A ball with mass M, moving horizontally at 4.00 m>s, collides elastically with a block with mass 3M that is initially hanging at rest from the ceiling on the end of a 50.0-cm wire. Find the maximum angle through which the block swings after it is hit.

θ = 53.7°

Explanation:

Given:

- The mass of ball = M

- The mass of object = 3M

- The wire length L = 0.5 m

- The velocity of ball vi = 4.0 m/s

- The velocity of ball vf

- The velocity of object Vf

Find:

Find the maximum angle through which the block swings after it is hit.

Solution:

- When two objects collide with no external force acting on the system the linear momentum of the system is conserved. The initial (Pi) and final (Pf) linear momentum are equal:

                                  Pi = Pf

                                  M*vi = M*vf + 3M*Vf

                                  vi = vf + 3*Vf

                                  4 = vf + 3*Vf

- For elastic collision between two particles the relative velocities before and after collision have the same magnitude but opposite sign; so,

                                   vi - 0 = Vf - vf

                                   4 = Vf - vf

- Solve the above two equation simultaneously.

                                   8 = 4*Vf

                                   Vf = 2 m/s

                                    vf = -2 m/s

- When the ball hits the object it swing under the influence of gravity only. Hence, no external force acts on the object so we can apply the conservation of energy as the object attains a height h.

                                   ΔK.E = ΔP.E

                                   0.5*(3M)*Vf^2 = (3M)*(g)*(h)

                                   h = Vf^2 / 2*g

- Plug in the values:

                                   h = 2^2 / 2*9.81

                                   h = 0.2039 m

- We can see that the maximum angle can be given as θ according trigonometric relation as follows:

                                  θ = arccos [ ( L - h ) / L ]

                                  θ = arccos [ ( 0.5 - 0.2039 ) / 0.5 ]

                                  θ = 53.7°

answer: strength i guess

explanation:

0.55x745.7j and a calculator