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: 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

batolisis batolisis · Answer 2 · 2021-11-17T14:10:43+01:00

A) The correct equation expressing the conservation of momentum in x direction ; ( B ) mAvA = mAv′Acosθ′A + mBv′Bcosθ′B

B) The correct equation expressing the conservation of momentum in y- direction ; ( C ) 0 = mAv′Asinθ′A − mBv′Bsinθ′B

C) The angle θ′B = 46.94°

D) The speed v′B = 1.246 m/s

Given data :

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

A) Taking X- axis as the original direction of ball in motion

The correct equation expressing the conservation of momentum in x direction along the x axis will be

mAvA = mAv′Acosθ′A + mBv′Bcosθ′B

Part B : correct equation expressing the conservation of momentum in y- direction should be expressed as

0 = mAv′Asinθ′A − mBv′Bsinθ′B ( given that vAy = 0 )

Part C : Determine The angle θ′B

we will make use of these equations below :

mAvA = mAv′Acosθ′A + mBv′Bcosθ′B ----- ( 1 )

Input the given data into the equation ( 1 )

Equation ( 1 ) becomes

v′Bcosθ′B = 0.8505

∴ v′B = [ 0.8505 / cosθ′B ] ----- ( i )

Next ;

0 = mAv′Asinθ′A − mBv′Bsinθ′B ----- ( 2 )

Input the given data into equation ( 2 )

Equation ( 2 ) beomes

v′Bsinθ′B = 0.91

∴ v′B = [ 0.91 / sinθ′B ] ------ ( ii )

Equating equations ( i ) and ( ii )

[ 0.8505 / cosθ′B ] = [ 0.91 / sinθ′B ]

∴ θ′B = tan⁻¹(1.0699) = 46.94°

Part D ; Determine the value of v′B

Given that v′B = [ 0.91 / sinθ′B ]

where : θ′B = 46.94°

∴ v′B = [ 0.91 /sin 46.94° ] = 1.246 m/s.

Hence we can conclude that the correct equations and answers to your questions are as listed above.

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