Chapter 13, Problem 60AP

Two spheres having masses M and 2M and radii R and 3R, respectively, are simultaneously released from rest when the distance between their centers is 12R. Assume the two spheres interact only with each other and we wish to find the speeds with which they collide. (a) What two isolated system models are appropriate for this system? (b) Write an equation from one of the models and solve it for ${\stackrel{\to }{v}}_1$, the velocity of the sphere of mass M at any time after release in terms of ${\stackrel{\to }{v}}_2$, the velocity of 2M. (c) Write an equation from the other model and solve it for speed v₁ in terms of speed v₂ when the spheres collide. (d) Combine the two equations to find the two speeds v₁ and v₂ when the spheres collide.

To determine

To determine: The two isolated system models that are appropriate for this system.

Answer to Problem 60AP

Answer: Conservation of energy and conservation of momentum are the two models that are appropriate for this system.

Explanation of Solution

Explanation:

Given information: The masses of the two spheres are $M$ and $2M$ respectively, the radius of spheres are $R$ and $3R$ respectively and the distance between their centers is $12R$.

The two isolated system models that are appropriate for this system are conservation of energy and conservation of momentum. In these situations, the energy and momentum is conserved before and after the collision of the bodies. This is also called as the perfect elastic collision.

Conclusion:

Therefore, the conservation of energy and conservation of momentum are the two models that are appropriate for this system.

To determine

To determine: The equation from one of the model and estimate the speed $v_1$.

Answer to Problem 60AP

Answer: The equation from conservation of momentum is $M{v}_1+2M{v}_2=0$ and the expression for speed is $v_1=-2v_2$.

Explanation of Solution

Explanation:

Given information: The masses of the two spheres are $M$ and $2M$ respectively, the radius of spheres are $R$ and $3R$ respectively and the distance between their centers is $12R$.

By the conservation of momentum for elastic collision,

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

$m_1$ is the mass of first sphere.

$u_1$ is the initial velocity of the first sphere.

$m_2$ is the mass of second sphere.

$u_2$ is the initial velocity of second sphere.

$v_1$ is the initial velocity of first sphere.

$v_2$ is the final velocity of second sphere.

Substitute $M$ for $m_1$, $2M$ for $m_2$, $0$ for $u_1$, $0$ for $u_2$ in above expression.

$\begin{array}{c}M\times 0+2M\times 0=M{v}_1+2M{v}_2\\ M{v}_1+2M{v}_2=0\\ v_1=-2v_2\end{array}$ (I)

Conclusion:

Therefore, the equation from conservation of momentum is $M{v}_1+2M{v}_2=0$ and the expression for speed is $v_1=-2v_2$.

To determine

To determine: The equation from one of the other model and estimate the speed $v_1$.

Answer to Problem 60AP

Answer: The equation from conservation of energy is $\frac{v_1{}^2}{2}+v_2{}^2=\frac{GM}{3R}$ and the expression for the speed is $v_1=\sqrt{2\left(\frac{GM}{3R}-v_2{}^2\right)}$.

Explanation of Solution

Explanation:

Given information: The masses of the two spheres are $M$ and $2M$ respectively, the radius of spheres are $R$ and $3R$ respectively and the distance between their centers is $12R$.

By the conservation of energy is,

$K{E}_1+P{E}_1=K{E}_2+P{E}_2$ (II)

$K{E}_1$ is the kinetic energy of the spheres before collide.

$K{E}_2$ is the final kinetic energy of the spheres after collide.

$P{E}_1$ is the potential energy of the spheres before collide.

$P{E}_2$ is final potential energy of the spheres after collide.

Formula to calculate the kinetic energy of the spheres after collide is,

$K{E}_2=\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{v}_2{}^2$

Formula to calculate the potential energy of the spheres before collide is,

$P{E}_1=\frac{-G{m}_1{m}_2}{R_1}$

$R_1$ is the initial distance between the center of first sphere and second sphere.

$G$ is the universal gravitational constant.

Formula to calculate the potential energy of the spheres after collide is,

$P{E}_2=\frac{-G{m}_1{m}_2}{R_2}$

$R_2$ is the final distance between the center of first sphere and second sphere.

Substitute $\frac{-G{m}_1{m}_2}{R_1}$ for $P{E}_1$, $\frac{-G{m}_1{m}_2}{R_2}$ for $P{E}_2$, $0$ for $K{E}_1$, and $\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{v}_2{}^2$ for $K{E}_2$ in equation (II).

$0+\left(\frac{-G{m}_1{m}_2}{R_1}\right)=\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{v}_2{}^2+\left(\frac{-G{m}_1{m}_2}{R_2}\right)$

Substitute $M$ for $m_1$, $2M$ for $m_2$, $0$ for $u_1$, $0$ for $u_2$, $12R$ for $R_1$ and $4R$ for $R_2$ in above expression.

$\begin{array}{c}0+\left(\frac{-G\times M\times 2M}{12R}\right)=\frac{1}{2}\times M\times v_1{}^2+\frac{1}{2}\times 2M\times v_2{}^2+\left(\frac{-G\times M\times 2M}{4R}\right)\\ \frac{1}{2}\times v_1{}^2+\frac{1}{2}\times 2\times v_2{}^2=\left(\frac{G\times 2M}{4R}\right)-\left(\frac{G\times 2M}{12R}\right)\\ \frac{v_1{}^2}{2}+v_2{}^2=2GM\left(\frac{1}{4R}-\frac{1}{12R}\right)\\ v_1=\sqrt{2\left(\frac{GM}{3R}-v_2{}^2\right)}\end{array}$ (III)

Conclusion:

Therefore, the equation from one of the other model is $\frac{v_1{}^2}{2}+v_2{}^2=\frac{GM}{3R}$ and the speed $v_1$ is $v_1=\sqrt{2\left(\frac{GM}{3R}-v_2{}^2\right)}$.

To determine

To determine: The values of speeds $v_1$ and $v_2$.

Answer to Problem 60AP

Answer: The value of speed $v_1$ is $-2\sqrt{\frac{GM}{9R}}$ and value of $v_2$ is $\sqrt{\frac{GM}{9R}}$.

Explanation of Solution

Explanation:

Given information: The masses of the two spheres are $M$ and $2M$ respectively, the radius of spheres are $R$ and $3R$ respectively and the distance between their centers is $12R$.

From equation (III) is,

$v_1=\sqrt{2\left(\frac{GM}{3R}-v_2{}^2\right)}$

Substitute $-2v_2$ for $v_1$ in above expression.

$\begin{array}{c}\left(-2v_2\right)=\sqrt{2\left(\frac{GM}{3R}-v_2{}^2\right)}\\ {\left(-2v_2\right)}^2=2\left(\frac{GM}{3R}-v_2{}^2\right)\\ 6v_2{}^2=\frac{2GM}{3R}\\ v_2=\sqrt{\frac{GM}{9R}}\end{array}$

Substitute $\sqrt{\frac{GM}{9R}}$ for $v_2$ in equation (I).

$v_1=-2\sqrt{\frac{GM}{9R}}$

Conclusion:

Therefore, the value of speeds $v_1$ is $-2\sqrt{\frac{GM}{9R}}$ and value of $v_2$ is $\sqrt{\frac{GM}{9R}}$.