Answer:
The speed of body B after collision is 128 cm/s
Explanation:
Given;
initial speed of body A, [tex]U_a[/tex] = 80 cm/s (forward direction)
initial speed of body B, [tex]U_b[/tex] = - 40 cm/s (backward direction)
mass of body A, [tex]M_a[/tex] = 140 g
mass of body B, [tex]M_b[/tex] = 60 g
let the speed of body A after collision = [tex]V_a[/tex]
let the speed of body B after collision = [tex]V_b[/tex]
Apply the principle of conservation of linear momentum;
[tex]M_aU_a + M_bU_b = M_aV_a + M_bV_b\\\\ 140(80) + 60(-40) = 140V_a + 60V_b\\\\8800 = 140V_a + 60V_b\\\\440 = 7V_a + 3V_b ----equation(1)[/tex]
One direction velocity;
[tex]U_a + V_a = U_b + V_b\\\\80 + V_a = -40 + V_b\\\\V_a = V_b -120[/tex]
substituting the value of Va in the first equation, we will have;
[tex]7V_a +3V_b = 440\\\\7(V_b-120) + 3V_b = 440\\\\7V_b-840+3V_b = 440\\\\10V_b = 1280\\\\V_b = 128 \ cm/s \ \ \ \ (backward )\\\\V_a = V_b -120\\\\V_a = 128 -120\\\\V_b = 8 \ cm/s \ \ \ \ (forward)[/tex]
Therefore, the speed of body B after collision is 128 cm/s