A curling stone slides on ice with a speed of 1.70 m/s and collides elastically

Question

A curling stone slides on ice with a speed of 1.70 m/s and collides elastically with an identical, stationary curling stone. After the collision, the first stone has a velocity of 0.800 m/s in a direction that makes a counterclockwise angle of 61.9 with its original direction of travel. At what speed and what direction is the second stone traveling after the collision?

The answers are:
-A                            1.50 m/s at a clockwise angle of 28.1 degree
-B                            1.40 m/s at a clockwise angle of 19.1 degree
-C                            1.60 m/s at a clockwise angle of 34.2 degree
-D                            1.40 m/s at a clockwise angle of 32.3 degree

Explanation / Answer

Let m be the mass of the stone The speed of the stone1 is u1 = 1.7 m/s The mometum of the stone is Pi = mu1                                                   = 1.7m The speed of stone 1 after the collision is v1 = 0.8 m/s The direction of the speed is 1 = 61.9o   counter clockwise The horizontal component is v1x = v1cos1                                                      = 0.38 m/s The vertical component is v1y = v1sin1                                                 = 0.71 m/s Let v2 be the speed of the secon stone at angle clockwise Then according to the law of conservation of linear momentum mu1 = mv1cos1 + mv2cos mv1sin1 = mv2sin v2cos = u1 - v1cos1              = 1.7 m/s - 0.38 m/s              = 1.32 m/s v2sin = v1sin1             = 0.71 m/s
Therefore v2 = [(1.32 m/s)2+(0.71 m/s)2]      = 1.5 m/s The direction is = tan-1(0.71/1.32)                              = 28.1o Option A is correct                                                      = 0.38 m/s The vertical component is v1y = v1sin1                                                 = 0.71 m/s Let v2 be the speed of the secon stone at angle clockwise Then according to the law of conservation of linear momentum mu1 = mv1cos1 + mv2cos mv1sin1 = mv2sin v2cos = u1 - v1cos1              = 1.7 m/s - 0.38 m/s              = 1.32 m/s v2sin = v1sin1             = 0.71 m/s
Therefore v2 = [(1.32 m/s)2+(0.71 m/s)2]      = 1.5 m/s The direction is = tan-1(0.71/1.32)                              = 28.1o Option A is correct Option A is correct

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