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
