Answer:
The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.
Explanation:
Given that,
Mass of object = 5 kg
Speed = 3 m/s
Mass of stationary object = 3 kg
Moving object deflected = 30°
Stationary object deflected = 31°
We need to calculate the velocity of each ball after collision
Using conservation of momentum
Along x-axis
[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta[/tex]
Put the value into the fomrula
[tex]5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45[/tex]
[tex]15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}[/tex]
[tex]15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}[/tex]....(I)
Along y -axis
[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta[/tex]
Put the value into the formula
[tex]0+0=5\times v_{1}\sin30-3\times v_{2}\sin45[/tex]
[tex]\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0[/tex]...(II)
From equation (I) and (II)
[tex]v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}[/tex]
[tex]v_{1}=2.19\ m/s[/tex]
Put the value of v₁ in equation (I)
[tex]\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0[/tex]
[tex]v_{2}=\dfrac{5.475\times\sqrt{2}}{3}[/tex]
[tex]v_{2}=2.58\ m/s[/tex]
Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.
