Two bricks with masses m1=5 kg and m2=3 kg lie at rest on a frictionless horizon

Question

Two bricks with masses m1=5 kg and m2=3 kg lie at rest on a frictionless horizontal plane. The bricks are attached to each other by an ideal string of length d=0.035 m. An ideal mass-less spring with spring constant k=2500 N/m and relaxed length D=0.05 m is compressed and then inserted between the bricks as shown in Fig. 1 and as we did in class.

(a) Calculate the tension in the string.

(b) The string is now cut and consequently the bricks are pushed apart by the spring. Calculate the final velocity of each brick following this event.

Explanation / Answer

here,

m1 = 5 kg

m2 = 3 kg

length of string , d = 0.035 m

spring constant , k = 2500 N/m

relaxed lengtn , D = 0.05 m

(a)

the tension in the spring , T = k*( D - d)

T = 2500 * ( 0.05 - 0.035)

T = 37.5 N

the tension in the string is 37.5 N

(b)

let their velocity be v1 and v2

using conservation of energy

0.5 k*x^2 = 0.5 * m1 * v1^2 + 0.5 * m2 * v2^2

2500 * 0.015^2 = 5 * v1^2 + 3 * v2^2 ....(1)

using conservation of momentum

m1*v1 = m2*v2

5 * v1 = 3*v2 ....(2)

from equation (1) and (2)

v1 = 0.21 m/s

v2 = 0.34 m/s

the final velocity of brick 1 is 0.21 m/s and brick2 is 0.34 m/s

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