Question

Billiard ball A of mass mA = 0.125 kg moving with speed vA = 2.80 m/s strikes ball B, initially at rest, of mass mB = 0.140 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 CSolve these equations for the angle, θ′B, of ball B after the collision. Do not assume the collision is elastic.Express your answer to three significant figures and include the appropriate unitsPart DSolve these equations for the speed, v′B, of ball B after the collision. Do not assume the collision is elastic.Express your answer to three significant figures and include the appropriate units

Answer 1

V=1.309

β= -41.997

Step-by-step explanation:

Law Newton's conservation motion

Axis x

`m_(1)*v_(x1)+m_(2)*v_(x2)=m_(1)*v_(fx1)+m_(2)*v_(fx2)\\v_(x1)=2.8(m)/(s)\\v_(x2)=0 (m)/(s)\\m_(1)=0.125kg(m)/(s)\\m_(2)=0.140kg(m)/(s)\\0.125kg*2.8(m)/(s)+0.14kg*0=0.125kg*2.10(m)/(s)*cos(30) +0.14kg*v_(fx2)\\0.35 (kg*m)/(s) =0.125kg*1.81(m)/(s)+0.14kg*v_(fx2)\\v_(fx2)=(0.12(kg*m)/(s) )/(0.14kg) \\v_(fx2)=0.876 (m)/(s)`

Axis y

`m_(1)*v_(y1)+m_(2)*v_(y2)=m_(1)*v_(fy1)+m_(2)*v_(fy2)`

`v_(y1) =0\\v_(y2) =0`

`0=m_(1)*v_(fy1) +m_(2)*v_(fy2) \\v_(fy2)=-(m_(1)*v_(fy1) )/(m_(2))\\ v_(fy2)=-(0.125kg*2.10*sen(30)(m)/(s))/(0.14kg)\\v_(fy2)= -0.937(m)/(s)`

So the velocity
`v_(f2)`

`v_(f2)=\sqrt{v_(fx2)^(2) +v_(fy2)^(2) } \\v_(f2)=\sqrt{0.876^(2) +0.983^(2) } \\v_(f2)=1.309(m)/(s)`

The angle can be find using both velocity factors

`\alpha =tanx^(-1)*(v_(fx2))/(v_(fy2)) \\\alpha =tanx^(-1)*(0.876)/(-0.973)\\ \alpha =tanx^(-1)*\\ \alpha =-41.997`

Check:

`m_(1)*v_(x1)+m_(2)*v_(x2)=m_(1)*v_(fx1)+m_(2)*v_(fx2)`

`0.125*2.80=0.125*2.1*cos(30)+0.14*1.03*cos(-41.997)\\0.35=0.227+0.107\\`

0.35≅0.3489