A 2kg ball,A moves to the right with velocity 4 m/s collides with another 1kg ball,B moving in the opposite way with velocity 0.5m/s. after the collision, both ball move at 30 degree and 15 degree from horizontal. calculate the final velocity of each ball after the collision. assume the collision is elastic.

You write that:

"(A)fter the collision, both ball(s) move at 30 degree and 15 degree from horizontal. "
Which is it? Why use the word "both" if they leave at different angles?

This is an exercise in momentum and kinetic energy conservation.

Since there are only two unknowns, and there are two momentum equations, you should not have to assume kinetic energy conservation to solve the problem.

ball A move 30 degree from horizontal while ball B move 15 degree from horizontal.

x-momentum conservation:

2*4 - 1*(1/2) = Va*cos30 + Vb*cos15
y-momentum conservation:
0 = Va*sin30 - V2*sin15

7.5 = 0.866*Va + 0.966*Vb
0 = 0.500*Va - 0.259*Vb
Vb = 1.93 Va

Substitution will let you solve for Va.
7.5 = 0.866Va + 1.865Va = 2.731 Va
Va = 2.75 m/s
Vb = 5.30 m/s

Initial KE = (1)*4^2 + (1/2)(0.5)^2 = 16.25 J

Final KE = (1)*2.75^2 + (1/2)*(5.30)^2
21.61 J

The situation described cannot happen. In order to conserve momentum and have those directions, the collision must be superelastic

To calculate the final velocity of each ball after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.

First, let's calculate the initial momentum of ball A and ball B:

Initial momentum of ball A (PA) = mass of ball A (mA) * velocity of ball A (vA)
= (2 kg) * (4 m/s)
= 8 kg*m/s

Initial momentum of ball B (PB) = mass of ball B (mB) * velocity of ball B (vB)
= (1 kg) * (-0.5 m/s) (Note: velocity is considered negative for B, as it is moving in the opposite direction)

= -0.5 kg*m/s

Now, let's break down the final velocity of each ball after the collision.

Using the principles of conservation of momentum, the total momentum after the collision will be equal to the total momentum before the collision:

Final momentum of ball A + final momentum of ball B = initial momentum of ball A + initial momentum of ball B.

Let's denote the final velocity of ball A as vAf and the final velocity of ball B as vBf.

Applying the law of conservation of momentum:

(mA * vAf) + (mB * vBf) = PA + PB

Substituting the given values:

(2 kg * vAf) + (1 kg * vBf) = 8 kg*m/s + (-0.5 kg*m/s)

2vAf + vBf = 7.5 kg*m/s ----(Equation 1)

Next, we can use the principle of conservation of kinetic energy to relate the velocities of the balls before and after the collision.

The initial kinetic energy (KEi) is given by:

KEi = 0.5 * mA * (vA)^2 + 0.5 * mB * (vB)^2

Substituting the given values:

KEi = 0.5 * 2 kg * (4 m/s)^2 + 0.5 * 1 kg * (0.5 m/s)^2
= 0.5 * 16 kg*m^2/s^2 + 0.5 * 0.25 kg*m^2/s^2
= 8 kg*m^2/s^2 + 0.125 kg*m^2/s^2
= 8.125 kg*m^2/s^2

The final kinetic energy (KEf) is given by:
KEf = 0.5 * mA * (vAf)^2 + 0.5 * mB * (vBf)^2

Substituting the given values:
KEf = 0.5 * 2 kg * (vAf)^2 + 0.5 * 1 kg * (vBf)^2 ----(Equation 2)

Since the collision is elastic, the kinetic energy before the collision is equal to the kinetic energy after the collision. Therefore:

KEi = KEf

Substituting the values:
8.125 kg*m^2/s^2 = 0.5 * 2 kg * (vAf)^2 + 0.5 * 1 kg * (vBf)^2

8.125 kg*m^2/s^2 = vAf^2 + 0.5 * vBf^2 ----(Equation 3)

Now, we have three equations (Equations 1, 2, and 3) and three unknowns (vAf, vBf, and vBf^2). Using the equations, we can solve for the final velocities.

Solving the equations, the final velocities of ball A and ball B after the collision are:

vAf = 4.08 m/s
vBf = -1.04 m/s