a ball A collides elastically with another ball B initially at rest. A is moving with a velocity of 10m/s at an angle 60' from the line joining the center. select the correct alternative:

a)velocity of ball A after collision is 5 m/s
b)velocity of ball B after collision is 5√3 m/s
c)velocity of ball A after collision is 7.5 m/s
d)velocity of ball B after collision is 5 m/s

x axis from center A to center B

Ball A before
Vx = 10 cos 60 = 5 sqrt 3
Vy = -10 sin 60 = -5
let m = 1 kg
Ke = (1/2)(100) = 50 Joules

Ball B before Vx = 0 and Vy = 0 and Ke = 0

Ball 1 after
Vx1
Vy1
Ke1 =(1/2)(Vx1^2+Vy1^2)

Ball 2 after
Vx2
Vy2 (has no Vy because hit in x direction
Ke2 = (1/2)(Vx2^2)

momentum before = momentum after
Vx = Vx1+Vx2 = 5
Vy = Vy1 = 8.66 = 5 sqrt 3

energy before = energy after
50*2 = (Vx1^2+Vy1^2)+(Vx2^2)
100 = Vx1^2 + 75 + (5-Vx1)^2
100 = Vx1^2 + 75 +25 -10Vx1 + Vx1^2

2 Vx1^2 - 10 Vx1 = 0
Vx1 = 5 or 0
try 0
then Vx2 = 5 ---- that is d)

To find the correct alternative, we can apply the laws of conservation of momentum and the laws of conservation of kinetic energy.

Let's break down the problem step by step:

1. Calculate the initial velocity components of ball A:
Since ball A is moving at an angle of 60° from the line joining the center, we can determine the horizontal and vertical components of its initial velocity.
Horizontal component: vAx = vA * cos(60°) = 10 m/s * cos(60°) = 5 m/s
Vertical component: vAy = vA * sin(60°) = 10 m/s * sin(60°) = 8.66 m/s (approx.)

2. Calculate the momentum of each ball before the collision:
Momentum (p) is given by p = mass * velocity
Since ball A is given to have a velocity of 10 m/s and ball B is initially at rest, its momentum is 0.

3. Apply the laws of conservation of momentum:
According to the law of conservation of momentum, the total momentum before the collision should equal the total momentum after the collision.
p(A)before = p(A)after + p(B)after
mass(A) * vA(before) = mass(A) * vA(after) + mass(B) * vB(after)

Since ball A collides elastically with ball B, the total kinetic energy is conserved:
(1/2) * mass(A) * (vA(before))^2 = (1/2) * mass(A) * (vA(after))^2 + (1/2) * mass(B) * (vB(after))^2

4. Now substitute the known values into the equations:
mass(A) = mass(B) = m (let's assume both balls have the same mass for simplicity)

mass(A) * vA(before) = mass(A) * vA(after) + mass(B) * vB(after)
m * 10 = m * vA(after) + m * vB(after) -> Equation (1)

(1/2) * m * (10)^2 = (1/2) * m * (vA(after))^2 + (1/2) * m * (vB(after))^2
50 = (1/2) * (vA(after))^2 + (1/2) * (vB(after))^2 -> Equation (2)

5. Solving the equations:
To solve for the velocities, we need to find vA(after) and vB(after).

Equation (1) gives us:
10 = vA(after) + vB(after) -> Equation (3)

Equation (2) can be simplified:
50 = (vA(after))^2 + (vB(after))^2

Since the answer choices are given in terms of vA(after) and vB(after), we can square the equation (3) and substitute it into the simplified Equation (2).

100 = (vA(after) + vB(after))^2
100 = (10)^2
100 = 100

Both sides of the equation are equal, which means Equation (2) is satisfied. This validates that the answer choices are available after the collision.

Comparing the answer choices, we can determine that the correct alternative is:

c) The velocity of ball A after the collision is 7.5 m/s

Therefore, the answer is option c)