How Does Relative Motion Affect Block R's Travel Distance?

[norcal]

Homework Statement

"Relative" is an important word. Block L of mass mL = 1.90 kg and block R of mass mR = 0.500 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 1.80 m/s relative to the floor, how far does block R travel in the next 0.800 s?
correct check mark m
(b) If, instead, the spring gives block L a release speed of 1.80 m/s relative to the velocity that the spring gives block R, how far does block R travel in the next 0.800 s?

Homework Equations

m1v1=m2v2

v=d/t

The Attempt at a Solution

1.9*1.8=.5*velocityR
velocityR = 6.84 m/s

d=v*time
d=(6.84)(.8)
d=5.472m

Ok, so 5.472m is the correct answer for part A but I don't know how to get part B. Any suggestions?

 

[PhanthomJay]

Homework Statement

"Relative" is an important word. Block L of mass mL = 1.90 kg and block R of mass mR = 0.500 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 1.80 m/s relative to the floor, how far does block R travel in the next 0.800 s?
correct check mark m
(b) If, instead, the spring gives block L a release speed of 1.80 m/s relative to the velocity that the spring gives block R, how far does block R travel in the next 0.800 s?

Homework Equations

m1v1=m2v2

v=d/t

The Attempt at a Solution

1.9*1.8=.5*velocityR
velocityR = 6.84 m/s

d=v*time
d=(6.84)(.8)
d=5.472m

Ok, so 5.472m is the correct answer for part A but I don't know how to get part B. Any suggestions?

You can use the same equation as in (a), realizing that if m1 moves left, m2 must move right. But you have 2 unknowns, v1 and v2, so you need a 2nd equation that relates v1 with v2. What is that relationship?

 

[m2003]

I would first commend the student for correctly using the conservation of momentum equation and the velocity equation to solve for the distance traveled by block R in part A. I would also highlight the importance of understanding the concept of relative motion in this scenario.

In part B, the release speed of block L is now given relative to the velocity of block R. This means that block L is already moving with a certain velocity, and the spring adds an additional velocity to it. Therefore, the total velocity of block L relative to the floor would be 1.80 m/s + velocity of block R.

Using the conservation of momentum equation again, we can set up the following equation:

mL(vL1 + vR1) = mL(vL2) + mR(vR2)

Where vL1 is the initial velocity of block L relative to the floor, vR1 is the initial velocity of block R relative to the floor, vL2 is the final velocity of block L relative to the floor, and vR2 is the final velocity of block R relative to the floor.

We know that vL1 = 1.80 m/s and vR1 = 0 m/s (since block R is initially at rest). We also know that vL2 = 1.80 m/s + vR (since block L's final velocity is the sum of its initial velocity and the velocity added by the spring).

Solving for vR, we get vR = 1.80 m/s.

Now, using the velocity equation and the given time of 0.800 s, we can calculate the distance traveled by block R:

d = v*t
d = (1.80)(0.800)
d = 1.44 m

Therefore, in part B, block R travels 1.44 m in the next 0.800 s. This is a shorter distance compared to part A, as the initial velocity of block L relative to the floor is higher than the initial velocity of block L relative to block R.

I would also encourage the student to continue practicing and understanding the concept of relative motion in order to master the application of conservation of momentum in different scenarios.