle of conservation of mechanical energy, we have E1 = E2.
- Substituting the values of E1 and E2, we get: 0.98 Joules = 0.1 kg * 9.8 m/s^2 * h' + (1/2) * 0.1 kg * (0.8v)^2.
4. Solving for h':
- Rearranging the equation, we have: 0.98 Joules - 0.1 kg * 9.8 m/s^2 * h' = (1/2) * 0.1 kg * (0.8v)^2.
- Simplifying further, we get: 0.98 - 0.98h' = 0.04 * v^2.
- Dividing both sides by 0.04, we have: 24.5 - 24.5h' = v^2.
- Since the velocity reduces by 20%, we can write: v = 0.8v.
- Substituting this value in the equation, we get: 24.5 - 24.5h' = (0.8v)^
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A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 m. Ball hits the ground andbounces off the ground. Upon impact with the ground, the velocity reduces by 20%. Theheight (in m) to which the ball will rise is ______(Important - Enter only the numerical value in the answer)Correct answer is '0.64'. Can you explain this answer?
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A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 m. Ball hits the ground andbounces off the ground. Upon impact with the ground, the velocity reduces by 20%. Theheight (in m) to which the ball will rise is ______(Important - Enter only the numerical value in the answer)Correct answer is '0.64'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 m. Ball hits the ground andbounces off the ground. Upon impact with the ground, the velocity reduces by 20%. Theheight (in m) to which the ball will rise is ______(Important - Enter only the numerical value in the answer)Correct answer is '0.64'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 m. Ball hits the ground andbounces off the ground. Upon impact with the ground, the velocity reduces by 20%. Theheight (in m) to which the ball will rise is ______(Important - Enter only the numerical value in the answer)Correct answer is '0.64'. Can you explain this answer?.
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