# hw8s - Zmﬁi_M_2 P8.37(a VCM" M ‘ M(2.00 kg(2.00i...

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Unformatted text preview: - Zmﬁi _M_2_ P8.37 (a) VCM " M ‘ M (2.00 kg)(2..00i m/s — 3.00} urn/s) + (3.00 kg)(1.00i m/s + 6.00 j m/s) 5.00 kg em = (1.4oi+2.40}) m/s (b) p = MvCM = (5.00 kg)(1.40i + 2.403) m/s = (7.003 + 12.0 j) kg - m/s P8.40 (a) Conservation of momentum for the two-ball system gives us: 0.200 kg(1.50 m/s) +0300 kg(—O.400 m/s) = 0.200 kg U], + 0.300 kg :22). Relative velocity equation: vzf ‘vlf Ill/S Then 0.300 — 0.120 = 0.2002;1 f + 0.300(1.90 + 01,) 01, =—0.780 111/5 02): =1.12 111/3 61). =~0.780i m/s 02f =1.12i m/s (1?) Before, VCM = 9m =(0.360 m/s)l Afterwards, the Center of mass must move at the same velocity, as momentum of the system is conserved. *P8.48 Using conservation of momentum from just before to just after the impact of the bullet With the block mvi = (M+m)vf or vi=(M+m)vf. (1) m The speed of the block and embedded bullet just after impact may be found using kinematic equations: 1 d=vft and h=—2'gtz. 2h -d- {£4112 mus,t=Eandvf—?~d Zh- 2h- . M + m gdz Substituting into (1) from above gives 0; = E . m FIG. P8.48 P8.56 x-component of momentum for the system of the two objects: Plix + Pzix = P1 fx + szxi —mv,> + 3mvi = 0 + Bmvz,‘ y-component of momentum of the system: 0 + 0 = —mvly + 3mv2y . 1 2 1 2 1 2 1 2 2 by conservahon of energy of the system: +§mvi +53mvi = Emv1y +E3m(v2, + vzy) 21)- we have 023‘ = _t 3 also 011”: svzy - 2 2 4v? 3 2 So the energy equauon becomes 40'. = gyzy +_3_ + 02y 80-2 T‘ = 12v;y or v ._ 27’.“ 2y 3 (a) The object of mass m has ﬁnal speed on = 3va = Jivi , 2 2 and the object of mass 3 m moves at ‘lvgx + vgy = %+ 23’— r— 2 _ —1 vzy _ _1 «EU,- 3 __ O 222,: . P10.10 Given r =1.00 m, a: 4.00 rad/s2 , wi =0 and Qi = 573°: 1.00 rad (a) wf=wi+at=0+w At t= 2.00 s, a), = 400 rad/s2 (2.00 S): (b) v = M) = 1.00 m(8.00 rad/s) = [(1,] =ac =ra)z =1.00 m(8.00 rad/s)2 = 64.0 m/s2 a, =ra= 1.00 m(4.00 rad/52) = 400 m/s2 \ The magnitude of the total acceleration is: e="a3+a,2 =‘/(64.0 m/sz)Z+(4.00 m/sz)z = The direction of the total acceleration vector makes an angle ¢with respect to the radius to point P: ¢=tan-l[:—r]=mn-l(%)= C __ 1 z _ 1 2 2 _ (c) 0,—9i+a),-t+§at _(1.00 rad)+E(4.00rad/s )(200s) _M P1036 (a) I=%M(R12 + 12%) =%(0.35 kg)[(0.02 m)2 +(0.03 m)2]= 2.28 x104 kg-m2 (K1 +K2 +Krot +ug2)i ‘fkd =(K1 +K2 +Krot)f 1 2 1 2 1 H, 2 0.82 m/s 2 5(0350 kg)(0.82 m/s) +£0.42 kg)(0.82 m/s) +E(2.28x10 kg-m 0.03m +0.42 kg(9.8 m/s2)(o.7 m)— 0.25(0.85 kg)(9.8 m/sz)(0.7 m) 2 _ 1 2 1 2 1 ‘4 2 vi 0.512 J+ 2.88 J—l.46 I: (0.761 kg)v; 1.94 J = = 1.59 vf 0.761 kg __ v _ 1.59 m/s _ (b) (0—7— 0.03 m — 53.1 rad/s P1037 (a) ng — T2 = mza T2 =m2(g—u)=20.0 kg(9.80 m/s2—2.oo m/sz)= T1 — m1 gsin37.0°= mla T1 =(15.o kg)(9.80sin37.0°+2.00) m/s2 = (b) (Tz—n)R=Ia=I(%) (T2 — T1)R2 (156 N—118 N)(0.250 m)2 z I =——— = ‘= 1.17 . a zoo m/sz FIG. P1037 ...
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## This note was uploaded on 04/02/2008 for the course PHYS 6A taught by Professor Mahaashour-abdalla during the Winter '07 term at UCLA.

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hw8s - Zmﬁi_M_2 P8.37(a VCM" M ‘ M(2.00 kg(2.00i...

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