Question

Problem 13.175 A 1-kg block B is moving with a velocity v0 of magnitude as it hits the 0.5-kg sphere A, which is at rest and hanging from a cord attached at O. Knowing that between the block and the horizontal surface and between the block and the sphere, determine after impact 0 v  2 m/s 0.6 k  e  0.8 (a) the maximum height h reached by the sphere, (b) the distance x traveled by the block

Answer 1

Explanation:

Mass of the Block, Mb = 1.0 kg Initial Velocity of Block, Vb = 2.0 m/s

Mass of the Ball, Ma = 0.5 kg Coefficient of Kinetic Friction, μ = 0.6

Coefficient of Restitution, e = 0.8 Gravity = 9.81 m/s^2

Sum of the forces of the X-axis components and Y-axis components are:

∑ Fx = Ff = Mb × a

Equation for frictional force,

Ff = μ × N

∑ Fy = N - mb × g = 0

Note:

N = mb × g

Therefore, to solve for the acceleration, we have:

μ × Mb × g = Mb × a

a = μ × g

= 0.6 × 9.81

= 5.88 m/s^2

Therefore, ratio of velocities using the coefficient of restitution, e:

e = (Vb2 – Va1)/ (Va – Vb)

Note: Va = zero (initially at rest)

Va2 – Vb2 = 0.8 × ( 0 - 2 m/s)

Vb2 – Va2 = -1.6 m/s

Vb2 = a2 – 1.6

Using Conservation of Momentum Equation:

Ma × Va + Mb × Vb = Ma × Va2 + Mb × Vb2

0 + 1kg × 2 m/s = 0.5 kg × Va2 + 1 kg × Vb2

Vb2 = 2 - 0.5 × Va2

Substitute in Vb2,

1.5 × Va2 = 3.6

Va2 = 2.4 m/s

Vbs = 0.8 m/s

.

mgh = 0.5 × Mv2

h = v2/ 2g

h = 0.294 m

B.

Using equation of motion,

Vf^2 = Vo^2 + 2a × S

Given:

Vf = 0 m/s

a = 5.88 m/s^2

0 = 0.82 + 2(5.88) × ΔS

Δx = 0.0544 m