Billiard ball A of mass mA = 0.115 kg moving with speed vA = 2.80 m/s strikes ball B , initially at rest, of mass mB = 0.144 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.

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.

0=mAv′Asinθ′A−mBv′Bsinθ′B
mAvA=mAv′Acosθ′A+mBv′Bcosθ′B
mAvA=mAv′Acosθ′A−mBv′Bsinθ′B
0=(mAvA+mBv′B)sinθ′B

Part 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.

mAvA=mAv′Acosθ′A−mBv′Bsinθ′B
0=(mAvA+mBv′B)sinθ′B
0=mAv′Asinθ′A−mBv′Bsinθ′B
mAvA=mAv′Acosθ′A+mBv′Bcosθ′B

Part 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.

Respuesta :

Answer:

Part a)

[tex]m_Av_A = m_Av_A' cos\theta_A' + m_Bv_B'cos\theta_B'[/tex]

Part b)

[tex]0 = m_Av_A' sin\theta_A' - m_Bv_B'sin\theta_B'[/tex]

Part c)

[tex]\theta_B' = 46.9 degree[/tex]

Part d)

[tex]v_B' = 1.15 m/s[/tex]

Explanation:

Part a)

As we know that momentum is conserved as there is no external force on the system of two balls

So here we will have

Along X direction we have

[tex]m_Av_A = m_Av_A' cos\theta_A' + m_Bv_B'cos\theta_B'[/tex]

Part b)

Now momentum conservation along Y direction is given as

[tex]0 = m_Av_A' sin\theta_A' - m_Bv_B'sin\theta_B'[/tex]

Part c)

Now we have

[tex]0.115(2.80) = 0.115(2.10) cos30 + 0.144v_B'cos\theta_B'[/tex]

[tex]v_B'cos\theta_B' = 0.784[/tex]

[tex]0 = 0.115(2.10)sin30 - 0.144v_B'sin\theta_B'[/tex]

[tex]v_B'sin\theta_B' = 0.839[/tex]

now from above two equations we have

[tex]\theta_B' = 46.9 degree[/tex]

Part d)

Now from any of the above equation we have

[tex]v_B' = 1.15 m/s[/tex]