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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

A \(0.70 \mathrm{~kg}\) ball is moving horizontally with a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) when it strikes a vertical wall. The ball rebounds with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the change in translational momentum of the ball?

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 its 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.

When jumping straight down, you can be seriously injured if you land stiff- legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. Suppose you have a mass \(m\) and you jump off a wall of height \(h\). (a) Use what you learned about constant acceleration motion to find the speed with which you hit the ground. Assume you simply step off the wall, so your initial \(y\) velocity is zero. Ignore air resistance. (Express your answer in terms of the symbols given.) (b) Suppose that the time interval starting when your feet first touch the ground until you stop is \(\Delta t .\) Calculate the (average) net force acting on you during that interval. (Again, express your answer in terms of the symbols given.) (c) Suppose \(h=1 \mathrm{~m}\). If you land stiff-legged, the time it takes you to stop may be as short as \(2 \mathrm{~ms}\), whereas if you bend your knees, it might be as long as \(0.1\) s. Calculate the average net force that would act on you in the two cases. (d) The net force on you while you are stopping includes both the force of gravity and the force of the ground pushing up. Which of these forces do you think does you the injury? Explain your reasoning. (e) For the two cases in part (c), calculate the upward force the ground exerts on you.

The last stage of a rocket, which is traveling at a speed of \(7600 \mathrm{~m} / \mathrm{s}\), consists of two parts that are clamped together: a rocket case with a mass of \(290.0 \mathrm{~kg}\) and a payload capsule with a mass of \(150.0 \mathrm{~kg}\). When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of \(910.0 \mathrm{~m} / \mathrm{s}\). What are the speeds of (a) the rocket case and (b) the payload after they have separated? Assume that all velocities are along the same line.

It is well known that bullets and other missiles fired at Superman simply bounce off his chest (Fig.7-22). Suppose that a gangster sprays Superman's chest with \(3 \mathrm{~g}\) bullets at the rate of 100 bullets/min, and the speed of each bullet is \(500 \mathrm{~m} / \mathrm{s}\). Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest from the stream of bullets?
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