Question

Pendulum impacting an inclined surface of a block attached to a spring-Dependent multi-part problem assign all parts NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. When the rope is at an angle of α = 30°, the 1-lb sphere A has a speed v0 = 4 ft/s. The coefficient of restitution between A and the 2-lb wedge B is 0.7 and the length of rope l = 2.6 ft. The spring constant has a value of 1.8 lb/in. and θ = 20°. Problem 13.188.a Pendulum impacting an inclined surface of a block attached to a spring-determine velocities after impact Determine the velocities of A and B immediately after the impact. The velocity of A immediately after the impact is ft/s ⦩ °. The velocity of B immediately after the impact is ft/s →.

Answer 1

vA = -2.55 m/s

vB = 0.947 m/s

Step-by-step 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