91影视

The given data can be listed below as:

• The unstretched length of the spring is /=0.32m.
• The mass of the block is m=0.2 kg.
• The initial stretched length of the spring is ${/}_1=0.4\mathrm{m}$.
• The final stretched length of the spring is ${/}_2=0.43\mathrm{m}$.
• The length of another spring is $s_1=0.4\mathrm{m}$.
• The mass of the bullet is $m_1=0.003\mathrm{kg}$.
• The velocity of the bullet is v=200 m/s.

The maximum height reached by the block ish

The collision is described as the direct forceful contact of two or more objects. It has two types such as elastic and inelastic collision.

Whenever the two particles tries to occupy the same space then collision will occur.

The equation of the time taken by the block is expressed as:

$$\begin{gathered} T=2蟺\sqrt{\frac{m\left(l_1-l\right)}{mg}} \\ =2蟺\sqrt{\frac{\left(l_1-l\right)}{g}} \end{gathered}$$

Here, T is the time taken by the block, m is the mass of the block, g is the acceleration due to gravity,/ is the unstretched length of the spring and$l_1$is the initial stretched length of the spring.

Substitute the values in the above equation.

$$\begin{gathered} T=2\left(3.14\right)\sqrt{\frac{\left(0.4\mathrm{m}-0.32\mathrm{m}\right)}{9.8\mathrm{m}/{\mathrm{s}}^2}} \\ =\left(6.28\right)\sqrt{\frac{\left(0.08\mathrm{m}\right)}{9.8\mathrm{m}/{\mathrm{s}}^2}} \\ =\left(6.28\right)\sqrt{8.16脳{10}^{-3}{\mathrm{s}}^2} \\ =\left(6.28\right)\left(0.0903\mathrm{s}\right) \\ =0.567\mathrm{s} \end{gathered}$$

Thus, the block needs 0.567 s to return from where it was released.

According to the law of conservation of momentum, the equation of the speed of the block is expressed as:

$m{v}_1+m_{1}v=\left(m+m_1\right)v_2$

Here,$m_1$ is the mass of the bullet,$v_1$ is the initial and v is the final velocity of the bullet.$v_2$ is the speed of the block.

As initially the gun was at rest, then the initial velocity of the bullet is zero.

Substitute 0 for $v_1$in the above equation.

$$\begin{gathered} m\left(0\right)+m_{1}v=\left(m+m_1\right)v_2 \\ {v}_2=\frac{m_{1}v}{\left(m+m_1\right)} \end{gathered}$$

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{v}}_2=\frac{\left(0.003\mathrm{kg}\right)\left(200\mathrm{m}/\mathrm{s}\right)}{\left(0.003\mathrm{kg}+0.2\mathrm{kg}\right)} \\ =\frac{\left(0.6\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)}{\left(0.203\mathrm{kg}\right)} \\ =2.95\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the block is 2.95 m/s.

According to the law of the conservation of energy, the equation of the kinetic energy is expressed as:

$K{E}_f-K{E}_i\left(m_1+m_2\right)gh+\frac{1}{2}k{\left(鈭�x\right)}^2=0$

Here,$K{E}_f$is the final and $K{E}_i$is the initial kinetic energy.k is the spring constant$鈭�x$and is the change in the length of the block.

$$\begin{gathered} \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{diagram},\mathrm{substitute}0.4\mathrm{m}-\mathrm{h}\mathrm{for}鈭�\mathrm{x},0\mathrm{for}{\mathrm{KE}}_{\mathrm{f}}\mathrm{and}\frac{1}{2}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right){\mathrm{v}}_{\mathrm{f}}^2 \\ \mathrm{where}{\mathrm{v}}_{\mathrm{f}}\mathrm{is}\mathrm{the}\mathrm{final}\mathrm{velocity}\mathrm{for}{\mathrm{KE}}_{\mathrm{i}}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ 0-\frac{1}{2}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right){\mathrm{v}}_{\mathrm{f}}^2+\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{gh}+\frac{1}{2}\mathrm{k}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right){\mathrm{v}}_{\mathrm{f}}^2=0 \\ \left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{gh}+\frac{1}{2}\mathrm{k}{\left(0.4\mathrm{m}-\mathrm{h}\right)}^2-\frac{1}{2}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right){\mathrm{v}}_{\mathrm{f}}^2=0 \\ \mathrm{Thus},\mathrm{the}\mathrm{equation}\mathrm{is}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{gh}+\frac{1}{2}\mathrm{k}{\left(0.4\mathrm{m}-\mathrm{h}\right)}^2-\frac{1}{2}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right){\mathrm{v}}_{\mathrm{f}}^2=0. \end{gathered}$$