A block of mass m1 = 3 kg slides along a frictionless table with a velocity of +

Question

A block of mass m1 = 3 kg slides along a frictionless table with a velocity of +10 m/s. Directly in front of it, and moving with a velocity of +3 m/s, is a block of mass m2 = 8 kg. A massless spring with spring constant k = 1120 N/m is attached to the second block as in the figure below. 1) (a) Before m1 runs into the spring, what is the velocity of the center of mass of the system? 4.9 m/s Submit Your submissions: 4.9 Computed value:4.9Submitted:Sunday, July 2 at 10:41 PM Feedback:Correct! 2) (b) After the collision, the spring is compressed by a maximum amount x. What is the value of x? cm Submit 3) (c) The blocks will eventually separate again. What is the final velocity of each block measured in the reference frame of the table? m/s (for m1) Submit 4) m/s (for m2) Submit

Explanation / Answer

(a)using linear momentum conservation-

m1*v1+m2*v2=(m1+m2)*v

v=m1*v1+m2*2/(m1+m2)=4.90m/s

(b) using energy conservation-

1/2m1v1^2+1/2m2v^2=1/2(m1+m2)*v^2+1/2Kx^2

150+36=132.055+560*x^2

x^2=0.096

x=0.31m

(c) using momentum conservation-

m1v1+m2v2=m1v1’+m2v2’

30+24=3v1’+8v2’

54=3v1’+8v2’.....................(1)

using relativistic principle

3-10=v1’+v2’

-7=v1’+v2’.........................(2)

solving equation 1 and 2

v1’= -22 m/s

v2’=15 m/s

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