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Ajay Makwana · Answer 1 · 2023-03-20T14:09:23+0000

The momentum of the blocks must be preserved, therefore, we have the following relationship:

$P_(A1)+P_(B1)=P_(A2)+P_(B2)_{}$

Where PA1 and PB1 are the initial momentum of the blocks and PB2, PA2 are the momentum of the blocks after release. Since the system is initially at rest this means that the initial momentum is zero:

Replacing the definition of momentum:

Where "m" is the mass and "v" the velocity, we get:

Form this we can solve for the velocity of block A:

$\begin{gathered} -m_Av_A=m_Bv_B_{} \\ v_A=-(m_Bv_B)/(m_A) \end{gathered}$

Replacing the values:

$v_A=-\frac{(3\operatorname{kg})(1.2\text{ m/s)}}{1\operatorname{kg}}$

Solving the operations:

$v_A=-3.6\text{ m/s}$

The negative sign means that the block moves in the opposite direction to block B.

Now, since the spring drops, this means that it transfers all of its kinetic energy to the blocks, therefore, the initial energy of the spring must be equal to the sum of the kinetic energies of the blocks. This means:

Using the formula for kinetic energy:

Replacing we get:

$U_s=(1)/(2)m_Av^2_A+(1)/(2)m_Bv^2_B_{}$

Replacing the values:

$U_s=(1)/(2)(1\operatorname{kg})(3.6(m)/(s))^2+(1)/(2)(3\operatorname{kg})(1.2(m)/(s))^2$

Solving the operations we get:

Therefore, the energy of the spring was 8.64 Joules.

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