A ball is attached to one end of a wire, the other end beingfastened to the ceil

Question

A ball is attached to one end of a wire, the other end beingfastened to the ceiling. The wire is held horizontal, and the ballis released from rest (see the drawing). It swings downward andstrikes a block initially at rest on a horizontal frictionlesssurface. Air resistance is negligible, and the collision iselastic. The masses of the ball and block are, respectively, 1.7 kgand 2.5 kg, and the length of the wire is 1.30 m. Find the velocity(magnitude and direction) of the ball (a) justbefore the collision, and (b) just after thecollision. Please show me how you get each step also, I have posted thisbefore and have no idea how to do it with diff. #'s Thanks A ball is attached to one end of a wire, the other end beingfastened to the ceiling. The wire is held horizontal, and the ballis released from rest (see the drawing). It swings downward andstrikes a block initially at rest on a horizontal frictionlesssurface. Air resistance is negligible, and the collision iselastic. The masses of the ball and block are, respectively, 1.7 kgand 2.5 kg, and the length of the wire is 1.30 m. Find the velocity(magnitude and direction) of the ball (a) justbefore the collision, and (b) just after thecollision. Please show me how you get each step also, I have posted thisbefore and have no idea how to do it with diff. #'s Thanks

Explanation / Answer

speed of ball immediately before collision is found by cons ofenergy: . initial PE of ball = final KE of ball .    m g h   = (1/2) mv2 .    v = 2gh = 2*9.8*1.30 =    5.048m/s . Now... . For the speed of the ball after the collision... call thespeed of the ball "s"   and the speed of the block (aftercollision) w. . Then we have, by conservation of momentum: .           mv    =   m s + M w  or .     1.7 * 5.048   = 1.7 s + 2.5 w .      85.812 = 17 s  + 25 w        (oneequation)                 w = . Now we also have conservation of kinetic energy, becausecollision is elastic: .     (1/2) m v2 = (1/2) ms2   + (1/2) M w2 .         1.7 *5.0482   = 1.7 s2  + 2.5 w2 .        433.16   = 17 s2   + 25w2    (second equation) .    now you have two equations and two unknowns (sand w) so we eliminate w and solve for s: .     433.16 = 17 s2 + 25 (3.4325 -   0.68 s )2 .     433.16   = 17s2    + 11.68s2   -   116.705 s  +   294.55 . Simplify:         0 = 28.68 s2   - 116.705s   -   138.61 , Use quadratic formula: .              s =   -0.961    or  5.03   . Only the negative makes sense, because the ball bouncesback... . So the final speed of the ballis     0.961 m/s .     433.16   = 17s2    + 11.68s2   -   116.705 s  +   294.55 . Simplify:         0 = 28.68 s2   - 116.705s   -   138.61 , Use quadratic formula: .              s =   -0.961    or  5.03   . Only the negative makes sense, because the ball bouncesback... . So the final speed of the ballis     0.961 m/s

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