Starting with an initial speed of 5.00 m/s at a height of 0.305 m, a 1.45-kg bal

Question

Starting with an initial speed of 5.00 m/s at a height of 0.305 m, a 1.45-kg ball swings downward and strikes a 4.50-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 1.45-kg ball just before impact. _______________m/s (b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision. __________m/s (1.45 kg-ball) __________m/s (4.50 kg-ball) (c) How high does each ball swing after the collision, ignoring air resistance? _________m (1.45 kg-ball) _________m (4.50 kg-ball) http://www.webassign.net/CJ/07-13alt.gif

Explanation / Answer

E = mgh + 1/2 mv² E = 1.55kg*9.81m/s³*0.325m+ 1/2*1.55kg*25m²/s² E= 24.3J now we will use that amount of energy to get the speed at height = 0m E=1/2 m v2² v2 = sqrt(E/(1/2m)) v2 = 5.60m/s

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