Two hockey players (mA = 80 kg, mB = 90 kg) converging on the puck at x = 0, y =

Question

Two hockey players (mA = 80 kg, mB = 90 kg) converging on the puck at x = 0, y = 0 become entangled and fall. Before the collision, vA = 9i + 4j (m/s) and VB - 3i + 6j (m/s) If the coefficient of kinetic friction between the players and the ice is mu k

Explanation / Answer

Totalm*ResultantV=m1*v1+m2*v2===>(80+90)*v=80*(va)+90*vb==>v=4.23i+1.88j-1.588i+3.17j==>v=2.35i+5.05j,So due to friction the decceleration will be mu*g=0.1*10=1,So using kinematic relation v^2=2*a*s ==>2.35^2+5.05^2=2*a*s,So s=15.26 m along the direction of resultant v (s_=s*vcap) vcap=v_/|v| v_=2.35i+5.05j |v|=30.522

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