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.

Answers (5)

sajid 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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ekanath 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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