Answer:
s_max = 0.8394m
Explanation:
From equilibrium of block, N = W = mg
Frictional force = μ_k•N = μ_k•mg
Since μ_k = 0.3,then F = 0.3mg
To determine the velocity of Block A just before collision, let's apply the principle of work and energy;
T1 + ΣU_1-2 = T2
So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²
Plugging in the relevant values to get ;
(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²
750 - 176.58 = 7.5(v_a1)²
v_a1 = 8.744 m/s
Using law of conservation of momentum;
Σ(m1v1) = Σ(m2v2)
Thus,
m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2
Thus;
15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)
Divide through by 5;
3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)
Thus,
3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)
Coefficient of restitution has a formula;
e = (v_b2 - v_a2)/(v_a1 - v_b1)
From the question, e = 0.6.
Thus;
0.6 = (v_b2 - v_a2)/(8.744 - 0)
0.6 x 8.744 = (v_b2 - v_a2)
(v_b2 - v_a2) = 5.246 - - - (eq2)
Solving eq(1) and 2 simultaneously, we have;
v_b2 = 8.394 m/s
v_a2 = 3.148 m/s
Now, to find maximum compression, let's apply conservation of energy on block B;
T1 + V1 = T2 + V2
Thus,
(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²
(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²
500(s_max)² = 352.29618
(s_max)² = 352.29618/500
(s_max)² = 0.7046
s_max = 0.8394m
