1) See diagram in attachment
2) Equations of momentum:
x-axis: [tex]m_1 u_1 = m_1 v_1 cos \phi + m_2 v_2 cos \theta[/tex]
y-axis: [tex]0=m_1 v_1 sin \phi + m_2 v_2 sin \theta[/tex]
3) Final velocity: 0.84 m/s at [tex]-33.3^{\circ}[/tex]
Explanation:
1)
The diagram of the situation is in attachment.
The "before" situation is represented on the left of the y-axis: the 1st ball is moving to the left, with a mass of
[tex]m_1 = 0.500 kg[/tex]
and a velocity of
[tex]u_1 = 1.00 m/s[/tex]
along the positive x-axis.
The "after" situation is represented on the right of the y-axis: the 2nd ball is hit by the 1st ball. The mass of the 2nd ball is
[tex]m_2 = 0.500 kg[/tex]
and it travels with velocity
[tex]v_2 = 0.543 m/s[/tex]
in a direction
[tex]\theta=57.2^{\circ}[/tex]
above the positive x-axis. Instead, the 1st ball travels at velocity [tex]v_1[/tex] in a direction [tex]\phi[/tex].
2)
The momentum equations in both directions are the following:
- Along the x-axis, the initial momentum is just the momentum of the first ball, while the final momentum is the sum of the x-momenta of the two balls. Since the momentum along each direction must be conserved, we have:
[tex]m_1 u_1 = m_1 v_1 cos \phi + m_2 v_2 cos \theta[/tex]
- Along the y-axis, the initial momentum is zero, because there is no motion along this direction; while the final momentum is the sum of the y-momenta of the two balls:
[tex]0=m_1 v_1 sin \phi + m_2 v_2 sin \theta[/tex]
3)
We can now solve the problem. We can actually rewrite the equations of the momentum as follows:
[tex]p_1 = p_{1x} + p_{2x}\\0 = p_{1y} + p_{2y}[/tex]
So that we can find the components of the final momentum of ball 1:
[tex]p_{1x} = p-p_{2x} = (0.500)(1.00) - (0.500)(0.543)(cos 57.2^{\circ})=0.35 kg m/s[/tex]
[tex]p_{1y}= -p_{2y} = -(0.500)(0.543)(sin 57.2^{\circ})=-0.23 kg m/s[/tex]
From this, we can find the angle of the direction of the 1st ball:
[tex]\theta=tan^{-1} (\frac{p_{1y}}{p_{1x}})=tan^{-1}(\frac{-0.23}{0.35})=-33.3^{\circ}[/tex]
And now we can also find the speed: since the magnitude of the final momentum of ball 1 is
[tex]p_1' = \sqrt{p_{1x}^2+p_{1y}^2}=\sqrt{(0.35)^2+(-0.23)^2}=0.41 kg m/s[/tex]
And so the final speed is
[tex]v_1 = \frac{p_1'}{m_1}=\frac{0.42}{0.500}=0.84 m/s[/tex]
Learn more about momentum:
brainly.com/question/7973509
brainly.com/question/6573742
brainly.com/question/2370982
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