Question

A spring has an unstretched length of 0.32 m. a block with mass 0.2 kg is hung at rest from the spring, and the spring becomes 0.4 mlong.Next the spring is stretched to a length of 0.43 mand the block is released from rest. Air resistance is negligible.

(a) How long does it take for the block to return to where it was released? (b) Next the block is again positioned at rest, hanging from the spring (0.4 m long) as shown in Figure 10.43. A bullet of mass 0.003 kg traveling at a speed of 200 m/s straight upward buries itself in the block, which then reaches a maximum height above its original position. What is the speed of the block immediately after the bullet hits? (c) Now write an equation that could be used to determine how high the block goes after being hit by the bullet (a height h), but you need not actually solve for h.

Short answer

(a) The block needs 0.567 s to return to where it was released.

(b) The speed of the block is 2.95 m/s.

(c) The equation is $\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{gh}+\frac{1}{2}\mathrm{k}{\left(0.4\mathrm{m}-\mathrm{h}\right)}^2-\frac{1}{2}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right){\mathrm{v}}_{\mathrm{f}}^2=0.$.

Step-by-step solution

Identification of the given data

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

Significance of the collision

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.

(a) Determination of the time

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

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

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.

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

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

(b) Determination of the speed of the block

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.

(c) Determination of the equation

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

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

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

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