Re: PHYSICS

"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?
(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?

For Further Reading

* Physics - bobpursley, Sunday, February 25, 2007 at 5:22pm

Momentum L has is equal to the momentum R has.

1.90*1.80=.500*veloictyR

Relative to each other, velocityLrelative=VelocityRrelative

Now to compute the distance here, obviously the relative distances are of no value....they equal each other. Here you have to convert to an absolute relative to floor velocity. AGain, use the principle that the center of gravity is constant, or, the momentums are equal.

* Physics - COFFEE, Sunday, February 25, 2007 at 11:47pm

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

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

This doesn't seem right...am I doing something wrong with my calculations??? PLEASE HELP :)

For Further Reading

* Re: PHYSICS - bobpursley, Tuesday, February 27, 2007 at 7:17am

yOu need to use average velocity, not initial velocity. The average velocity is 1/2 the initial velocity.

distance=averagevelocity*time

So I did...

d=(-1.8+6.84)(.8)
d=2.016m

This is wrong. What am I doing wrong?

"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?
(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?

"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?
****Momentum is conserved: mL*vL=mR*Vr implies that
Vr=mL/mR *vL= 1.90/.500 (1.80) = 6.84 m/s
distanceR=6.84*.800=5.47 meters with respect to the floor.
(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?
****Momentum is conserved: mL*vL=mR*Vr implies that
Vr=mL/mR *vL= 1.90/.500 (1.80) = 6.84 m/s with respect to the velocity given L

At this point, consider the words "relative to the velocity that the spring gives block R." That means to me, L is moving at 1.84 less than R is moving to the right. If R is fixed, then L is moving 1.84 to the left. If R is moving 10m/s to the right, then L is moving 10-1.84 to the RIGHT.
Here, vL=vR-1.84 where vL, vR are relative to floor.
and vR= 6.84-vL where vL,vR are relative to floor.
So vR= 6.84-vL = 6.84-vR+1.84
vR= 4.34 m/s
check my thinking and math.

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