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2 years ago

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“Relative” is an important word. Block L of mass mL = 1.90 kg and block R of mass mR = 0.460 kg are held in place with a compres

sed 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 1.10 m/s relative to the floor, how far does block R travel in the next 0.740 s? (b) If, instead, the spring gives block L a release speed of 1.10 m/s relative to the velocity that the spring gives block R, how far does block R travel in the next 0.740 s?

1 answer:

Evgesh-ka [11]2 years ago

4 0

Answer:m

R

ν

R

+m

L

ν

L

=0 ⇒ (0.500kg)ν

R

+(1.00kg)(−1.20m/s)=0

which yields ν

R

=2.40m/s . Thus , Δx=ν

R

t=(2.40m/s)(0.800s)=1.92m .

(b) Now we have m

R

ν

R

+m

L

(ν

R

−1.20m/s)=0 which yields

ν

R

=

m

L

+m

R

(1.2m/s)m

L

=

1.00kg+0.500kg

(1.20m/s)(1.00kg)

=0.800m/s .

Consequently , Δx=ν

R

t=0.640m .

Explanation:

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