Consider one object, made up of {mass 1 =6 kilograms}, and {mass 2 = 2 kilograms

Question

Consider one object, made up of {mass 1 =6 kilograms}, and {mass 2 = 2 kilograms}, both are ‘boxes’. A spring that does not have mass is inbetween them, the spring is compressed between the boxes, and in the springs compressed state, it’s {length/distance =2.5 cm}. As the boxes with the spring in between them(it’s all one object at this time) slides on a table top horizontally (no friction) with Vel. Initial of 2.5 m/s, and all of a sudden, the spring rapidly releases its compression, and the boxes (mass 1 and mass 2) are “propelled” as two components.

a) calc. the post.spring.release velocity Of mass 1 in respect to mass 2 in the reference perspective of the table.
b) calc. the postrelease vel. Of mass 2 in respect to mass 1 in the reference perspective of the centerofmass.

[mass 1]—[mass 2] [mass 1] ---- [mass 2}
Initial after spring relases compression

(please explain your steps and equations used)

Explanation / Answer

The mass of the first block is m1 = 6 kg The mass of the second block is m2 = 2 kg The compressed length of the spring is dx = 2.5 cm = 0.025 m The initial speed of the system is u = 2.5 m/s Let the spring constant of the spring be k (no value given) The initial kinetic energy of the system is KEint = 0.5*(6 kg+2 kg)*(2.5 m/s)^2 KEint = 25 J The potential energy stored in the compressed spring is U = 0.5*k*(0.025 m)^2 U = k*3.125*10^-4 J After the spring is released, let v1-2 and v2-1 be the relative velocities of the blocks respectively. Then the kinetic energy of the system after the spring is released is KEfin = 0.5*(6 kg)*v1^2 + 0.5*(2 kg)*v2^2 KEfin = 3v1-2^2 + v2-1^2 This energy should be equal to the compressed potential enegy of the spring k*3.125*10^-4 J = 3v1-2^2 + v2-1^2 -----------> (1) Let v1 and v2 be the individual velocities of the blocks respectively. Then according to the conservation of energy KEint + U = 0.5*6*v1^2 + 0.5*2*v2^2 25 J + k*3.125*10^-4 J = 3v1^2 + v2^2 -----------> (2) And the relative velocities are given by v1-2 = v2 - v1 v2-1 = v1 - v2 Solving all the above equations we gat v1-2 and v2-1

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