A) Before the 36 kg block makes contact with the spring, what is the velocity of

Question

A) Before the 36 kg block makes contact with the spring, what is the velocity of the center of the mass of the system? Answer in units of m/s. B) during the collision, the spring is compressed by a maximum amount delta x. What is the max compression? Answer in units of m. C) the blocks will eventually separate again. What is the final velocity of the 36 kg block? Answer in units of m/s. D) what is the final velocity of the 85kg block? Answer in units of m/s. For my answers I got 21.6, 2.35, 13.14, and 25.14 but they're all wrong. Please help.

Explanation / Answer

a)    

from conservation of momentum

m1v1 +m2v2 = (m1+m2) Vcm

36x30 + 85x18 = (36+85)Vcm

Vcm = (1080+1530) / 121 m/s

Vcm =21.57m/s

b)

When the spring is compressed at its maximum amount, the velocity of the two blocks will be the same as each other. It will also be the same as the answer to a. The kinetic energy that is lost in the collision must be stored in the spring, so
1/2m1v12 +1/2m2v22 = 1/2kx2+ ½ (m1+m2)(Vcm)2

0.5 (36) (30)2 + 0.5 (85)(18)2 = 0.5 (1482)x2 +0.5 (36+85) (21.57)2

16200+13770 =741x2+28148.53

29970 -28147.53 =741x2

x2 =1822.47 / 741

x=1.57m

c) & d)

assume the collision is perfectly elastic, in which case, you use momentum conservation
m1v1 +m2v2 = (m1+m2) Vcm

36(30) + 85(18) = 36(V1A) + 85(V2A)

AND the elastic collision relative velocity equation:

V2B - V1B = V1A - V2A
18-36= V1A - V2A

Solve the above equations and u will get the answers for c & d part

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