A ball of mass M is moving horizontally at 6.00 m/s. Itcollides inelastically wi

Question

A ball of mass M is moving horizontally at 6.00 m/s. Itcollides inelastically with a block with mass 3 M that is initiallyhanging at rest from the ceiling on the end of a 70.0 cm wire. Findthe maximum angle through which the block swings after it is hit bythe ball. Draw free body diagrams.
HOW DID THIS PERSON GET 33.27 DEGREES?!?! im getting 80.6 ...thanks for the help in advance! :) masses m = M           m ' = 3M Initial speed of M is u = 6 m / s Initial velocity of the 3M is u ' = 0 from law of conservation of momentum , m u + m ' u ' = ( m + m' ) v                                                               6M + 0 = 4Mv                                                                        6 = 4 v from this final velocity v = 6/4 = 1.5 m / s we know kinetic energy of the system just ater colliiosn =potential energy at maximum height                                             ( 1/ 2) ( m + m ' ) v ^ 2 = ( m + m ' ) gh                                                                            h = v ^ 2 / 2g                                                                               = 0.1147 m we know h = L ( 1- cos )    1- cos = h / L                  =0.1147 / 0.7         sinceL = length = 70 cm = 0.7 m                 = 0.16399              = 33.27 degrees A ball of mass M is moving horizontally at 6.00 m/s. Itcollides inelastically with a block with mass 3 M that is initiallyhanging at rest from the ceiling on the end of a 70.0 cm wire. Findthe maximum angle through which the block swings after it is hit bythe ball. Draw free body diagrams.
HOW DID THIS PERSON GET 33.27 DEGREES?!?! im getting 80.6 ...thanks for the help in advance! :) masses m = M           m ' = 3M Initial speed of M is u = 6 m / s Initial velocity of the 3M is u ' = 0 from law of conservation of momentum , m u + m ' u ' = ( m + m' ) v                                                               6M + 0 = 4Mv                                                                        6 = 4 v from this final velocity v = 6/4 = 1.5 m / s we know kinetic energy of the system just ater colliiosn =potential energy at maximum height                                             ( 1/ 2) ( m + m ' ) v ^ 2 = ( m + m ' ) gh                                                                            h = v ^ 2 / 2g                                                                               = 0.1147 m we know h = L ( 1- cos )    1- cos = h / L                  =0.1147 / 0.7         sinceL = length = 70 cm = 0.7 m                 = 0.16399              = 33.27 degrees HOW DID THIS PERSON GET 33.27 DEGREES?!?! im getting 80.6 ...thanks for the help in advance! :) masses m = M           m ' = 3M Initial speed of M is u = 6 m / s Initial velocity of the 3M is u ' = 0 from law of conservation of momentum , m u + m ' u ' = ( m + m' ) v                                                               6M + 0 = 4Mv                                                                        6 = 4 v from this final velocity v = 6/4 = 1.5 m / s we know kinetic energy of the system just ater colliiosn =potential energy at maximum height                                             ( 1/ 2) ( m + m ' ) v ^ 2 = ( m + m ' ) gh                                                                            h = v ^ 2 / 2g                                                                               = 0.1147 m we know h = L ( 1- cos )    1- cos = h / L                  =0.1147 / 0.7         sinceL = length = 70 cm = 0.7 m                 = 0.16399              = 33.27 degrees masses m = M           m ' = 3M Initial speed of M is u = 6 m / s Initial velocity of the 3M is u ' = 0 from law of conservation of momentum , m u + m ' u ' = ( m + m' ) v                                                               6M + 0 = 4Mv                                                                        6 = 4 v from this final velocity v = 6/4 = 1.5 m / s we know kinetic energy of the system just ater colliiosn =potential energy at maximum height                                             ( 1/ 2) ( m + m ' ) v ^ 2 = ( m + m ' ) gh                                                                            h = v ^ 2 / 2g                                                                               = 0.1147 m we know h = L ( 1- cos )    1- cos = h / L                  =0.1147 / 0.7         sinceL = length = 70 cm = 0.7 m                 = 0.16399              = 33.27 degrees

Explanation / Answer

33.27 is right

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