Question

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

Answer 1

The speed of body B after collision is 128 cm/s

Step-by-step explanation:

Given;

initial speed of body A,
`U_a` = 80 cm/s (forward direction)

initial speed of body B,
`U_b` = - 40 cm/s (backward direction)

mass of body A,
`M_a` = 140 g

mass of body B,
`M_b` = 60 g

let the speed of body A after collision =
`V_a`

let the speed of body B after collision =
`V_b`

Apply the principle of conservation of linear momentum;

`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)`

One direction velocity;

`U_a + V_a = U_b + V_b\\\\80 + V_a = -40 + V_b\\\\V_a = V_b -120`

substituting the value of Va in the first equation, we will have;

`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)`

Therefore, the speed of body B after collision is 128 cm/s