Answer:
vA = -2.55 m/s
vB = 0.947 m/s
Explanation:
Given:-
- The initial angle of rope, α = 30°
- The angle of rope just before impact or wedge angle, θ = 20°
- The weight of sphere, Ws = 1-lb
- The initial position velocity, vi = 4 ft/s
- The coefficient of restitution, e = 0.7
- The weight of the wedge, Ww = 2-lb
- The spring constant, k = 1.8 lb/in
- The length of rope, L = 2.6 ft
Find:-
Determine the velocities of A and B immediately after the impact.
Solution:-
- We can first consider the ball ( acting as a pendulum ) to be isolated for study.
- There are no unbalanced fictitious forces acting on the sphere ball. Hence, we can reasonably assume that the energy is conserved.
- According to the principle of conservation for the initial point and the point just before impact.
Let,
vA : The speed of sphere ball before impact
Change in kinetic energy = Change in potential energy
ΔK.E = ΔE.P
0.5*ms* ( uA^2 - vi^2 ) = ms*g*L*( cos ( θ ) - cos ( α ) )
uA^2 = 2*g*L*( cos ( θ ) - cos ( α ) ) + vi^2
uA = √ [ 2*32*2.6*( cos ( 20 ) - cos ( 30 ) ) + 4^2 ] = √28.25822
uA = 5.316 ft/s
- The coefficient of restitution (e) can be thought of as a measure of the extent to which mechanical energy is conserved when an object bounces off a surface:
e^2 = ( K.E_after impact / K.E_before impact )
- The respective Kinetic energies are:
K.E_after impact = K.E_sphere + K.E_block
= 0.5*ms*vA^2 + 0.5*mb*vB^2
K.E_before impact = K.E = Ws*L*( cos ( θ ) - cos ( α ) )
= 1*2.6*( cos ( 20 ) - cos ( 30 ) )
= 0.1915 J
32*2*0.1915*0.7^2 = Ws*vA^2 + Wb*vB^2
6.00544 = vA^2 + 2*vB^2 ... Eq1
- From conservation of linear momentum we have:
vB = e*( uA - uB )*cos ( 20 ) + vA
vB = 0.7*( 5.316 - 0 )*cos ( 20) + vA
vB = 3.49678 + vA .... Eq 2
- Solve two equation simultaneously:
6.00544 = vA^2 + 2*(3.49678 + vA)^2
6.00544 = 3vA^2 + 13.98*vA + 24.455
3vA^2 + 14.8848*vA + 18.4495 = 0
vA = -2.55 m/s
vB = 0.947 m/s
