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andre [41]
2 years ago
15

imagine a bouncing ball that does not lose any energy as it bounces. could it ever bounce to a greater height than it was droppe

d from? explain your answer

Physics
2 answers:
iren [92.7K]2 years ago
8 0
Depends on if it was thrown or dropped
Allisa [31]2 years ago
3 0

Answer:

No, due to conservation of energy, the bouncing ball would need to create energy in order to bounce higher.

Explanation:

The question says that the ball was dropped, so we suppose that it will start at an initial velocity of zero. (If the ball was tossed, then it's initial velocity would be greater than zero and it would bounce accordingly, but it would never bounce higher than it's maximum height)

Due to conservation of energy, the energy of the ball will remain the same throughout its trip. As the ball travels down to the ground, its potential energy decreases while its kinetic energy will increase, its sum will remain constant.  When the ball hits the ground, it will have its maximum kinetic energy, which is equal to its potential energy (the energy the ball started with). The energy is transmited to the ground and back to the ball (supposing the bounce is completely elastic and no energy was lost as heat or vibration).

So the ball starts its way back up. The height it can reach will depend on its potential energy, since energy remains constant (energy is not created nor destroyed), the ball can only get as high as it was originaly dropped from sine its mass and the acceleration of gravity remains constant.

Take a look at the uploaded diagram which represents the given situation.  

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A soccer ball with a mass of 0.434 kg approaches a player horizontally with a speed of 13.0 m/s. The player kicks the ball with
zaharov [31]

Answer:

a) 15.49

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Explanation:

a)

\Delta p=m \Delta v\\\\\Delta p= (0.434)(13.0-(-22.7)) \\\\\Delta p=(0.434)(35.7)\approx 15.49 kg \cdot m/s

b)

Since the player is kicking the ball in the opposite direction to which it came, the impulse is being directed opposite to the ball's initial velocity.

c)

\Delta p= F \Delta t \\\\\\F=\dfrac{\Delta p}{\Delta t} \\\\\\F=\dfrac{15.49 kg \cdot m/s}{0.06 s}\approx 258.16N

Hope this helps!

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