# In order to convert a tough split in bowling, it is necessary to strike the pin a glancing blow as shown in th?

In order to convert a tough split in bowling, it is necessary to strike the pin a glancing blow as shown in the figure. Assume that the bowling ball, initially traveling at 13.0 m/s, has five times the mass of a pin and that the pin goes off at 75 degrees from the original direction of the ball. Assume the collision is elastic and ignore any spin of the ball.

18-02-2013

06-03-2013
"Momentum and energy will be conserved 5*m=M 5*m*13=cosA*5*m*vM+cos75*m*vm or 5*13=cosA*5*vM+cos75*vm and 5*m*sinA*vM-sin75*vm*m=0 or 5*sinA*vM-sin75*vm=0 and .5*5*m*13^2=.5*5*m*vM^2+.5*m*vm^2 or 5*13^2=5*vM^2+vm^2 5*13-cos75*vm=cosA*5*vM sin75*vm=5*sinA*vM divide TanA=vm/(5*13-cos75*vm)"
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16-05-2013
"from the law of momentum we have : MVo = M.Vb.cosθ + m.Vp.cosα ...........(1) M.Vb.sinθ - m.Vp.sinα = 0 ....................(2) from the law of energy kinetik we have : ½ MVo² = ½ MVb² + ½ mVp² ...............(3) from equation (1) & (2) we get : M.Vb.sinθ - m.Vp.sinα = 0 ----> MVb = mVp(sinα / sinθ) MVo = MVb.cosαθ + mVp.cosα MVo = mVp(sinα / sinθ) + mVp.cosα (M/m)Vo = Vp(sinα/sinθ + cosα) 6Vo = Vp(sinα/sinθ + cosα) .....................(4) from equation (2) & (3) we get : ½ MVo² = ½ MVb² + ½ mVp² (M/m)Vo² = (M/m)Vb² + Vp² 6Vo² = 6Vb² + Vp² 6Vo² = 6(1/6 Vp(sinα/sinθ))² + Vp² 36Vo² = (sinα/sinθ)² Vp² + Vp² 6Vo = (1 + sinα/sinθ )Vp sinθ = sinα/(6Vo/Vp - 1) ..................(5) from equation (4) & (5) we get : 6Vo = Vp(sinα/sinθ + cosα) 6Vo = Vp(sinα/[sinα/(6Vo/Vp - 1)] + cosα) 6Vo = Vp[(6Vo/Vp - 1) + cosα] 6VoVp - 60Vo = 6VoVp - Vp²cosα + Vpcosα Vp²cosα - Vpcosα = 60Vo Vp² + Vp = 60Vo.secα (Vp + ½)² = ¼ + 60Vo.secα Vp = -½ +√(¼ + 60Vo.secα) ....................(6) from equation (6) we get : Vo = 12.1 m/s , α = 82.8° Vp = -½ +√(¼ + 60*12.1*sec 82.8) Vp ≈ 75.61 m/s "
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