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

Answer:

V=1.309

β= -41.997

Explanation:

Law Newton's conservation motion

Axis x

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

Axis y

$m_{1}*v_{y1}+m_{2}*v_{y2}=m_{1}*v_{fy1}+m_{2}*v_{fy2}$

$v_{y1} =0\\v_{y2} =0$

$0=m_{1}*v_{fy1} +m_{2}*v_{fy2} \\v_{fy2}=-\frac{m_{1}*v_{fy1} }{m_{2}}\\ v_{fy2}=-\frac{0.125kg*2.10*sen(30)\frac{m}{s}}{0.14kg}\\v_{fy2}= -0.937\frac{m}{s}$

So the velocity $v_{f2}$

$v_{f2}=\sqrt{v_{fx2}^{2} +v_{fy2}^{2} } \\v_{f2}=\sqrt{0.876^{2} +0.983^{2} } \\v_{f2}=1.309\frac{m}{s}$

The angle can be find using both velocity factors

$\alpha =tanx^{-1}*\frac{v_{fx2}}{v_{fy2}} \\\alpha =tanx^{-1}*\frac{0.876}{-0.973}\\ \alpha =tanx^{-1}*\\ \alpha =-41.997$

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