A horizontal spring of elastic constant k=60N/cm is compressed by 10 cm then rel

Question

A horizontal spring of elastic constant k=60N/cm is compressed by 10 cm then released. As it un-compresses the spring pushes on a block of mass m=0.5 kg over a frictionless horizontal table. The mass loses contact with the spring as the spring reaches its natural length. As it continues on its path the mass collides head-on and elastically with a second mass of 6kg initially at rest. If the 0.5 kg mass would bounce back , it would go towards the spring. If that is the case by how much will the mass compress the spring? A horizontal spring of elastic constant k=60N/cm is compressed by 10 cm then released. As it un-compresses the spring pushes on a block of mass m=0.5 kg over a frictionless horizontal table. The mass loses contact with the spring as the spring reaches its natural length. As it continues on its path the mass collides head-on and elastically with a second mass of 6kg initially at rest. If the 0.5 kg mass would bounce back , it would go towards the spring. If that is the case by how much will the mass compress the spring? A horizontal spring of elastic constant k=60N/cm is compressed by 10 cm then released. As it un-compresses the spring pushes on a block of mass m=0.5 kg over a frictionless horizontal table. The mass loses contact with the spring as the spring reaches its natural length. As it continues on its path the mass collides head-on and elastically with a second mass of 6kg initially at rest. If the 0.5 kg mass would bounce back , it would go towards the spring. If that is the case by how much will the mass compress the spring?

Explanation / Answer

here,

spring constant , K = 60 N/cm = 6000 N/m

initially , as the spring compresses

let the initial speed of m1 be u1

using conservtaion of energy

0.5 * k * x^2 = 0.5 * m1 * u1^2

0.5 * 6000 * 0.1^2 = 0.5 * 0.5 * u1^2

10.95 m/s = u1

mass , m2 = 6 kg

let the final speed after the collison be v1 and v2

using conservation of momentum

m1 * u1 = m1 * v1 + m2 * v2

0.5 * 10.95 = 0.5 * v1 + 6 * v2 ....(1)

and

using conservation of energy

0.5 * m1 * u1^2 = 0.5 * m1 * v1^2 + 0.5 * m2 * v2^2

0.5 * 10.95^2 = 0.5 * v1^2 + 6 * v2^2 ....(2)

from (1) and (2)

v1 = - 9.26 m/s

v2 = 1.68 m/s

let the spring compresses x' m

using conservtaion of energy

0.5 * k * x'^2 = 0.5 * m1 * v1^2

0.5 * 6000 * x'^2 = 0.5 * 0.5 * 9.26^2

x' = 0.084 m = 8.4 cm

the spring compresses 8.4 cm

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