Question

A block of mass m moving with velocity vo collides
with another stationary block of mass Mas
shown. The maximum compression in the spring is
[Assume all surfaces to be smooth and spring
constant is Kl
m" room
mM
"
V2K(M+ m)
O
mM
(2) VKU
M+m) to
| 2mM
(3) KM+m) XVO
2 mM
(4) KiM+m) XVO​

Answer 1

The spring attached with block of mass M will compress untill both the particles starts to move together.

as external force acting on the system = 0,

so, initial momentum = final momentum

or, $mv_0=(m+M)v$

or, $v=\frac{mv_0}{(m+M)}$

let maximum compression is x

from law of conservation of energy,

initial kinetic energy = spring potential energy + kinetic energy of system of blocks.

$or,\frac{1}{2}mv_0^2=\frac{1}{2}Kx^2+\frac{1}{2}(m+M)\left(\frac{mv_0}{(m+M)}\right)^2$

$or,v_0^2\left[m-\frac{m^2}{(m+M)}\right]=Kx^2$

or, $x^2=\frac{mM}{(m+M)K}v_0^2$

or, $x=\sqrt{\frac{mM}{(m+M)K}}v_0$

hence, option (2) is correct choice.