An elastic collision occurs between two air hockey pucks inwhich one puck is at

Question

An elastic collision occurs between two air hockey pucks inwhich one puck is at rest and the other is moving with a speed of0.9 m/s. After the collision, the puck initially in motion makes anangle of 20.00 o with its original direction, and thestruck puck moves at an angle of 70.00 o on the otherside of the original direction. What is the final speed of thefirst puck? What is the final speed of the second? THANKS! An elastic collision occurs between two air hockey pucks inwhich one puck is at rest and the other is moving with a speed of0.9 m/s. After the collision, the puck initially in motion makes anangle of 20.00 o with its original direction, and thestruck puck moves at an angle of 70.00 o on the otherside of the original direction. What is the final speed of thefirst puck? What is the final speed of the second? THANKS!

Explanation / Answer

Let A & B are two  air hockey pucks . mass of these two are same and equal to m. va = 0.9 m/s vb = 0 a' = 200 b' = - 700 now we have to find va' & vb' Apply conservation of momentum for the x & ycomponents (for x) mva = mva' cos 20 +mvb' cos(-70)            va= va' cos 20 - vb'cos(70)   -----------------(1) (for y) 0 = mva' sin 20 + mvb' sin(-70)                               0 =va' sin 20 - vb' sin (70)            vb'=  va' (sin 20/sin 70)                =  va' (0.342/0.9396)                 =0.3639 va' Substitute the above value in equation(1) we get           0.9 = 0.9396 va' - 0.1244 va'            0.9 = 0.8152 va'           va' = 1.1040 m/s AND      vb' = 0.3639 *1.1040 = 0.4017 m/s now we have to find va' & vb' Apply conservation of momentum for the x & ycomponents (for x) mva = mva' cos 20 +mvb' cos(-70)            va= va' cos 20 - vb'cos(70)   -----------------(1) (for y) 0 = mva' sin 20 + mvb' sin(-70)                               0 =va' sin 20 - vb' sin (70)            vb'=  va' (sin 20/sin 70)                =  va' (0.342/0.9396)                 =0.3639 va' Substitute the above value in equation(1) we get           0.9 = 0.9396 va' - 0.1244 va'            0.9 = 0.8152 va'           va' = 1.1040 m/s AND      vb' = 0.3639 *1.1040 = 0.4017 m/s

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