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A ball of mass 50gm is dropped from a height h = 10m. It rebounds losing 75 percent of its kinetic energy. If it remains in contact with the ground for Δt = 0.01 sec., the impulse of the impact force (in N-s) is : (take g =10 m/s2)
Correct answer is '1.05'. Can you explain this answer?
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A ball of mass 50gmis dropped from a heighth= 10m. It rebounds losing ...
The correct answer is: 1.05
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A ball of mass 50gmis dropped from a heighth= 10m. It rebounds losing ...
We can solve this problem using the principle of conservation of mechanical energy. The initial mechanical energy of the ball is equal to the sum of its potential energy and kinetic energy when it is at a height of 10m. The final mechanical energy of the ball is equal to the sum of its potential energy and kinetic energy when it rebounds after hitting the ground.
Initial mechanical energy:
Ei = PE + KE
PE = mgh
KE = 0 (since the ball is at rest initially)
Ei = mgh = 0.05kg * 9.8m/s^2 * 10m = 4.9J
Final mechanical energy:
Ef = PE + KE
PE = 0 (since the ball is at the ground level)
KE = (1 - 0.75) * KE_initial = 0.25 * KE_initial
Ef = 0 + 0.25 * KE_initial = 0.25 * 4.9J = 1.225J
Since the ball remains in contact with the ground, its potential energy is zero. Therefore, the final kinetic energy is equal to the final mechanical energy.
KE_final = Ef = 1.225J
We can now use the formula for kinetic energy to find the final velocity of the ball:
KE = 0.5mv^2
1.225J = 0.5 * 0.05kg * v^2
v^2 = 1.225J / (0.5 * 0.05kg) = 49m^2/s^2
v = sqrt(49m^2/s^2) = 7m/s
Therefore, the final velocity of the ball after rebounding is 7m/s.
Initial mechanical energy:
Ei = PE + KE
PE = mgh
KE = 0 (since the ball is at rest initially)
Ei = mgh = 0.05kg * 9.8m/s^2 * 10m = 4.9J
Final mechanical energy:
Ef = PE + KE
PE = 0 (since the ball is at the ground level)
KE = (1 - 0.75) * KE_initial = 0.25 * KE_initial
Ef = 0 + 0.25 * KE_initial = 0.25 * 4.9J = 1.225J
Since the ball remains in contact with the ground, its potential energy is zero. Therefore, the final kinetic energy is equal to the final mechanical energy.
KE_final = Ef = 1.225J
We can now use the formula for kinetic energy to find the final velocity of the ball:
KE = 0.5mv^2
1.225J = 0.5 * 0.05kg * v^2
v^2 = 1.225J / (0.5 * 0.05kg) = 49m^2/s^2
v = sqrt(49m^2/s^2) = 7m/s
Therefore, the final velocity of the ball after rebounding is 7m/s.
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A ball of mass 50gmis dropped from a heighth= 10m. It rebounds losing ...
Use kimematice find out for velocity than use imluse concept
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A ball of mass 50gmis dropped from a heighth= 10m. It rebounds losing 75percentof its kinetic energy. If it remains in contact with the ground forΔt = 0.01 sec.,the impulse of the impact force(inN-s) is : (takeg=10m/s2)Correct answer is '1.05'. Can you explain this answer?
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A ball of mass 50gmis dropped from a heighth= 10m. It rebounds losing 75percentof its kinetic energy. If it remains in contact with the ground forΔt = 0.01 sec.,the impulse of the impact force(inN-s) is : (takeg=10m/s2)Correct answer is '1.05'. Can you explain this answer? for Physics 2025 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A ball of mass 50gmis dropped from a heighth= 10m. It rebounds losing 75percentof its kinetic energy. If it remains in contact with the ground forΔt = 0.01 sec.,the impulse of the impact force(inN-s) is : (takeg=10m/s2)Correct answer is '1.05'. Can you explain this answer? covers all topics & solutions for Physics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of mass 50gmis dropped from a heighth= 10m. It rebounds losing 75percentof its kinetic energy. If it remains in contact with the ground forΔt = 0.01 sec.,the impulse of the impact force(inN-s) is : (takeg=10m/s2)Correct answer is '1.05'. Can you explain this answer?.
A ball of mass 50gmis dropped from a heighth= 10m. It rebounds losing 75percentof its kinetic energy. If it remains in contact with the ground forΔt = 0.01 sec.,the impulse of the impact force(inN-s) is : (takeg=10m/s2)Correct answer is '1.05'. Can you explain this answer? for Physics 2025 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A ball of mass 50gmis dropped from a heighth= 10m. It rebounds losing 75percentof its kinetic energy. If it remains in contact with the ground forΔt = 0.01 sec.,the impulse of the impact force(inN-s) is : (takeg=10m/s2)Correct answer is '1.05'. Can you explain this answer? covers all topics & solutions for Physics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of mass 50gmis dropped from a heighth= 10m. It rebounds losing 75percentof its kinetic energy. If it remains in contact with the ground forΔt = 0.01 sec.,the impulse of the impact force(inN-s) is : (takeg=10m/s2)Correct answer is '1.05'. Can you explain this answer?.
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