PROBLEM 3 & 4 March 31 2010

PROBLEM 3 & 4 March 31 2010 - (B) What is the...

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SOLVING PROBLEM 3 March 31, 2010 A 5 kg ball falls 1.8 m onto the surface of an elastic sheet. The sheet is stretched down 0.2 m by the ball. At this point the velocity of the ball is reduced to zero, the ball reverses its motion and flies up into the air. (A) What is the time from first contact with the sheet until the velocity of the ball is reduced to zero? What do you know? Conservation of energy: The potential energy of the ball at rest just before release is the same as the potential energy of the ball when it is momentarily at rest on the stretched trampoline sheet. Thus Potential energy #1 mgh U 1 equals Potential energy #2 2 2 1 2 kx U . Equating the right hand sides of these two equations (because 2 1 U U ) we have 2 2 1 kx mgh . Small trick: The total height h includes both the 1.8 m and the 0.2 m, so h = 2 m. Solve for k : m N m m s m kg x mgh k / 5000 ) 2 . 0 /( 2 / 10 5 2 / 2 2 2 2 The angular frequency is 1 1000 / s m k . The period is s s T 2 . 0 ) 5 . 31 / 2 ( / 2 . Trick: The time to stop is one quarter of a period or 0.05s.
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Unformatted text preview: (B) What is the maximum acceleration? The maximum acceleration occurs when the ball is at rest at its lowest height when the force is N m m N kx F 1000 2 . / 5000 . Since F=ma we then have g ms kg N m F a 20 200 5 / 1000 / 2 SOLVING PROBLEM 4 A 5 g M16 bullet traveling horizontally at 1 km/s hits a 0.995 kg mass hanging at rest from a 10 m string. How high does the mass rise after the impact? What do you know? Momentum will be conserved even though most of the kinetic energy of the bullet is turned into splinters. So Momentum before ( ) equals momentum after ( ). s kgm mv P / 5 1 ' 2 Mv P After the bullet stops in the wood, the kinetic energy of the log + bullet is J kg s m kg M v m M P M P K 5 . 12 2 / 25 2 / 2 / 2 / 2 2 2 2 2 2 1 2 2 . The log swings till it rises to a height h such that Mgh =12.5 J. Thus h = 12.5 J/ mg =1.25m....
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