A ball of mass 5kg is dropped from a cliff of height 75m. What is the ratio of the kinetic energy to the potential energy of the ball when it reaches the halfway point of the fall?
A
2
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B
1
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C
12
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D
34
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Solution
The correct option is B1 According to the law of conservation of energy, the total energy of the ball remains constant. ∴ PE at the top of the cliff is equal to the sum of kinetic and potential energies half way through the fall. (KE at top of the cliff =0)
Let h= total height of the cliff hhalf= Half of the total height of cliff
Then, (PE)Top=KEhalf+(PE)half ⇒mgh=12mv2half+mghhalf ⇒2gh=v2half+2ghhalf
Here h=2hhalf ⇒v2half=2g(2hhalf)−2ghhalf ⇒vhalf=√4ghhalf−2ghhalf=√2ghhalf
The ratio of kinetic to potential energy half way through the fall =(KE)half(PE)half=12mv2halfmghhalf=m×(√2ghhalf)22×mghhalf ⇒(KE)half(PE)half=1