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Chapter 7: Problem 24

Two blocks of masses \(1.0 \mathrm{~kg}\) and \(3.0 \mathrm{~kg}\) on a frictionless surface are connected by a stretched spring and initially are held at rest. Then the two blocks are simultaneously released from rest. Shortly after the spring starts contracting we find that the \(1.0\) kg block is traveling toward the other at \(1.7 \mathrm{~m} / \mathrm{s}\). What is the velocity of the other block at that moment?

Short Answer

Expert verified
The velocity of the 3.0 kg block is -0.5667 m/s.

Step by step solution

01

Identify the Given Information

We have two blocks connected by a spring on a frictionless surface. One block has a mass of 1.0 kg and is moving towards the other block at a velocity of 1.7 m/s. The other block has a mass of 3.0 kg, and we need to find its velocity.
02

Apply Conservation of Momentum

Since no external forces act on the system of the two blocks and the spring, the total linear momentum of the system is conserved. Therefore, the initial momentum must equal the final momentum. Initially, both blocks are at rest, so the initial total momentum is zero.
03

Write the Momentum Conservation Equation

The total momentum before and after the blocks are released must be equal. Thus, \[ (m_1 \times v_{1i} + m_2 \times v_{2i}) = (m_1 \times v_{1f} + m_2 \times v_{2f}) \]Where: \( m_1 = 1.0 \text{ kg} \)\( m_2 = 3.0 \text{ kg} \)v_{1i} and v_{2i} are the initial velocities which are zero.v_{1f} is 1.7 m/s (the velocity of the 1 kg block)v_{2f} is the velocity of the 3 kg block which we need to determine.
04

Solve the Conservation Equation

Since the initial momentum is zero (both blocks are at rest initially), we have:\[ 0 = (1.0 \text{ kg} \times 1.7 \text{ m/s}) + (3.0 \text{ kg} \times v_{2f}) \] Which simplifies to:\[ 0 = 1.7 \text{ kg} \text{ m/s} + 3.0 \text{ kg} \times v_{2f} \] \[ -1.7 \text{ kg} \text{ m/s} = 3.0 \text{ kg} \times v_{2f} \] Dividing both sides by 3.0 kg:\[ v_{2f} = - \frac{1.7 \text{ kg} \text{ m/s} }{ 3.0 \text{ kg}} \] \[ v_{2f} = -0.5667 \text{ m/s} \]
05

Interpret the Result

The negative sign indicates that the 3.0 kg block is moving in the opposite direction to the 1.0 kg block.

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics that describes the quantity of motion an object has. It is a product of an object's mass and velocity, expressed in the formula:
\ p = m \times v \
Where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. Momentum is a vector quantity, meaning it has both magnitude and direction. In closed systems, the total momentum remains constant if no external forces act. This principle, known as the conservation of momentum, is crucial in solving problems involving collisions or interactions between objects.
Frictionless Surface
A frictionless surface is an idealized concept where no frictional force opposes the relative motion of objects across it. In real-world scenarios, friction can never be completely negated, but modeling surfaces as frictionless helps simplify problems. In the given exercise, the frictionless surface means:
  • No energy is lost as heat due to friction.
  • The blocks move without any resistance once released.
This assumption allows us to apply the conservation of momentum since no external horizontal forces interfere with the system's total momentum.
Linear Motion
Linear motion refers to movement in a straight line. It is the simplest form of motion and can be described using parameters like velocity, acceleration, and displacement. For the blocks in the exercise:
  • The 1.0 kg block moves towards the 3.0 kg block with a velocity of 1.7 m/s.
  • Since they are on a frictionless surface, their motion remains linear after they are released.
Their velocities can be determined by using the conservation of momentum since their initial and final linear momenta must be equal.
Spring-Mass System
A spring-mass system consists of masses connected by a spring, where the spring can either stretch or compress. In such systems, the restoring force exerted by the spring follows Hooke's Law, given by:
\ F = - k \times x \
Where:
  • \( F \) is the restoring force.
  • \( k \) is the spring constant.
  • \( x \) is the displacement from the equilibrium position.
When the spring in the exercise contracts, it exerts equal and opposite forces on the two blocks, propelling them at different velocities. The conservation of momentum ensures that these velocities are related and that the total momentum of the system remains unchanged.

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Most popular questions from this chapter

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Two balls \(A\) and \(B\), having different but unknown masses, collide. Initially, \(A\) is at rest and \(B\) has speed \(v_{B}\). After the collision, \(B\) has speed \(v_{B} / 2\) and moves perpendicularly to its original motion. (a) Find the direction in which ball \(A\) moves after the collision. (b) Show that you cannot determine the speed of \(A\) from the information given.

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Vessel at Rest Explodes A vessel at rest explodes, breaking into three pieces. Two pieces, having equal mass, fly off perpendicular to one another with the same speed of \(30 \mathrm{~m} / \mathrm{s}\). The third piece has three times the mass of each other piece. What are the magnitude and direction of its velocity immediately after the explosion?
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