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Chapter 8: Problem 44

Combining Conservation Laws. A 15.0 -kg block is attached to a very light horizontal spring of force constant 500.0 \(\mathrm{N} / \mathrm{m}\) and is resting on a frictionless horizontal table. (Fig. E8.44). Suddenly it is struck by a 3.00 -kg stone traveling horizontally at 8.00 \(\mathrm{m} / \mathrm{s}\) to the right, whereupon the stone rebounds at 2.00 \(\mathrm{m} / \mathrm{s}\) horizontally to the left. Find the maximum distance that the block will compress the spring after the collision.

Short Answer

Expert verified
Maximum compression of the spring is approximately 0.35 meters.

Step by step solution

01

Understand the Problem

We need to find how far the 15.0-kg block compresses the spring after being hit by a 3.00-kg stone. The key concepts involved are conservation of momentum and conservation of mechanical energy.
02

Apply Conservation of Momentum

Apply the conservation of momentum to find the velocity of the 15.0-kg block after the collision. Initial momentum: \[ (3.00 \, \text{kg})(8.00 \, \text{m/s}) = 24.00 \, \text{kg} \cdot \text{m/s} \] Final momentum of stone: \[ (3.00 \, \text{kg})(-2.00 \, \text{m/s}) = -6.00 \, \text{kg} \cdot \text{m/s} \] Using conservation of momentum: \[ 24.00 + 0 = 15.0 \times V_b + (-6.00) \] Solve for \(V_b\): \[ V_b = \frac{30.00}{15.0} = 2.00 \, \text{m/s} \] (where \(V_b\) is the velocity of the block after collision)
03

Convert Kinetic Energy to Potential Energy

Once the block starts moving, all its kinetic energy will be converted into potential energy stored in the spring. Initial kinetic energy of block: \[ KE = \frac{1}{2}mv^2 = \frac{1}{2}(15.0 \, \text{kg})(2.00 \, \text{m/s})^2 = 30.00 \, \text{J} \] Potential energy in spring: \[ PE = \frac{1}{2}kx^2 \] Where \(k = 500.0 \, \text{N/m}\). Set \(KE = PE\) and solve for \(x\): \[ 30.00 = \frac{1}{2}(500.0)x^2 \] \[ 30.00 = 250.0x^2 \] Solve for \(x\): \[ x^2 = \frac{30.00}{250.0} = 0.12 \] \[ x = \sqrt{0.12} = 0.346 \approx 0.35 \, \text{m} \]

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Key Concepts

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

Conservation of Momentum
Conservation of momentum is a fundamental principle of physics, stating that the total momentum of a closed system remains constant if no external forces act upon it. In simpler terms, the momentum before an event is equal to the momentum after the event.
This concept helps predict how objects will move after colliding. In the example, a 3 kg stone strikes a 15 kg block.
The initial momentum of the stone (mass times velocity) is 24 kg鈰卪/s. After the collision, the stone moves backwards, i.e., it has a negative momentum of -6 kg鈰卪/s.
To find the block's velocity after the collision, we use conservation of momentum:
  • Initial momentum: 24.00 kg鈰卪/s (stone) + 0 (block)
  • Final momentum: (-6.00 kg鈰卪/s from the stone) + (momentum of block)
By setting them equal, we find the block's velocity is 2 m/s. The momentum shifts from the stone to the block.
Spring Constant
The spring constant, denoted as \( k \), tells us how stiff a spring is. It is measured in Newtons per meter (N/m) and determines how much force is needed to compress or extend the spring by a certain amount.
A higher \( k \) means a stiffer spring, which requires more force to change its length. In our problem, the spring constant is 500 N/m.
This means that for every meter the spring is compressed, a force of 500 Newtons is exerted.
The spring constant is crucial for calculating potential energy stored in a spring, as we'll see in the potential energy section.
Understanding \( k \) helps us comprehend how energy is transferred and stored in spring systems.
Kinetic Energy
Kinetic energy (\( KE \)) is the energy an object possesses due to its motion. It is calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.
In our scenario, the block of mass 15 kg moves at 2 m/s after the stone hits it, so we calculate its kinetic energy:
\[ KE = \frac{1}{2}(15)(2)^2 = 30 \, J \]
This 30 Joules represent all the motion energy of the block and is crucial in understanding how this energy transforms into potential energy when the block compresses the spring.
The conservation of kinetic energy in systems helps us determine how motion is affected by external interactions like collisions.
Potential Energy
Potential energy (\( PE \)) is the stored energy of an object due to its position or state. When talking about springs, potential energy relates to how compressed or stretched the spring is.
The formula for the potential energy stored in a spring is \( PE = \frac{1}{2}kx^2 \), where \( k \) is the spring constant, and \( x \) is the displacement from its original length.
When the moving block compresses the spring, its kinetic energy is converted to potential energy.
In this case, we set the kinetic energy equal to the potential energy to find out the maximum compression of the spring:
\[ 30 = \frac{1}{2}(500)x^2 \]
Solving gives us \( x = 0.35 \) meters, signifying the distance by which the spring is compressed when the block has stopped.
This interplay of energy types beautifully illustrates the principles of energy conservation.

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Most popular questions from this chapter

One 110 -kg football lineman is running to the right at 2.75 \(\mathrm{m} / \mathrm{s}\) while another 125 -kg lineman is running directly toward him at 2.60 \(\mathrm{m} / \mathrm{s}\) . What are (a) the magnitude and direction of the net momentum of these two athletes, and (b) their total kinetic energy?

Just before it is struck by a racket, a tennis ball weighing 0.560 \(\mathrm{N}\) has a velocity of \((20.0 \mathrm{m} / \mathrm{s}) \hat{\imath}-(4.0 \mathrm{m} / \mathrm{s}) \hat{\boldsymbol{J}}\) . During the 3.00 \(\mathrm{ms}\) that the racket and ball are in contact, the net force on the ball is constant and equal to \(-(380 \mathrm{N}) \hat{\boldsymbol{\imath}}+(110 \mathrm{N}) \hat{\boldsymbol{J}}\) . (a) What are the \(x\) - and \(y\) -components of the impulse of the net force applied to the ball? (b) What are the \(x\) - and \(y\) -components of the final velocity of the ball?

A 0.150 -kg frame, when suspended from a coil spring, stretches the spring 0.070 m. A 0.200 -kg lump of putty is dropped from rest onto the frame from a height of 30.0 \(\mathrm{cm}\) (Fig. P8.82). Find the maximum distance the frame moves downward from its initial position.

CALC At time \(t=0,\) a 2150 -kg rocket in outer space fires an engine that exerts an increasing force on it in the \(+x\) -direction. This force obeys the equation \(F_{x}=A t^{2},\) where \(t\) is time, and has a magnitude of 781.25 \(\mathrm{N}\) when \(t=1.25 \mathrm{s}\) . (a) Find the SI value of the constant \(A,\) including its units. (b) What impulse does the engine exert on the rocket during the 1.50 -s interval starting 2.00 \(\mathrm{s}\) after the engine is fired? (c) By how much does the rocket's velocity change during this interval?

Force of a Baseball Swing. A baseball has mass 0.145 \(\mathrm{kg}\) . (a) If the velocity of a pitched ball has a magnitude of 45.0 \(\mathrm{m} / \mathrm{s}\) and the batted ball's velocity is 55.0 \(\mathrm{m} / \mathrm{s}\) in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. (b) If the ball remains in contact with the bat for 2.00 \(\mathrm{ms}\) , find the magnitude of the average force applied by the bat.
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