This question is from the book \"Physics, For Scientists andEngineers\"sixth edi

Question

This question is from the book "Physics, For Scientists andEngineers"sixth edition Author Tipler. Chapter 8 question number 59. A block of mass m1 = 2.0 kg slides along a frictionless tablewith a speed of 10m/s. Directly in front of it, and moving in thesame direction with a speed of 3.0m/s, is a block of mass m2 = 5.0kg. A massless spring that has a force constant k = 1120 N/m isattached to the second block, as in the figure in the book given onpage 283. A.) What is the velocity of the center of mass of thesystem? B.) During the collision, the spring is compressed by amaximum amount delta x. What is the value of delta x? C.) The blocks will eventually separate again. What are theveloctities of the two blocks measured in the reference frame ofthe table, after they separate? Hopefuly this will not confuse you. Underneath is a textualschematic of the pictorial in the book. [(Block 1) m1 --10m/s-->]    [(Spring attached to block)-----(Block 2)m2--3m/s-->] This question is from the book "Physics, For Scientists andEngineers"sixth edition Author Tipler. Chapter 8 question number 59. A block of mass m1 = 2.0 kg slides along a frictionless tablewith a speed of 10m/s. Directly in front of it, and moving in thesame direction with a speed of 3.0m/s, is a block of mass m2 = 5.0kg. A massless spring that has a force constant k = 1120 N/m isattached to the second block, as in the figure in the book given onpage 283. A.) What is the velocity of the center of mass of thesystem? B.) During the collision, the spring is compressed by amaximum amount delta x. What is the value of delta x? C.) The blocks will eventually separate again. What are theveloctities of the two blocks measured in the reference frame ofthe table, after they separate? Hopefuly this will not confuse you. Underneath is a textualschematic of the pictorial in the book. [(Block 1) m1 --10m/s-->]    [(Spring attached to block)-----(Block 2)m2--3m/s-->]

Explanation / Answer

1/2 m v12 + 1/2MV12   = 1/2 mv22 + 1/2 MV22
m (v12 - v22) = M (V22 -V12)                 (I)
m v1 + M V1 = m v2 + M V2
m (v1 - v2) = M (V2 - V1)    (II)
m v1 + M V1 = m v2 + M V2
v1 + v2 = V2 + V1       dividing (I) by (II)       (III)
v1 - v2 = M / m (V2 - V1)     from(II)            (IV)
2 v1 = V1 (1 - M / m) + V2 (1 + M /m)        adding (III) &(IV)
Using the given values solve for   V2 = 7 m / s
Use (III) to get v2 = 0
The spring has no mass so the forces on either of the spring mustbe equal or the spring would undergo
infinite acceleration
Since m is motionless after the collision, all of its energywas transmitted thru the spring to M
Then 1/2 m v12 = 1/2 k x2
x = (m / k) v1 = .423 m
Using the velocity of the center of mass does not appear to beuseful in solving the problem
Vcm = (m v1 + M V1) / (m + M) = (2 * 10 + 5 * 3) / (5 + 2) =7 m/s
Vcm = M V2 / (m + M) = 5 * 7 / (5 + 2) = 7 m/s Sorry about the mixup in subscripts &exponents (glitch) v12 = v12 etc. Sorry about the mixup in subscripts &exponents (glitch) v12 = v12 etc.

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