Billiard ball a of mass ma = 0.117 kg moving with speed va = 2.80 m/s strikes ball b, initially at rest, of mass mb = 0.135 kg . as a result of the collision, ball a is deflected off at an angle of θ′a = 30.0∘ with a speed v′a = 2.10 m/s, and ball b moves with a speed v′b at an angle of θ′b to original direction of motion of ball a. part a taking the x axis to be the original direction of motion of ball a , choose the correct equation expressing the conservation of momentum for the components in the x direction. a. 0=mav′asinθ′a−mbv′bsinθ′bb. mava=mav′acosθ′a+mbv′bcosθ′bc. mava=mav′acosθ′a−mbv′bsinθ′bd. 0=(mava+mbv′b)sinθ′bpart b taking the x axis to be the original direction of motion of ball a , choose the correct equation expressing the conservation of momentum for the components in the y direction. a. mava=mav′acosθ′a−mbv′bsinθ′bb. 0=(mava+mbv′b)sinθ′bc. 0=mav′asinθ′a−mbv′bsinθ′bd. mava=mav′acosθ′a+mbv′bcosθ′bpart c solve these equations for the angle, θ′b , of ball b after the collision. do not assume the collision is elastic. part d solve these equations for the speed, v′b , of ball b after the collision. do not assume the collision is elastic.

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

answer would be d.) i hope this

explanation:

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