Respuesta :
Answer:
Golf ball ([tex]0.1\; {\rm kg}[/tex]): approximately [tex](-19.506)\; {\rm m\cdot s^{-1}}[/tex] (backwards).
Steel ball ([tex]8\; {\rm kg}[/tex]): approximately [tex]0.49383\; {\rm m\cdot s^{-1}}[/tex] (forward.)
Explanation:
Apply unit conversion and ensure that the unit of all mass are in kilograms: [tex]100\; {\rm g} = 0.1\; {\rm kg}[/tex].
In an elastic collision, both momentum [tex]p = m\, v[/tex] and kinetic energy [tex]\text{KE} = (1/2)\, m\, v^{2}[/tex] are conserved. Momentum of the two balls before the collision are:
- [tex]0.1\; {\rm kg} \times 20\; {\rm m\cdot s^{-1}} = 2\; {\rm kg \cdot m \cdot s^{-1}}[/tex] for the golf ball, and
- [tex]8\; {\rm kg} \times 0\; {\rm m\cdot s^{-1}} = 0\; {\rm kg \cdot m\cdot s^{-1}}[/tex] for the steel ball initially at rest.
Hence, the total momentum of the two balls was [tex]2\; {\rm kg \cdot m\cdot s^{-1}}[/tex] before the collision and (by conservation) will still be [tex]2\; {\rm kg \cdot m\cdot s^{-1}}\![/tex] after the collision.
Kinetic energy of the two balls before the collision are:
- [tex](1/2)\times 0.1\; {\rm kg} \times (20\; {\rm m\cdot s^{-1}})^{2} = 200\; {\rm kg \cdot m^{2} \cdot s^{-2}}[/tex] for the golf ball, and
- [tex](1/2) \times 8\; {\rm kg} \times (0\; {\rm m\cdot s^{-1}})^{2} = 0\; {\rm kg \cdot m\cdot s^{-1}}[/tex] for the steel ball initially at rest.
Thus, the total kinetic energy of the two balls will be [tex]200\; {\rm kg \cdot m^{2} \cdot s^{-2}}[/tex] before and after the collision.
Let [tex]m_{a}[/tex] and [tex]v_{a}[/tex] denote the mass and velocity of the golf ball after collision; [tex]m_{a} = 0.1\; {\rm kg}[/tex].
Let [tex]m_{b}[/tex] and [tex]v_{b}[/tex] denote the mass and velocity of the steel ball after collision; [tex]m_{b} = 8\; {\rm kg}[/tex].
Total momentum after the collision shall be [tex]2\; {\rm kg \cdot m\cdot s^{-1}}\![/tex]. Thus:
[tex]m_{a}\, v_{a} + m_{b}\, v_{b} = 2\; {\rm kg \cdot m\cdot s^{-1}}[/tex].
Total kinetic energy after the collision shall be [tex]200\; {\rm kg \cdot m^{2} \cdot s^{-2}}[/tex]. Thus:
[tex]\displaystyle \frac{1}{2} \, m_{a}\, v_{a} + \frac{1}{2}m_{b}\, v_{b} = 200\; {\rm kg \cdot m^{2}\cdot s^{-2}}[/tex].
Assume that the unit of [tex]v_{a}[/tex] and [tex]v_{b}[/tex] are both "meters per second" ([tex]{\rm m\cdot s^{-1}}[/tex].) Combine and solve this system of two equations and two variables:
[tex]\left\lbrace\begin{aligned}&0.1\, v_{a} + 8\; v_{b} = 2 \\ &\frac{1}{2}\, {v_{a}}^{2} + 4\, {v_{b}}^{2} = 200\end{aligned}\right.[/tex].
Rewrite the first equation to obtain [tex]v_{a} = 20 - 80\, v_{b}[/tex]. Substitute this equation into the second one to eliminate [tex]v_{a}[/tex]:
[tex]\displaystyle \frac{1}{2}\, (20 - 80\, v_{b})^{2} + 4\, v_{b}^{2} = 200[/tex].
Solve this equation for [tex]v_{b}[/tex]:
[tex]324\, {v_{b}}^{2} - 160\, v_{b} = 0[/tex].
There are two solutions to this quadratic equation: [tex](40 / 81)[/tex] and [tex]0[/tex]. Note that the velocity of the steel ball must be non-zero right after the collision. Hence, [tex]v_{b} \ne 0[/tex]. Therefore, the only possible value for [tex]v_{b}[/tex] will be [tex](40 / 81)\![/tex], which is approximately [tex]0.49383\; {\rm m\cdot s^{-1}}[/tex].
Substitute [tex]v_{b} = (40 / 81)[/tex] back into the first equation of the system and solve for [tex]v_{a}[/tex]: [tex]v_{a} = 20 - 80\, v_{b} \approx (-19.506)\; {\rm m\cdot s^{-1}}[/tex]. Note that the velocity of the golf ball [tex]v_{a}\![/tex] is negative since the golf ball is travelling backwards, opposite to its initial direction of motion.
In other words, the velocity right after collision will be approximately [tex](-19.506)\; {\rm m\cdot s^{-1}}[/tex] (backwards) for the golf ball and approximately [tex]0.49383\; {\rm m\cdot s^{-1}}[/tex] (forwards) for the steel ball.
