A block of mass m1 = 2 kg slides along a frictionless table with a velocity of +10 m/s.? - m 1120 mfp printer
Directly in front of her, and moves at a speed of 3 m / s, is a block of mass m2 = 9 kg. A massless spring, spring constant R = 1120 N / m is inserted in the second block in Figure 8-59.
a) Before entering the M1 spring, is what the speed of the center of mass of the system?
(b) After the collision, the spring to a maximum of x. Compressed What is the value of x?
(c) The time blocks are separated again. What is the terminal velocity of each block measures under the table?
M 1120 Mfp Printer A Block Of Mass M1 = 2 Kg Slides Along A Frictionless Table With A Velocity Of +10 M/s.?
1 comments:
a) 2 (10) + 9 (3) = 11Vcm
VCM = 4.2 m / s
b) If the pen is pressed to the maximum speed of the two blocks as the other. Moreover, the same will be like the answer to A. The kinetic energy is lost in the collision must be stored in the spring when ...
1 / 2 (2) 10 ^ 2 + 1 / 2 (9) 3 ^ 2 = 1 / 2 (1120) x ^ 2 + 1 / 2 (11) (4.2) ^ 2
KE1Before + KE2Before = PEspring + KEafter
Solving this for X, and you will have the maximum spring compression during the collision.
c) I think they want us to assume the collision is perfectly elastic to use in this case, the conservation of momentum
2 (10) + 9 (3) = 2 (V1A) + 9 (V2A)
And the elastic collision equation of relative velocity:
V2B - V1B = V1A - V2A
V1A = 3.10 - V2A
You now have two equations and two unknowns.
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