Answer :
Answer:
[tex]v_{2f}=2m/s[/tex] in the positive x-direction.
Explanation:
Conservation of momentum tells us that the momentum before the collision must be equal to the momentum after the collision: [tex]p_i=p_f[/tex].
Before the collision we have:
[tex]p_i=m_1v_{1i}+m_2v_{2i}=mv_{1i}+mv_{2i}=m(v_{1i}+v_{2i})[/tex]
And after the collision:
[tex]p_f=m_1v_{1f}+m_2v_{2f}=mv_{1f}+mv_{2f}=m(v_{1f}+v_{2f})[/tex]
Since they must be equal we have:
[tex]m(v_{1i}+v_{2i})=m(v_{1f}+v_{2f})[/tex]
Which is the same as:
[tex]v_{1i}+v_{2i}=v_{1f}+v_{2f}[/tex]
And can be written as:
[tex]v_{1i}-v_{1f}=v_{2f}-v_{2i}[/tex]
Since the collision is elastic, kinetic energy is conserved: [tex]K_i=K_f[/tex]
Before the collision we have:
[tex]K_i=\frac{m_1v_{1i}^2}{2}+\frac{m_2v_{2i}^2}{2}=\frac{mv_{1i}^2}{2}+\frac{mv_{2i}^2}{2}=\frac{m}{2}(v_{1i}^2+v_{2i}^2)[/tex]
And after the collision:
[tex]K_f=\frac{m_1v_{1f}^2}{2}+\frac{m_2v_{2f}^2}{2}=\frac{mv_{1f}^2}{2}+\frac{mv_{2f}^2}{2}=\frac{m}{2}(v_{1f}^2+v_{2f}^2)[/tex]
Since they must be equal we have:
[tex]\frac{m}{2}(v_{1i}^2+v_{2i}^2)=\frac{m}{2}(v_{1f}^2+v_{2f}^2)[/tex]
Which is the same as:
[tex]v_{1i}^2+v_{2i}^2=v_{1f}^2+v_{2f}^2[/tex]
And can be written as:
[tex]v_{1i}^2-v_{1f}^2=v_{2f}^2-v_{2i}^2[/tex]
Which is the same as:
[tex](v_{1i}-v_{1f})(v_{1i}+v_{1f})=(v_{2f}-v_{2i})(v_{2f}+v_{2i})[/tex]
But since we already proved that [tex]v_{1i}-v_{1f}=v_{2f}-v_{2i}[/tex], we can cancel those terms and we have that:
[tex]v_{1i}+v_{1f}=v_{2f}+v_{2i}[/tex]
So we have the system:
[tex]v_{1i}-v_{1f}=v_{2f}-v_{2i}[/tex]
[tex]v_{1i}+v_{1f}=v_{2f}+v_{2i}[/tex]
Adding both sides between them we obtain:
[tex]v_{1i}-v_{1f}+v_{1i}+v_{1f}=v_{2f}-v_{2i}+v_{2f}+v_{2i}[/tex]
Which is:
[tex]2v_{1i}=2v_{2f}[/tex]
Which means:
[tex]v_{2f}=v_{1i}=2m/s[/tex]
Where we used a positive velocity since ball 1 travels in the positive x-direction at the beginning.
The final velocity of the first ball is 1 m/s in negative x-direction and the final velocity of the second ball is 2 m/s in positive x-direction.
The given parameters;
- mass of the balls, = m
- initial speed of the first ball, = u₁ = 2 m/s
- initial speed of the second ball, = u₂ = 1 m/s
The final speed of the two balls is calculated by applying principles of linear momentum.
[tex]m_1 u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\2m -m = mv_1 + mv_2\\\\m = m(v_1+ v_2)\\\\1 = v_1 + v_2[/tex]
Apply one-dimensional velocity equation;
[tex]u_1 + v_1 = u_2 + v_2\\\\2 + v_1 = -1 + v_2\\\\v_1 + 3 = v_2[/tex]
Substitute the value of [tex]v_2[/tex] in the first equation;
[tex]1 = v_1 + v_2\\\\1 = v_1 + (v_1 + 3) \\\\1 = 2v_1 + 3\\\\-2 = 2v_1\\\\v_1 = -1 \ m/s[/tex]
The final velocity of the second ball is calculated as follows;
[tex]v_2 = 3 + v_1\\\\v_2 = 3 - 1\\\\v_2 = 2 \ m/s[/tex]
Thus, the final velocity of the first ball is 1 m/s in negative x-direction and the final velocity of the second ball is 2 m/s in positive x-direction.
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