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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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padmajai 20-05-2013
    Let us allocate designs to everyone the appropriate portions, write down the equations revealing efficiency regarding impetus and energy after which net ball a algebra automatic robot clear up these.
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steve.jobs 28-05-2013
    "Within the straight down isle path we now have exactly the bowling ball inwards motion at first, so in the future the musket ball and also bowling pin. Conservation regarding impetus inside the forward path says           Thousand*V0 = Thousand*V1cos(T) + meters*sixth v*cosine(some sort of)           Conservation regarding traction within the 'sideway' direction provides           nought = -Thousand*V1*hell(t) + meters*five*sin(a)           Lastly, energy efficiency states           i/a couple of*L*V0enqa_2 = one particular/ii*Thousand*V1enqa_a couple of + i/a couple of*michael*venqa_only two           "
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somo 31-05-2013
    "We should remember that Meters= 6m and in the offered valuations--           a= eighty two.eight degrees = eighty-two.8-10*pi/one hundred and eighty radians.           Accumulating all the equations in readiness with regard to definitely the actual QuickMath's equality problem solver,           V0 = 12.1     M= 6m     a= 82.8*pi/180     M*V0 = M*V1*cos(t) + m*v*cos(a)     0 = -M*V1*sin(t) + m*v*sin(a)     1/2*M*V0enqa_2 = 1/2*M*V1enqa_2 + 1/2*m*venqa_2           I recently cut and also glued these kind of to the Advanced manner with the Convergent thinker.           QuickMath provides a few alternatives! 3 of which possess damaging data transfer speeds, which can be extremely hard. Your next is simply the predicament prior to ball visitors the actual personal identification number. The fifth is this fact:           V1 = xii.0534 thousand/S     five =3.59977 mirielle/azines     T =nought.0356722 radians.           Therefore the bowl slows a bit, deflects a little, plus the personal identification number actions off of with a speeding of about only two.half a dozen thousand/ersus. These kinds of appear like sensible valuations if you ask me.           "
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