Ablock, m1=10kg, slides down a smooth, curved incline of height 5m. it collides elastically with another block, m2=5kg, which is sitting at the bottom of the incline. assume the first block comes to rest after the collision. the second box travels over a rough surface (µk = 0.5) for 2m. at the end of the rough surface there is a spring of spring constant 250n/m. the spring is on a frictionless surface. by how much would the second block compress the spring?

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

= 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem,  we have

Work done by the friction force = change in kinetic energy of the mass 2

or

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring =  Kinetic energy of the mass 2

or

on substituting the values, we get

or

x = 2.86 m

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