Question

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.0230 kg and is moving along the x axis with a velocity of +7.74 m/s. It makes a collision with puck B, which has a mass of 0.0460 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the speed of (a) puck A and (b) puck B. At rest Before collision After collision

Answer 1

VAsinest / UA=774mis s NACE 65° AX V UB=0B Atrest VB GS 37° Y VB sin 37' ® Before collision After collision. MA = 0.0230 ky , MB= 0.0460 kg. UA = 7.74 mis, up=o. ,VA=?, UB=? Using the law of conservation of linear momentum, Along horizontal axis - - MAUA + MB. B = MAVACOS 6S + MB UB 6S 37°. 7 0.0230x7.70 + 0 = 0.0230 X VAX 0,4226 +0.0460xVB Xo.7986 = 0.0097 VA + 0.0367.VB = 0.1771 - 01) Se allong vertical axis – Oto = MA VA sin 65° - MBVB Sin 37. O = 0.0230 X VAX0.9063 -0.0460 X VBX0.6018 0 = 0.020BVA – 0.0277 VB VB = 0.0208 v = 0.75 VA — 2) 0.0277 putting the value of VB in equation (I), we get- 0.0097 VA +0.0367 X 0.75 VA = 0.1771 + Bio 372 VA = 0.1771 → VA= 01771_54.76 ☺.0372. or VA = 4.76 mis &202075*4.76= 3.57 mis.