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Two bodies A and B move toward each other with speeds of 80 cm/s and 40 cm/s, respectively. The mass of A is 140 g and that of B is 60 g. After a head-on, perfectly elastic collision, the speed of B is

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onyebuchinnaji

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