A small ball of mass m, attached to a spring with stiffness coefficient $k$, makes harmonic oscillations in the direction of the spring axis with amplitude $A$. When the spring was compressed, a second spring pendulum of the same mass is placed in the path of the ball, both springs having a common axis and the equilibrium positions of the balls coinciding. Describe the further motion of the system in the following cases:
(a) The shock is central and perfectly elastic;
b) the impact is central and absolutely inelastic (i.e. the balls "stick together").
At the initial moment the second ball rests in the equilibrium position.
Decision:
a) Since the impact is absolutely elastic and the masses of the balls are equal, the first ball will stop at the equilibrium position and the second ball will start moving with the same velocity as the first one was moving before the impact. Obviously, the amplitude of oscillation of the second pendulum is also equal to $A$. The second ball will move a distance $A$ from the equilibrium position and return back, colliding in the equilibrium position with the first ball. As a result of this collision, it will remain in place at the equilibrium position, and the first ball will come in motion, first moving a distance $A$ from the equilibrium position, then returning to the first ball. from the equilibrium position, then returning to it and hitting again the second ball, etc.
b) At the initial moment the second ball rests in the position of equilibrium. The first ball will approach the equilibrium position with velocity $v_{0}=A \sqrt{k/m}$, calculated from the law of conservation of energy in harmonic oscillation. After a completely inelastic impact, the balls stick together and form a body of mass $2m$, whose velocity immediately after the impact is $v_{0}/2$, which follows from the law of conservation of momentum.
When this body is displaced by distance x from the equilibrium position, the force $F = -2 kx$ acts on it. Therefore, the period of oscillation $T = 2 \pi \sqrt{2m/(2k)}$ will remain the same as that of a single spring pendulum. The amplitude of the oscillations of the system of stuck balls can be found from the law of conservation of energy:
$A^{\prime}=\frac{v_{0}}{2}\sqrt{\frac{m}{k}}=\frac{A}{2}$.
Note. When the balls hit, half of the energy of the first ball is transferred to heat:
$Q=m \frac{v_{0}^{2}}{2} -2m \frac{(v_{0}/2)^{2}}{2} = m \frac{v_{0}^{2}}{4}$.
(a) The shock is central and perfectly elastic;
b) the impact is central and absolutely inelastic (i.e. the balls "stick together").
At the initial moment the second ball rests in the equilibrium position.
Decision:
a) Since the impact is absolutely elastic and the masses of the balls are equal, the first ball will stop at the equilibrium position and the second ball will start moving with the same velocity as the first one was moving before the impact. Obviously, the amplitude of oscillation of the second pendulum is also equal to $A$. The second ball will move a distance $A$ from the equilibrium position and return back, colliding in the equilibrium position with the first ball. As a result of this collision, it will remain in place at the equilibrium position, and the first ball will come in motion, first moving a distance $A$ from the equilibrium position, then returning to the first ball. from the equilibrium position, then returning to it and hitting again the second ball, etc.
b) At the initial moment the second ball rests in the position of equilibrium. The first ball will approach the equilibrium position with velocity $v_{0}=A \sqrt{k/m}$, calculated from the law of conservation of energy in harmonic oscillation. After a completely inelastic impact, the balls stick together and form a body of mass $2m$, whose velocity immediately after the impact is $v_{0}/2$, which follows from the law of conservation of momentum.
When this body is displaced by distance x from the equilibrium position, the force $F = -2 kx$ acts on it. Therefore, the period of oscillation $T = 2 \pi \sqrt{2m/(2k)}$ will remain the same as that of a single spring pendulum. The amplitude of the oscillations of the system of stuck balls can be found from the law of conservation of energy:
$A^{\prime}=\frac{v_{0}}{2}\sqrt{\frac{m}{k}}=\frac{A}{2}$.
Note. When the balls hit, half of the energy of the first ball is transferred to heat:
$Q=m \frac{v_{0}^{2}}{2} -2m \frac{(v_{0}/2)^{2}}{2} = m \frac{v_{0}^{2}}{4}$.