Answer:
v₁ = 3.9 m/s
v₂ = 5.4 m/s
The loss in the kinetic energy = 1.05 J
Explanation:
Given:
mass, m₁ = 2 kg
m₂ = 3 kg
initial speed of mass m₁, u₁ = 6 m/s
Initial speed of the mass m₂, u₂ = 4 m/s
coefficient of restitution, e = (3/4)
let, the final speed of mass m₁ and m₂ be v₁ and v₂ respectively
Now,
![e=\frac{v_2-v_1}{u_1-u_2}\\](https://tex.z-dn.net/?f=e%3D%5Cfrac%7Bv_2-v_1%7D%7Bu_1-u_2%7D%5C%5C)
on substituting the values, we get
![\frac{3}{4}=\frac{v_2-v_1}{6-4}\\](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B4%7D%3D%5Cfrac%7Bv_2-v_1%7D%7B6-4%7D%5C%5C)
or
1.5 = v₂ - v₁
or
v₂ = 1.5 + v₁
also,
from the conservation of momentum, we have
( 2 × 6 ) + ( 3 × 4 ) = ( 2 × v₁ ) + ( 3 × v₂ )
or
24 = 2 × v₁ + ( 3 × ( 1.5 + v₁ ) )
or
24 = 2 × v₁ + 4.5 + 3 × v₁
19.5 = 5 × v₁
or
v₁ = 3.9 m/s
and
v₂ = 1.5 + v₁
or
v₂ = 1.5 + 3.9
or
v₂ = 5.4 m/s
now,
the loss in the kinetic energy = initial kinetic energy - Final kinetic energy
or
= ![(\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2)-(\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2)](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B2%7Dm_1u_1%5E2%2B%5Cfrac%7B1%7D%7B2%7Dm_2u_2%5E2%29-%28%5Cfrac%7B1%7D%7B2%7Dm_1v_1%5E2%2B%5Cfrac%7B1%7D%7B2%7Dm_2v_2%5E2%29)
or
= ![(\frac{1}{2}\times2\times6^2+\frac{1}{2}\times3\times4^2)-(\frac{1}{2}\times2\times3.9^2+\frac{1}{2}\times3\times5.4^2)](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B2%7D%5Ctimes2%5Ctimes6%5E2%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes3%5Ctimes4%5E2%29-%28%5Cfrac%7B1%7D%7B2%7D%5Ctimes2%5Ctimes3.9%5E2%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes3%5Ctimes5.4%5E2%29)
or
the loss in the kinetic energy = 1.05 J