Answer:
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
![0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)](https://tex.z-dn.net/?f=0.5%5Ctimes5%5Ctimes9.8%5Ctimes2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes5%5Ctimes%28%20v_3%5E2-19.79%5E2%29)
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
![\frac{1}{2}kx^2=\frac{1}{2}mv_3^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dkx%5E2%3D%5Cfrac%7B1%7D%7B2%7Dmv_3%5E2)
on substituting the values, we get
![\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes250%5Ctimes%20x%5E2%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes5%5Ctimes20.27%5E2)
or
x = 2.86 m