$3.3.20.$ Two balls of equal mass $m$, connected by an undeformed spring of stiffness $k$, slide along a horizontal plane with a velocity $v$. The balls strike a vertical elastic wall. Describe the subsequent motion of the balls. Will a second collision with the wall occur?
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### Solution
At the initial moment, both balls and the undeformed spring are moving to the right with a velocity $v$.
When the right ball (let us call it ball 1) strikes the vertical elastic wall, a perfectly elastic collision occurs. Its velocity instantly reverses direction and becomes $-v$. The left ball (ball 2) at this moment continues to move to the right by inertia with a velocity $v$.
@@ -17,7 +13,7 @@Solution
Let us find the velocity of the center of mass (CM) of the system immediately after this first collision:
The center of mass of the system **comes to a halt**. The subsequent motion of the balls until the next contact with the wall will occur relative to the stationary center of mass.
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The center of mass of the system comes to a halt. The subsequent motion of the balls until the next contact with the wall will occur relative to the stationary center of mass.
Since the center of mass is at rest, and the balls are moving towards each other with velocities $v$, they will begin to compress the spring. Harmonic oscillations of the two masses on the spring will occur.
The period of oscillation of such a two-body system is determined by its reduced mass $\mu = \frac{m \cdot m}{m + m} = \frac{m}{2}$:
Ball 1 (located at the wall) starts moving to the left, compressing the spring. It will reach its leftmost point, and then, under the action of the expanding spring, it will begin to move back to the right, towards the wall.
After half a period of oscillation ($t = T/2$), the spring will completely go through the stages of compression and expansion, returning to its initial undeformed state:
$$t = \frac{T}{2} = \pi\sqrt{\frac{m}{2k}}$$
@@ -32,21 +28,17 @@Solution
At this moment, ball 1 will return to its initial coordinate — exactly at the wall.
Since the system is conservative (there is no energy dissipation) and the spring is undeformed again, the kinetic energy of the system is equal to the initial one. In the center of mass frame, each ball simply reversed the direction of its velocity.
Consequently, immediately before the second contact with the wall:
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* The velocity of the right ball 1 is directed to the right: $v_1 = v$.
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* The velocity of the left ball 2 is directed to the left: $v_2 = -v$.
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1. The velocity of the right ball 1 is directed to the right: $v_1 = v$.
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2. The velocity of the left ball 2 is directed to the left: $v_2 = -v$.
The right ball again undergoes a perfectly elastic collision with the wall. Its velocity $v_1 = v$ instantly changes to $v_1' = -v$ (to the left).
Let us look at the state of the system immediately after the second collision:
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* Velocity of ball 1: $-v$
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* Velocity of ball 2: $-v$
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* Deformation of the spring: $x = 0$
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1. Velocity of ball 1: $-v$
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2. Velocity of ball 2: $-v$
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3. Deformation of the spring: $x = 0$
Both balls now have strictly identical velocities in magnitude and direction. The relative motion between them has disappeared.
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#### Answer:
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Yes, a second collision will occur. After the second collision, the oscillations will cease, and the system will move to the left (away from the wall) as a single rigid body with a constant velocity $v$, while the spring will remain undeformed.
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#### Answer
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[Insert a concise answer or boxed result]
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Yes, a second collision will occur. After the second collision, the oscillations will cease, and the system will move to the left (away from the wall) as a single rigid body with a constant velocity $v$, while the spring will remain undeformed.
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### Statement
### Statement
$3.3.20.$ [Insert the problem statement]
### Solution
### Statement
$3.3.20.$ Two balls of equal mass $m$, connected by an undeformed spring of stiffness $k$, slide along a horizontal plane with a velocity $v$. The balls strike a vertical elastic wall. Describe the subsequent motion of the balls. Will a second collision with the wall occur?
$3.3.20.$ Two balls of equal mass $m$, connected by an undeformed spring of stiffness $k$, slide along a horizontal plane with a velocity $v$. The balls strike a vertical elastic wall. Describe the subsequent motion of the balls. Will a second collision with the wall occur?
### Solution
### Solution
At the initial moment, both balls and the undeformed spring are moving to the right with a velocity $v$.
At the initial moment, both balls and the undeformed spring are moving to the right with a velocity $v$.
When the right ball (let us call it ball 1) strikes the vertical elastic wall, a perfectly elastic collision occurs. Its velocity instantly reverses direction and becomes $-v$. The left ball (ball 2) at this moment continues to move to the right by inertia with a velocity $v$.
When the right ball (let us call it ball 1) strikes the vertical elastic wall, a perfectly elastic collision occurs. Its velocity instantly reverses direction and becomes $-v$. The left ball (ball 2) at this moment continues to move to the right by inertia with a velocity $v$.
@@ -17,7 +13,7 @@Solution
Let us find the velocity of the center of mass (CM) of the system immediately after this first collision:
Let us find the velocity of the center of mass (CM) of the system immediately after this first collision:
The center of mass of the system **comes to a halt**. The subsequent motion of the balls until the next contact with the wall will occur relative to the stationary center of mass.
The center of mass of the system comes to a halt. The subsequent motion of the balls until the next contact with the wall will occur relative to the stationary center of mass.
Since the center of mass is at rest, and the balls are moving towards each other with velocities $v$, they will begin to compress the spring. Harmonic oscillations of the two masses on the spring will occur.
Since the center of mass is at rest, and the balls are moving towards each other with velocities $v$, they will begin to compress the spring. Harmonic oscillations of the two masses on the spring will occur.
The period of oscillation of such a two-body system is determined by its reduced mass $\mu = \frac{m \cdot m}{m + m} = \frac{m}{2}$:
The period of oscillation of such a two-body system is determined by its reduced mass $\mu = \frac{m \cdot m}{m + m} = \frac{m}{2}$:
Ball 1 (located at the wall) starts moving to the left, compressing the spring. It will reach its leftmost point, and then, under the action of the expanding spring, it will begin to move back to the right, towards the wall.
Ball 1 (located at the wall) starts moving to the left, compressing the spring. It will reach its leftmost point, and then, under the action of the expanding spring, it will begin to move back to the right, towards the wall.
After half a period of oscillation ($t = T/2$), the spring will completely go through the stages of compression and expansion, returning to its initial undeformed state:
After half a period of oscillation ($t = T/2$), the spring will completely go through the stages of compression and expansion, returning to its initial undeformed state:
$$t = \frac{T}{2} = \pi\sqrt{\frac{m}{2k}}$$
$$t = \frac{T}{2} = \pi\sqrt{\frac{m}{2k}}$$
@@ -32,21 +28,17 @@Solution
At this moment, ball 1 will return to its initial coordinate — exactly at the wall.
At this moment, ball 1 will return to its initial coordinate — exactly at the wall.
Since the system is conservative (there is no energy dissipation) and the spring is undeformed again, the kinetic energy of the system is equal to the initial one. In the center of mass frame, each ball simply reversed the direction of its velocity.
Since the system is conservative (there is no energy dissipation) and the spring is undeformed again, the kinetic energy of the system is equal to the initial one. In the center of mass frame, each ball simply reversed the direction of its velocity.
Consequently, immediately before the second contact with the wall:
Consequently, immediately before the second contact with the wall:
* The velocity of the right ball 1 is directed to the right: $v_1 = v$.
1. The velocity of the right ball 1 is directed to the right: $v_1 = v$.
* The velocity of the left ball 2 is directed to the left: $v_2 = -v$.
2. The velocity of the left ball 2 is directed to the left: $v_2 = -v$.
The right ball again undergoes a perfectly elastic collision with the wall. Its velocity $v_1 = v$ instantly changes to $v_1' = -v$ (to the left).
The right ball again undergoes a perfectly elastic collision with the wall. Its velocity $v_1 = v$ instantly changes to $v_1' = -v$ (to the left).
Let us look at the state of the system immediately after the second collision:
Let us look at the state of the system immediately after the second collision:
* Velocity of ball 1: $-v$
1. Velocity of ball 1: $-v$
* Velocity of ball 2: $-v$
2. Velocity of ball 2: $-v$
* Deformation of the spring: $x = 0$
3. Deformation of the spring: $x = 0$
Both balls now have strictly identical velocities in magnitude and direction. The relative motion between them has disappeared.
Both balls now have strictly identical velocities in magnitude and direction. The relative motion between them has disappeared.
#### Answer:
Yes, a second collision will occur. After the second collision, the oscillations will cease, and the system will move to the left (away from the wall) as a single rigid body with a constant velocity $v$, while the spring will remain undeformed.
#### Answer
#### Answer
Yes, a second collision will occur. After the second collision, the oscillations will cease, and the system will move to the left (away from the wall) as a single rigid body with a constant velocity $v$, while the spring will remain undeformed.
