Irodov 1.150: Gravity Moves the Center Like One Particle

Problem Irodov 1.150

Two small spheres of masses $m_1$ and $m_2$ are joined by a weightless spring, free in space apart from the uniform gravitational field of the Earth (air drag neglected). At time $t=0$ the spheres are given initial velocities $\vect v_1$ and $\vect v_2$ and the system is released. Find, as functions of time: (a) the total momentum $\vect p(t)$ of the two-sphere system, and (b) the radius vector $\vect r_C(t)$ of its center of inertia, measured from the center’s position at $t=0$ .

The two spheres joined by the spring, launched with velocities v_1 and v_2 at t=0; thereafter the whole system falls freely under the uniform field g, the spring force acting only between the two spheres. — The two spheres joined by the spring, launched with velocities $\vect v_1$ and $\vect v_2$ at $t=0$ ; thereafter the whole system falls freely under the uniform field $\vect g$ , the spring force acting only between the two spheres.

Given

$m_1$
$m_2$
$\vect v_1$
$\vect v_2$
$\vect g$

Find

(a)

\vect p(t)

, (b)

\vect r_C(t)

Solution

Step 1The only external force is the weight of the whole system

The spring pulls the two spheres with equal and opposite forces, an internal action-reaction pair that cancels in the sum, so it never appears in the momentum theorem for the pair taken together; gravity is the sole external force, constant and equal to the total weight.

\frac{d\vect p}{dt} = (m_1+m_2)\vect g

Step 2Integrating the momentum theorem

Integrating from $t=0$ , where the momentum is fixed by the given launch velocities, to a general time $t$ gives $\vect p(t)$ .

\begin{align*}\int_{\vect p_0}^{\vect p(t)} d\vect p' &= \int_0^t (m_1+m_2)\vect g\,dt' \\ \vect p(t)-\vect p_0 &= (m_1+m_2)\vect g\,t\end{align*}

\begin{gathered}\vect p(t) = \vect p_0+(m_1+m_2)\vect g\,t,\\ \vect p_0=m_1\vect v_1+m_2\vect v_2\end{gathered}

(1)

Step 3The center of inertia moves like a single free particle

Dividing the momentum (1) by the total mass gives the center’s velocity, and integrating that from $t=0$ gives its displacement from the initial center.

\begin{align*}\vect V_C(t) &= \frac{\vect p(t)}{m_1+m_2} = \vect v_0+\vect g t, \qquad \vect v_0=\frac{m_1\vect v_1+m_2\vect v_2}{m_1+m_2} \\ \vect r_C(t) &= \int_0^t \vect V_C(t')\,dt' = \int_0^t \bigl(\vect v_0+\vect g t'\bigr)dt'\end{align*}

\vect r_C(t) = \vect v_0\,t+\dfrac{\vect g t^2}{2}

(2)

Check

Setting $\vect g=\vect 0$ recovers uninterrupted straight-line motion, $\vect p(t)=\vect p_0$ and $\vect r_C(t)=\vect v_0t$ , exactly the free two-body system with no external force. Dimensionally $[\vect g][t^2]=(\mathrm{m/s^2})(\mathrm s^2)=\mathrm m$ , matching $\vect r_C$ .

Remark

Neither result depends on the spring’s stiffness, natural length, or initial stretch at all: the internal force can only redistribute momentum and energy between the two spheres, never change the pair’s total momentum or the path of their common center. The two spheres individually trace complicated, generally non-parabolic paths as they oscillate against each other, yet their center of mass traces a perfect parabola, exactly as if all the mass sat at one point.