Question

A baseball (A, weight 0.33 lb) moves horizontally at 20 ft/s when it strikes a stationary block (B, weight 10 lb), supported by a frictionless surface. If the impact occurs over a time interval of 0.01 sec, with a coefficient of restitution of e=0.75, find the average normal force between A and B during impact.

Answer 1

Answer:

F = 154.85 N

Explanation:

Applying consevation of linear momentum for ball A and block B:

ma₁ × Va₁ + m b₁ × Vb₁ = ma₂ × Va₂ + mb₂ × Vb₂  

Block B is at rest → Vb₁ = 0  

0.33 × 20 + 10 (0) = 0.33 (Va₂) + 10 (Vb₂)        

6.6 = 0.33 Va₂ + 10 Vb₂    (1)

Using the ecuation of the coefficient of restitution:

e = (Vb₂ -  Va₂) / (Va₁ - Vb₁)

0.75 = (Vb₂ - Va₂) / (20 - 0)  

0.75 × 20 = Vb₂ - Va₂

15 = Vb₂ - Va₂     (2)

From (1), (2):

6.6 = 0.33 ×  (Vb₂ - 15) + 10Vb₂

6.6 = 0.33Vb₂ - 4.95 + 10Vb₂  

Vb₂ = 11.55 / 10.33  

Vb₂ = 1.12 ft/s    

Into (2):

15 = 1.12 - Va₂

Va₂ = -13.88 ft/s

Using the ecuation:

F = Δp / Δt,  F=Normal force between A y B

F =  [(mb₂ × Vb₂) - (m b₁ × Vb₁)] / Δt    

F = [(10 × 1.12) - (10 × 0)] / 0.01  

F =  1120 lb ft/s²  = 154.85 N

Have a nice day!