Starting with an initial speed of 5.57 m/s at a height of 0.185 m, a 2.37-kg bal

Question

Starting with an initial speed of 5.57 m/s at a height of 0.185 m, a 2.37-kg ball swings downward and strikes a 4.22-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 2.37-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 2.37-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.22-kg ball just after the collision. (d) How high does the 2.37-kg ball swing after the collision, ignoring air resistance? (e) How high does the 4.22-kg ball swing after the collision, ignoring air resistance?

(a) ___ m/s (b)_ m/s (c) ___ m/s (d)__ m (e)__ m

Explanation / Answer

by using principle of conservation of mechanical energy, a)v=5.8868 m/s b)velocity of 4.22 ball be V and 2.37 ball be vf after collision then mv =MV -mvf momentum conservation 1/2 m v^2 = 1/2 MV^2 +1/2 mvf^2 energy conservation solving these we get V = 4.234 m/s in the direction of initial velocity of 2.37 ball k= 1.6526 m/s in the direction of opposite to its initial velocity heights reached are 1/mv^2 = mgh h1= 0.914 m by 4.22 ball h2 =0.1392 m by 2.37 ball

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