91Ó°ÊÓ

block \(L\) of mass \(m_{L}=1.00 \mathrm{~kg}\) and block \(R\) of mass \(m_{R}=0.500 \mathrm{~kg}\) are held in place with a compressed spring between them. When the blocks are released, the spring sends them sliding across a frictionless floor. (The spring has negligible mass and falls to the floor after the blocks leave it.) (a) If the spring gives block \(L\) a release speed of \(1.20 \mathrm{~m} / \mathrm{s}\) relative to the floor, how far does block \(R\) travel in the next \(0.800 \mathrm{~s} ?\) (b) If, instead, the spring gives block \(L\) a release speed of \(1.20 \mathrm{~m} / \mathrm{s}\) relative to the velocity that the spring gives block \(R,\) how far does block \(R\) travel in the next \(0.800 \mathrm{~s} ?\)

Step by step solution

01

Determine the Velocity of Block L

In scenario (a), when the spring releases block L, its speed relative to the floor is given as \( v_L = 1.20 \, \text{m/s} \). This speed is straightforward because it is directly provided in the problem.

02

Use Conservation of Momentum (Scenario a)

Since both blocks are initially at rest and no external forces are acting on the system, the center of mass of the system will not change its position. By conservation of momentum and the fact that momentum is initially zero, we have: \[ m_L \cdot v_L = m_R \cdot v_R \]Substituting values, we get:\[ 1.00 \cdot 1.20 = 0.500 \cdot v_R \]Solving for \( v_R \), we find:\[ v_R = \frac{1.20}{0.500} = 2.40 \, \text{m/s} \].

03

Calculate the Distance Traveled by Block R (Scenario a)

Once we have the velocity of block R, we can calculate the distance it travels in the given time using the formula for distance, \( d = v_R \cdot t \). With \( v_R = 2.40 \, \text{m/s} \) and \( t = 0.800 \, \text{s} \), we calculate:\[ d = 2.40 \times 0.800 = 1.92 \, \text{m} \].

04

Set Up for Scenario b - Relative Speed

In scenario (b), block L's speed \( v_L \) is given relative to block R. Hence,\[ v_{L\,\text{relative to R}} = 1.20 \, \text{m/s} \].We need to set up equations for both blocks considering their speeds relative to each other. This implies,\[ v_L = v_R + 1.20 \].

05

Apply Conservation of Momentum (Scenario b)

Using the conservation of momentum, similar to what we did in Step 2, but with a new expression for \( v_L \):\[ m_L \cdot (v_R + 1.20) = m_R \cdot v_R \].Substitute \( m_L = 1.00 \), \( m_R = 0.500 \), and solve for \( v_R \):\[ 1.00 \cdot (v_R + 1.20) = 0.500 \cdot v_R \],\[ v_R + 1.20 = 0.500v_R \],Rearrange to find:\[ v_R = 2.40 \, \text{m/s} \].

06

Calculate the Distance Traveled by Block R (Scenario b)

Finally, we use previously calculated speed \( v_R = 2.40 \, \text{m/s} \) and calculate the distance as in Step 3. Therefore, the distance is:\[ d = v_R \cdot t = 2.40 \times 0.800 = 1.92 \, \text{m} \].

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Key Concepts

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

Kinematic Equations

Kinematic equations are key tools used to describe the motion of objects. They allow us to calculate different parameters like distance, velocity, and time, assuming constant acceleration. In the context of our exercise, we used a simple version of a kinematic equation:

• The equation \(d = v \cdot t\) relates distance \(d\) to velocity \(v\) and time \(t\).
• In this case, since there's no acceleration (because the surface is frictionless), the formula simplifies to multiplying the velocity of block R by the time given to find the distance.

It's crucial to understand that kinematic equations form the backbone of solving many physics problems involving motion. These equations help in predicting future positions or velocities given initial conditions, which was essential in finding the distance that block R traveled.

Relative Velocity

Relative velocity is an important concept when analyzing movements from different frames of reference. It tells us how fast one object is moving from the perspective of another.

• In scenario (b) of the exercise, the velocity of block L was given relative to block R. This means we had to consider how fast block L would be moving should block R also be in motion.
• The equation for relative velocity was \( v_L = v_R + 1.20 \) because block L is 1.20 m/s faster than block R in scenario (b).

By working out such a relationship, we can understand how different objects interact or line up their motions relative to one another, crucial for solving dynamics problems in physics.

Frictionless Surface

A frictionless surface is an idealization used in physics problems to simplify analysis. By assuming no friction, we can ignore forces that slow down or alter the movement of objects.

• In the exercise, the assumption of a frictionless floor allowed the conservation of momentum to be applied neatly, as no external forces acted on the system.
• Frictionless means there's no energy lost to heat, and the only forces at play (like the spring force) impact the objects without interference.

This simplification helps focus on primary forces acting in a system, like the spring force in this case, making it easier to understand the conservation laws in action without complicating factors like friction.