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Two blocks of masses \(1.0 \mathrm{~kg}\) and \(3.0 \mathrm{~kg}\) are connected by a spring and rest on a frictionless surface. They are given velocities toward each other such that the \(1.0 \mathrm{~kg}\) block travels initially at \(1.7 \mathrm{~m} / \mathrm{s}\) toward the center of mass, which remains at rest. What is the initial speed of the other block?

Short Answer

Expert verified
The initial speed of the 3 kg block is 0.5667 m/s.

Step by step solution

01

Analyze the System

The system consists of two blocks, one with mass \(1.0\, \text{kg}\) and the other with mass \(3.0\, \text{kg}\). They are connected by a spring and move on a frictionless surface. To conserve the center of mass, we know the total momentum of the system must be zero because the center of mass is at rest.
02

Apply Conservation of Momentum

According to the conservation of momentum, the initial momentum of the system is zero. Thus, the momentum of the \(1.0 \text{ kg}\) block should be equal and opposite to that of the \(3.0 \text{ kg}\) block. This can be expressed as:\[ m_1 v_1 + m_2 v_2 = 0 \]where \(m_1 = 1.0\, \text{kg}\), \(v_1 = -1.7\, \text{m/s}\) (negative since it moves toward the center), \(m_2 = 3.0\, \text{kg}\), and \(v_2\) is the speed we need to find.
03

Solve for the Speed of the Second Block

Substitute the known values into the momentum equation:\[ (1.0\, \text{kg})(-1.7\, \text{m/s}) + (3.0\, \text{kg})(v_2) = 0 \]\[ -1.7 \text{ kg m/s} + 3.0\, v_2 = 0 \]Rearranging gives:\[ 3.0\, v_2 = 1.7 \]\[ v_2 = \frac{1.7}{3.0} \]\[ v_2 = 0.5667\, \text{m/s} \]
04

Conclusion

The initial speed of the \(3.0 \text{ kg}\) block must be \(0.5667\, \text{m/s}\) in the opposite direction to maintain a stationary center of mass.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frictionless Surface
A frictionless surface is an idealized concept often used in physics to simplify problems. On such a surface, objects can move without any resistance, meaning no energy is lost to friction. This allows us to focus on other forces at play, like those from a spring or gravity. In the context of the exercise, the frictionless surface ensures that the only variables affecting the blocks are their masses and the velocity with which they are set into motion. This is crucial for accurately applying the conservation of momentum because it eliminates possible external forces that could interfere with the movement of the objects.
Additionally, without friction, the total momentum of a system is conserved as long as no external forces act on it. That's why in the initial condition, while the center of mass remains at rest, the blocks must move towards each other with equal and opposite momentum. This balance ensures that any forward motion by one block is countered by the backward motion of the other, maintaining zero overall momentum.
Center of Mass
The center of mass of a system is a crucial concept in understanding motion and momentum conservation. It is the point where the total mass of a system can be considered to be concentrated. For the given scenario, even though the individual blocks move, the center of mass remains at rest. This tells us that the overall distribution of mass doesn't change in the system's center.
  • Because the center of mass is not moving, the system's total momentum must be zero.
  • Each block's velocity is calculated relative to this center mass.
  • By observing the balance of opposing velocities, we can determine unknown values like the speed of the second block.
When we apply the principle of conservation of momentum, this stationary center of mass provides a reliable point of reference. It implies that changes or movements within the system can happen, but they must maintain the overall balance by having opposing movements that cancel each other out.
Spring Dynamics
Springs are fascinating components that follow Hooke’s Law, which states that the force exerted by a spring is proportional to its displacement, usually expressed as \( F = -kx \) where \( k \) is the spring constant and \( x \) is the displacement from its equilibrium. Yet, in the context of this problem, the spring acts chiefly as a connecting element that does not directly impact the calculation of initial velocities.
Even though we don't calculate the forces in the spring dynamics, understanding its role is vital:
  • It connects the two blocks, meaning their motions are interconnected.
  • The spring will extend or compress depending on the velocities of the blocks, but this doesn't affect the initial conditions because the focus is on initial velocities.
  • Knowing that the spring will not influence the center of mass or the overall momentum initially helps simplify the problem.
In examples like this, it's essential to simplify by considering only the most relevant forces and dynamics, ensuring we’re applying the appropriate laws of motion while ignoring elements that would otherwise complicate the picture unnecessarily.

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Most popular questions from this chapter

A \(1400 \mathrm{~kg}\) car moving at \(5.3 \mathrm{~m} / \mathrm{s}\) is initially traveling north along the positive direction of a \(y\) axis. After completing a \(90^{\circ}\) right-hand turn in \(4.6 \mathrm{~s}\), the inattentive operator drives into a tree, which stops the car in \(350 \mathrm{~ms}\). In unit-vector notation, what is the impulse on the car (a) due to the turn and (b) due to the collision? What is the magnitude of the average force that acts on the car (c) during the turn and (d) during the collision? (e) What is the direction of the average force during the turn?

A body of mass \(2.0 \mathrm{~kg}\) makes an clastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed of the \(2.0 \mathrm{~kg}\) body was \(4.0 \mathrm{~m} / \mathrm{s} ?\)

an \(80 \mathrm{~kg}\) man is on a ladder hanging from a balloon that has a total mass of \(320 \mathrm{~kg}\) (including the basket passenger). The balloon is initially stationary relative to the ground. If the man on the ladder begins to climb at \(2.5 \mathrm{~m} / \mathrm{s}\) relative to the ladder, (a) in what direction and (b) at what speed does the balloon move? (c) If the man then stops climbing, what is the speed of the balloon?

After a completely inelastic collision, two objects of the same mass and same initial speed move away together at half their initial speed. Find the angle between the initial velocities of the objects.

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at \(3.50 \mathrm{~m} / \mathrm{s}\) along a line making an angle of \(22.0^{\circ}\) with the cue ball's original direction of motion, and the second ball has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball. (c) Is kinetic energy (of the centers of mass, don't consider the rotation) conserved?

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