a 2 kg block A is pushed up against a spring compressing it a distance x=0.107 m. The block is then released from rest and slides down the 20° incline until it strikes a 1-kg sphere B that is suspended from a 1 m inextensible rope. The spring constant k=800 N/m, the coefficienct of friction between A and the ground is 0.2, the distance A slides from the unstretched length of the string is d=1.5, and the coefficient of restitution between A and B is 0.8
a) velocity of block A right before impact
b) velocities of both blocks right after impact
c) velocity of B and the tension of the rope at alpha=30°
Sum of forces in Y-direction
Frictional force
From Work-Energy theroem
(1/2)Kx12 + mAg(x+d)sin20 -Ff(x+d) =(1/2)mAVA2
(1/2)*800*0.1072+2*9.81*(0.107+1.5)*sin20 -3.687(0.107+1.5)=(1/2)*2*VA2
VA=3.072 m/s
By Conservation of momentum in x-direction
mAVACos20 =mAVA'Cos20 +mBVB
2*3.072*Cos20=2*VA'Cos20+1*VB
1.88VA'+VB' =5.773--------------------1
VB'Cos20-VA' =e(VA-0)
-VA'+0.94VB' =2.457-------------------2
solving 1 and 2 we get
VA'=1.073 m/s
VB' =3.755 m/s
From Work energy theorem to find velocity at alpha =30o
(1/2)mBVB'2 = (1/2)mBV22+mBgL(1-Cos30)
V22 =(VB')2-2gl(1-cos30)=3.7552-2*9.8*1*(1-Cos30)
V2 =3.386 m/s
acceleration
a=3.3862/1 =11.4694 m/s2
Tension
T=mB(a+gCos40) =1(11.464+9.81Cos30)
T=19.96 N
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