Question

An impulse of J= 30 Ns is given to the 3 kg block attached with 2kg block by a spring. Initially the spring is unstretched and the system is at rest and all surfaces are smooth. Find the velocity of the centre of mass at t= 5 s.
Given spring constant for k= 5N/m

Answer 1

Hint: Impulse if a force acting for a very short period of time and mathematically it is equivalent to the change in the momentum. We are given a spring and we know spring possesses elastic properties. In order to find the velocity of the centre of mass we have to consider all the forces acting.

Complete step by step answer:
We are given two blocks, one of mass 2 kg and the other of mass 3 kg. they are connected with the help of a spring whose spring constant is 5 N/m. When an impulse is applied to a 3 kg block, the system should move.
Impulse= change in momentum
Acting on a 3 kg block which was initially at rest.
\[J=mv-(mu)\]
\[\begin{align}
  & 30=3v-0 \\
 & v=10m/s \\
\end{align}\]
So, the velocity comes out to be 10m/s, now in order to find the velocity of centre of mass we use the following formula,
Velocity of centre of mass will be \[\frac{{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}}{{{m}_{1}}+{{m}_{2}}}\]
\[\begin{align}
  & v=10m/s \\
 & =\dfrac{3\times 10+0}{3+2} \\
 & =6m/s \\
\end{align}\]

So, the velocity of centre of mass is 6 m/s..

Note:
In such problems we have to keep in mind that velocity is a vector quantity and hence, such we have to also take into account its direction. If the top to down direction is taking positive then its opposite will be taken negative and vice-versa. Here it was mentioned that the floor was frictionless, so we do not have to consider the effect of frictional force.