A block of mass 2.30 {\ m kg} slides down a 30.0 ^\\circ incline which is 3.60 {

Question

A block of mass 2.30 { m kg} slides down a 30.0 ^circ incline which is 3.60 { m m} high. At the bottom, it strikes a block of mass 7.65 { m kg} which is at rest on a horizontal surface. (Assume a smooth transition at the bottom of the incline.)

(a) If the collision is elastic, and friction can be ignored, determine the speed of the block with mass 2.30 { m kg} after the collision.

(b) If the collision is elastic, and friction can be ignored, determine the speed of the block with mass 7.65 { m kg} after the collision.

(c) How far back up the incline the smaller mass will go.

Explanation / Answer

Mass of the first block m = 2.30 Kg Mass of the second block M = 7.65 Kg Angle of inclination = 300 Height of the incline H = 3.60 m Initial speeds : u = Sqrt[2gH] = Sqrt[70.56] = 8.4 m/s U = 0 (a) Final speed of the 2.30 Kg block after collision :                  v = [(m-M)/(m+M)]u + [2M/(m+M)]U                     = -4.516 + 0                     = -4.156 m/s (b) Final speed of the 7.65 Kg block after collision :                      V = [(2m)/(m+M)]u + [(M-m)/(m+M)]U                          = 3.883 + 0                          = 3.883 m/s (c) Vertical height rised by the lighterr body h = v^2/2g = 0.881 m Distance covered by the lighter body along the incline after collision :                                                S = h/sin30                                                   = 0.881/0.5                                                   = 1.762 m     (c) Vertical height rised by the lighterr body h = v^2/2g = 0.881 m Distance covered by the lighter body along the incline after collision :                                                S = h/sin30                                                   = 0.881/0.5                                                   = 1.762 m    

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