an object of mass 2kg is dropped from a certain height on rebounding from the ground, it rises vertically till 2/5th of its initial height. the ratio of magnitude of momentum of the object just before and after striking the ground is
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Velocity of object when it hit the ground can be calculated as
1/2 mv² = mgh
v = √2gh in downward direction
On striking the ground let the object rebound with velocity u in upward direction and rises vertically till 2/5 h
1/2 mu² = mg 2/5 h
u = √4/5gh in an upward direction
The ratio of magnitude of momentum of the object just before and after striking the ground is
P before/ Pafter = mv/mu = √2gh/ √4/5 gh = √2/ 2 √1/5 = 1.414/ 0.8944 = 1.58 is the Answer
1/2 mv² = mgh
v = √2gh in downward direction
On striking the ground let the object rebound with velocity u in upward direction and rises vertically till 2/5 h
1/2 mu² = mg 2/5 h
u = √4/5gh in an upward direction
The ratio of magnitude of momentum of the object just before and after striking the ground is
P before/ Pafter = mv/mu = √2gh/ √4/5 gh = √2/ 2 √1/5 = 1.414/ 0.8944 = 1.58 is the Answer
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Answer:
1.58
Explanation:
Velocity of object when it hit the ground can be calculated as
1/2 mv² = mgh
v = √2gh in downward direction
On striking the ground let the object rebound with velocity u in upward direction and rises vertically till 2/5 h
1/2 mu² = mg 2/5 h
u = √4/5gh in an upward direction
The ratio of magnitude of momentum of the object just before and after striking the ground is
P before/ P after = mv/mu = √2gh/ √4/5 gh = √2/ 2 √1/5 = 1.414/ 0.8944 = 1.58 is the Answer
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