a body British moving with a velocity of 20ms(power -1) collides with another body A, moving in the same direction on the same horizontal line with a velocity of 10ms (power -1).if the mass of the body A is three times the mass of body B, and the collision is perfectly elastic. calculate the velocity of A and B after the collision and the velocity of B relativemail to that of A after the collision

I am not going to do all the math.

Conservation of momentum, and energy applies
Momentum
A*20+B*10=A*Va + B*Vb
but A=3B
Energy:
1/2 A 20^2+1/2 B*10^2=1/2 A Va^2 + 1/2 B Vb^2

in equation 1, solve for Va in therms of all else, then substiture it in equation 2, and solve the quadratic.

To calculate the velocities of bodies A and B after the collision, we can apply the conservation of momentum and kinetic energy.

1. Conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision.
Momentum = Mass x Velocity

Let's assign the following variables:
Mass of body A = mA
Mass of body B = mB
Velocity of body A before the collision = VA_initial
Velocity of body B before the collision = VB_initial
Velocity of body A after the collision = VA_final
Velocity of body B after the collision = VB_final

Given information:
mA = 3mB (Mass of body A is three times the mass of body B)
VA_initial = 10 m/s
VB_initial = 20 m/s

Using the conservation of momentum:
(mA x VA_initial) + (mB x VB_initial) = (mA x VA_final) + (mB x VB_final)

Substituting the given values:
(3mB x 10) + (mB x 20) = (3mB x VA_final) + (mB x VB_final)

2. Conservation of kinetic energy:
The total kinetic energy before the collision is equal to the total kinetic energy after the collision.

Using the kinetic energy formula:
Kinetic Energy = (1/2) x Mass x Velocity^2
Total kinetic energy before = (1/2) x mA x VA_initial^2 + (1/2) x mB x VB_initial^2
Total kinetic energy after = (1/2) x mA x VA_final^2 + (1/2) x mB x VB_final^2

Since the collision is perfectly elastic, the kinetic energy is conserved. So we can equate the kinetic energy before and after the collision.

(1/2) x mA x VA_initial^2 + (1/2) x mB x VB_initial^2 = (1/2) x mA x VA_final^2 + (1/2) x mB x VB_final^2

Substituting the given values:
(1/2) x 3mB x 10^2 + (1/2) x mB x 20^2 = (1/2) x 3mB x VA_final^2 + (1/2) x mB x VB_final^2

Now we have two equations with two unknowns (VA_final and VB_final). We can solve these equations simultaneously to find the values of VA_final and VB_final.

Simplifying and rearranging the equations:
1. 30mB + 20mB = 3mB x VA_final + mB x VB_final
2. (3/2) x 100mB + 200mB = 3mB x VA_final^2 + mB x VB_final^2

Solving these equations will give us the values of VA_final and VB_final.

Note: The velocity of B relative to A after the collision can be calculated as the difference between their velocities:
Velocity of B relative to A = VB_final - VA_final

You can substitute the given values into these equations and solve them to get the final velocities.