A spring within a spring launcher is compressed a distance, x, and locked into position by a pin at point P. A spherical ball M is placed onto the compressed spring. When the pin is removed, the spring accelerates the ball upwards until the ball then reaches some maximum height above the spring, HM. Assume the compression of the spring is small in comparison to the maximum height reached by the launched ball. The ball is then replaced by a ball having twice the mass, 2M. The same spring is compressed the same distance, x, and the new ball is launched. •HM

Principles of Physics: A Calculus-Based Text
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Chapter6: Energy Of A System
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A spring within a spring launcher is compressed
a distance, x, and locked into position by a pin
at point P. A spherical ball M is placed onto the
compressed spring. When the pin is removed,
the spring accelerates the ball upwards until the
ball then reaches some maximum height above
the spring, HM. Assume the compression of the
spring is small in comparison to the maximum
height reached by the launched ball.
The ball is then replaced by a ball having twice
the mass, 2M. The same spring is compressed
the same distance, x, and the new ball is
launched.
( ) ) ) ) )
HM
In terms of HM, determine
height reached by the heavier ball.
the maximum
Transcribed Image Text:A spring within a spring launcher is compressed a distance, x, and locked into position by a pin at point P. A spherical ball M is placed onto the compressed spring. When the pin is removed, the spring accelerates the ball upwards until the ball then reaches some maximum height above the spring, HM. Assume the compression of the spring is small in comparison to the maximum height reached by the launched ball. The ball is then replaced by a ball having twice the mass, 2M. The same spring is compressed the same distance, x, and the new ball is launched. ( ) ) ) ) ) HM In terms of HM, determine height reached by the heavier ball. the maximum
In terms of HM, determine the maximum
height reached by the heavier ball.
Choose 1 answer:
✪ +HM
A
B
C
D
1
-Нм
HM
HM
2HM
Transcribed Image Text:In terms of HM, determine the maximum height reached by the heavier ball. Choose 1 answer: ✪ +HM A B C D 1 -Нм HM HM 2HM
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