Answer:
a) [tex](v_1)=0.3989m/s[/tex]
b) [tex]\theta_1=80.5 \textdegree[/tex]
c) [tex]K.E=0.036J[/tex]
Explanation:
From the question we are told that:
Initial speed of 1st ball [tex]u_{1}=0 m/s[/tex]
Mass of 1st ball [tex]m_1=0.585kg[/tex]
Mass of 2nd ball [tex]m_2=0.420kg[/tex]
Initial speed of 2nd ball [tex]u_{2}=0.270 m/s[/tex]
Final speed of 2nd ball [tex]v_{2}=0.220 m/s[/tex]
Angle of collision [tex]\angle=36.9 \textdegree[/tex]
a)
Generally the equation for law of conservation is mathematically given by
[tex]m_1u_1+m_2u_2=m_1v_1^2+m_2v_2^2[/tex]
The final velocity [tex]v_1[/tex] is given as
[tex]0+(0.420)(0.270)=(0.585)(v_1)^2+(0.420)(0.220)^2[/tex]
[tex](v_1)^2=\frac{(0.420)(0.270)-(0.420)(0.220)^2}{0.585}[/tex]
[tex](v_1)^2=0.1591[/tex]
[tex](v_1)=0.3989m/s[/tex]
b)
Generally the equation for law of conservation is mathematically given by
[tex]m_1u_1+m_2u_2=m_1v_1cos\theta_1+m_2\theta_2[/tex]
[tex]0+(0.420)(0.270)=(0.585)(1.511)cos\theta_1+(0.420)(0.220)cos36 \textdegree[/tex]
[tex]cos\theta_1= \frac{(0.420)(0.270)-(0.420)(0.220)cos36 \textdegree}{(0.585)(0.3989)}[/tex]
[tex]cos\theta_1=0.1656[/tex]
[tex]\theta_1=80.5 \textdegree[/tex]
c)
Generally the equation for kinetic energy is mathematically given by
[tex]K.E=\frac{1}{2} mv^2[/tex]
1st Ball
[tex]K.E=\frac{1}{2} (0.585)(0.3989)^2[/tex]
[tex]K.E=0.0465J[/tex]
2nd ball
[tex]K.E=\frac{1}{2} (0.420)(0.220)^2[/tex]
[tex]K.E=0.101J[/tex]
Therefore the change in the total kinetic energy of the two balls as a result of the collision is
[tex]0.101-0.0465[/tex]
[tex]K.E=0.036J[/tex]
