Question

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°

200 4

Answer 1

Sum of forces in Y-direction

$N-m_{A}gCos20=0$

$N=m_{A}gCos20=2\times 9.81\times Cos20=18.44 N$

Frictional force

$F_{f}=\mu _{K}N=0.2\times 18.44=3.687 N$

From Work-Energy theroem

(1/2)Kx₁² + m_Ag(x+d)sin20 -F_f(x+d) =(1/2)m_AV_A²

(1/2)*800*0.107²+2*9.81*(0.107+1.5)*sin20 -3.687(0.107+1.5)=(1/2)*2*V_A²

V_A=3.072 m/s

By Conservation of momentum in x-direction

m_AV_ACos20 =m_AV_A'Cos20 +m_BV_B

2*3.072*Cos20=2*V_A'Cos20+1*V_B

1.88V_A'+V_B' =5.773--------------------1

V_B'Cos20-V_A' =e(V_A-0)

-V_A'+0.94V_B' =2.457-------------------2

solving 1 and 2 we get

V_A'=1.073 m/s

V_B' =3.755 m/s

From Work energy theorem to find velocity at alpha =30^o

(1/2)m_BV_B'² = (1/2)m_BV₂²+m_BgL(1-Cos30)

V₂² =(V_B')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=m_B(a+gCos40) =1(11.464+9.81Cos30)

T=19.96 N