A solid ball of mass of 200 kg and from rest swings from the cliff by rigidly holding on a solid vertical rod through a cord, which is 10 m measured from the supporting limb A to its center of mass. Determine its speed just after the cord strikes the lower horizontal extension beam at B. Also, with what force is required for the ball to be intac with the cord just after the cord contacts the extension beam at B.

1) 5.12 m/s & 2.03 kW
2) 3.25 m/s & 3.45 kW
3) 4.25 m/s & 2.60 kW
4) 6.15 m/s & 5.03 kW

Assuming the datum at C and applying the principle of conservation of energy we have :
T₁ + V₁ = T₂ + V₂
0 + 0 = 1/2(200)(v_B)² - 200(9.81)10(1-cos30^o)
v_B = 5.12 m/s
So, the velocity of the ball at Bis v_B = 5.12 m/s

Also for point B just after striking the extension, applying the principle of equillibrium we have that :
S Fn = ma_n
T - 981 = 200 (5.12)² /5
T = 40(5.12)^{2 + 981
T = 2032.43 = 2.03 kN}

So the force required for the ball to be intact with the cord just after striking the extended beam
T = 2.03 kN