**Answer:**

Part A:

B) mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B

Part B:

C) 0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B

since vAy = 0 m/s

Part C:

θ′B = tan⁻¹(1.0699) = 46.94°

Part D:

v′B = 1.246 m/s

**Explanation:**

Given:

mA = 0.117 kg

vA = vAx = 2.80 m/s

mB = 0.135 kg

vB = 0 m/s

θ′A = 30.0°

v′A = 2.10 m/s

Part A: Taking the x axis to be the original direction of motion of ball A, the correct equation expressing the conservation of momentum for the components in the x direction is

B) mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B

Part B: Taking the x axis to be the original direction of motion of ball A, the correct equation expressing the conservation of momentum for the components in the y direction is

C) 0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B

since vAy = 0 m/s

Part C: Solving these equations for the angle, θ′B , of ball B after the collision and assuming that the collision is not elastic:

mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B

⇒ (0.117)(2.80) = (0.117)(2.10)Cos 30° + (0.135)*v′B*cosθ′B

⇒ v′B*cosθ′B = 0.8505 ⇒ v′B = 0.8505/cosθ′B

then

0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B

⇒ 0 = (0.117)(2.10)Sin 30° - (0.135)*v′B*sinθ′B

⇒ v′B*sinθ′B = 0.91 ⇒ v′B = 0.91/sinθ′B

if we apply

0.8505/cosθ′B = 0.91/sinθ′B

⇒ tanθ′B = 0.91/0.8505 = 1.0699

⇒ θ′B = tan⁻¹(1.0699) = 46.94°

Part D: Solving these equations for the speed, v′B , of ball B after the collision and assuming that the collision is not elastic:

if v′B = 0.91/sinθ′B

⇒ v′B = 0.91/sin 46.94°

⇒ v′B = 1.246 m/s