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Chapter 9: Problem 27

A \(3.00 \mathrm{~kg}\) block slides on a frictionless horizontal surface, first moving to the left at \(50.0 \mathrm{~m} / \mathrm{s}\). It collides with a spring whose other end is fixed to a wall, compresses the spring, and is brought to rest momentarily. Then it continues to be accelerated toward the right by the force of the compressed spring. The block acquires a final speed of \(40.0 \mathrm{~m} / \mathrm{s}\). It is in contact with the spring for \(0.020 \mathrm{~s}\). Find (a) the magnitude and (b) the direction of the impulse of the spring force on the block. (c) What is the magnitude of the spring's average force on the block?

Short Answer

Expert verified
Impulse: 270 Ns to the right; Average Force: 13500 N.

Step by step solution

01

Understanding Impulse

Impulse is the change in momentum of an object when a force is applied to it. It can be calculated using the formula: \( I = ext{Change in Momentum} = m(v_f - v_i) \), where \( m \) is the mass, \( v_f \) is the final velocity, and \( v_i \) is the initial velocity.
02

Calculate Initial and Final Velocities

Identify \( v_i = -50.0 \, \mathrm{m/s} \) (leftward direction) and \( v_f = 40.0 \, \mathrm{m/s} \) (rightward direction). Note that the negative sign indicates the initial direction is opposite to the final direction.
03

Calculate Change in Velocity

The change in velocity \( \Delta v \) is given by \( v_f - v_i = 40.0 - (-50.0) = 90.0 \, \mathrm{m/s} \).
04

Calculate Impulse

Using the impulse formula \( I = m \Delta v \), substitute \( m = 3.00 \, \mathrm{kg} \) and \( \Delta v = 90.0 \, \mathrm{m/s} \): \[ I = 3.00 \, \mathrm{kg} \times 90.0 \, \mathrm{m/s} = 270 \, \mathrm{Ns} \].
05

Determine Direction of Impulse

Since the final velocity is positive (to the right) and the initial velocity is negative (to the left), the impulse direction is to the right.
06

Calculate Average Force

The average force \( F_{\text{avg}} \) can be found by dividing the impulse by the contact time: \[ F_{\text{avg}} = \frac{I}{\Delta t} = \frac{270 \, \mathrm{Ns}}{0.020 \, \mathrm{s}} = 13500 \, \mathrm{N} \].

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Key Concepts

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

Frictionless Motion
When a block moves on a frictionless surface, it's an ideal scenario where no resistance opposes the block's motion. This means the only forces acting on the block are involved in interactions, such as contact with a spring. The mass of the block and its velocity play key roles in determining its momentum. Here, the block initially moves left at 50 m/s, collides with a spring, and is briefly stopped before it moves in the opposite direction at 40 m/s. Because there is no friction, we can focus solely on how the spring influences the block's motion.
  • The absence of friction means conservation of momentum is straightforward, except when external forces like the spring apply.
  • Frictionless surfaces allow us to focus entirely on forces applied directly to the block.
Understanding frictionless motion helps simplify calculations, making it easier to analyze the motion and forces involved.
Spring Force
Spring force is the restoring force exerted by a compressed or stretched spring upon any object attached to it. When the block compresses the spring, it stores potential energy, which is then released to push the block in the opposite direction when the spring expands. Springs follow Hooke's Law, where the force is directly proportional to the displacement, but for this problem, knowing the spring constant isn't necessary.
  • Springs exert a force that either accelerates or decelerates objects in contact.
  • The direction of spring force is always opposite to the direction of compression or stretch initially.
    • In our example, the spring compressed as the block moved to rest. Then, the force exerted by the spring accelerated the block to the right with a final velocity of 40 m/s. The knowledge of spring force is key when understanding how objects are brought to rest and then accelerated in frictionless systems.
    Average Force Calculation
    Calculating average force helps us understand the overall impact of a force across a period of time. In this exercise, the average force is calculated using impulse and the duration of contact between the block and the spring.To find the average force, we use the formula:\[ F_{\text{avg}} = \frac{I}{\Delta t} \]where:
    • \( I = 270 \, \mathrm{Ns} \) is the impulse, calculated from the change in momentum.
    • \( \Delta t = 0.020 \, \mathrm{s} \) is the contact time.
    This results in:\[ F_{\text{avg}} = \frac{270 \, \mathrm{Ns}}{0.020 \, \mathrm{s}} = 13500 \, \mathrm{N} \]The average force of 13500 N is exerted over the 0.020 seconds during which the block is in contact with the spring. This quantifies the force applied to change the block's velocity from left to right.

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    Most popular questions from this chapter

    Block 1 of mass \(m_{1}\) slides along a frictionless floor and into a one- dimensional elastic collision with stationary block 2 of mass \(m_{2}=3 m_{1}\). Prior to the collision, the center of mass of the twoblock system had a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). Afterward, what are the speeds of (a) the center of mass and (b) block \(2 ?\)

    A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass \(3.0 \mathrm{~kg}\), is moving upward at \(20 \mathrm{~m} / \mathrm{s}\) and the other ball, of mass \(2.0 \mathrm{~kg}\), is moving downward at \(10 \mathrm{~m} / \mathrm{s}\). How high do the combined two balls of putty rise above the collision point? (Neglect air drag.)

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