# Two blocks with masses m(1) =1.0 kg and m(2) = 3.0 kg, connected by a spring, rest on a...

## Question:

Two blocks with masses {eq}m_{1} {/eq} = 1.0 kg and {eq}m_{2} {/eq} = 3.0 kg, connected by a spring, rest on a friction-less surface. If the two are given velocities such that the first travels at {eq}v_{1} {/eq} = +1.7 m/sec toward the center of mass (which remains at rest), what is the velocity of the second?

## Conservation of linear momentum:

It states that if the external force acting on a system of particles is zero, its total momentum remains conserved. The given problem is a consequence of the conservation of linear momentum. As the centre of mass of the system remains at rest we can conclude that the final momentum of the two-particle system is zero as it was initially zero.

In other terms, the two masses are given velocities such that the center of mass remains at rest. One may think, that because the blocks were initially at rest, there must have been external forces to give them such velocities. But it appears that the external forces were applied to both blocks and they were adjusted to cancel each other, so that the net force on the entire system is zero.

### Solution 1

Initially, the two blocks are at rest. Therefore, the total momentum of the two block system is zero.

We are given,

{eq}m_{1}=1.0kg\\m_{2}=3.0kg\\v_{1}=1.7m/s {/eq}

Let the final velocities of the two blocks be {eq}\vec{v_{1}} {/eq} and {eq}\vec{v_{2}} {/eq} then final momentum of the two block system will be zero as the centre of mass remains at rest.

From the conservation of linear momentum we have

{eq}\vec{P_{f}}=\vec{P_{i}} {/eq}

{eq}m_{1}\vec{v_{1}}+m_{2}\vec{v_{2}}=0 {/eq}

{eq}\displaystyle \vec{v_{2}}=\frac{-m_{1}}{m_{2}}\vec{v_{1}} {/eq}

{eq}\displaystyle \vec{v_{2}}=\frac{-1}{3}\times1.7 {/eq}

{eq}\vec{v_{2}}=-0.57m/s {/eq}

Therefore the velocity of the second block is 0.57m/s in a direction opposite to the velocity of the first block.

### Solution 2

There is another way to solve this problem. As it was noted in the previous section, the external forces were applied in such a manner that the net force is zero.

that is {eq}\vec F_1 + \vec F_2 = 0 {/eq}

Then we can apply Newton's second law, and write:

{eq}F_1 = m_1 a_1 = - m_2 a_2 = - F_2 {/eq}

Because both masses start from rest, their final velocities can be found using

{eq}v_1 = a_1 t {/eq}

{eq}v_2 = a_2 t {/eq}

Which gives us the same equation as above.

{eq}v_2 = - \frac{m_1 v_1}{m_2} {/eq}