Question

A bullet of mass 0.102 kg traveling horizontally at a speed of 300 m/s embeds itself in a block of mass 3.5 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? (b) Calculate the kinetic energy of the bullet plus the block before the collision. (c) Calculate the kinetic energy of the bullet plus the block after the collision. (d) Was this collision elastic or inelastic? (e) Calculate the rise in internal energy of the bullet plus block as a result of the collision.

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)8.495\mathrm{m}/\mathrm{s} \\ \left(\mathrm{b}\right)4590\mathrm{J} \\ \left(\mathrm{c}\right)129.969\mathrm{J} \\ \left(\mathrm{d}\right)\mathrm{inelastic} \\ \left(\mathrm{e}\right)4460.031\mathrm{J} \end{gathered}$$

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The mass of the bullet is, $m_1=0.102\mathrm{kg}$
• The initial velocity of the bullet is,$v_1=300\mathrm{m}/\mathrm{s}$
• The mass of the block is, $m_{2_{}}=3.5\mathrm{kg}$
• The initial velocity of the block is,$v_2=0$as it was at rest.

Significance of the law of conservation, work-energy principle of momentum, and the first law of thermodynamics

The law of conservation of momentum states that the momentum of a system before and after collision remains constant if no external force acts.

The first law of thermodynamics states that the change in the internal energy is equal to the difference between the heat added to the system and the work done by the system.

The work energy principle states that the work done by a system is the change in the kinetic energy of the system.

Determination of the speed of the block the bullet gets inserted in it

(a)

The expression for the initial momentum of the bullet-block system can be expressed as,

$$\begin{gathered} {\mathrm{p}}_{\mathrm{initial}}={\mathrm{p}}_{\mathrm{bullet},\mathrm{initial}}+{\mathrm{p}}_{\mathrm{block}\mathrm{initial}} \\ ={\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2 \end{gathered}$$

Here, ${\mathrm{p}}_{\mathrm{bullet},\mathrm{initial}}$and ${\mathrm{p}}_{\mathrm{block}\mathrm{initial}}$are the initial momentum of the bullet and the block respectively, $m_1$and $m_2$are the mass of the bullet and the block respectively and $v_1$and $v_2$are the velocities of the bullet and the block respectively.

For ${\mathrm{m}}_1=\left(0.102\mathrm{kg}\right),{\mathrm{v}}_1=\left(300\mathrm{m}/\mathrm{s}\right),{\mathrm{m}}_{2_{}}=\left(3.5\mathrm{kg}\right)\mathrm{and}{\mathrm{v}}_2=\left(0\mathrm{m}/\mathrm{s}\right)$

$$\begin{gathered} {\mathrm{p}}_{\mathrm{initial}}=\left(0.102\mathrm{kg}\right)\times \left(300\mathrm{m}/\mathrm{s}\right)+\left(3.5\mathrm{kg}\right)\times \left(0\mathrm{m}/\mathrm{s}\right) \\ =30.6\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

According to the law of conservation of momentum, the initial and the final momentum of a system is same if no external force acts, thus, $p_{initial}=p_{final}$.

The expression for the final momentum of the system can be expressed as,

$$\begin{gathered} p_{final}=v\left(m_1+m_2\right) \\ {p}_{inital}=v\left(m_1+m_2\right) \end{gathered}$$

Here, v is the combined velocity of the bullet and the block.

For$m_1=\left(0.102\mathrm{kg}\right),m_2=\left(3.5\mathrm{kg}\right)\mathrm{and}{p}_{initial}=30.6\mathrm{kg}.\mathrm{m}/\mathrm{s}$

$$\begin{gathered} 30.6\mathrm{kg}.\mathrm{m}/\mathrm{s}=\mathrm{v}\left(0.102\mathrm{kg}+3.5\mathrm{kg}\right) \\ \mathrm{v}=\frac{30.6\mathrm{kg}.\mathrm{m}/\mathrm{s}}{3.602\mathrm{kg}} \\ =8.495\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the block after the bullet embeds itself in the block is $8.495\mathrm{m}/\mathrm{s}$.

Determination of the kinetic energy of the system before collision

(b)

The expression for the kinetic energy of the bullet plus block before the collision is expressed as,

$K.E.=\frac{1}{2}\left(M_1{V_1}^2+M_2{V_2}^2\right)$

For $$\begin{gathered} {\mathrm{m}}_1=\left(0.102\mathrm{kg}\right),{\mathrm{v}}_1=\left(300\mathrm{m}/\mathrm{s}\right),{\mathrm{m}}_2=\left(3.5\mathrm{kg}\right)\mathrm{and}{\mathrm{v}}_2=\left(0\mathrm{m}/\mathrm{s}\right) \\ \mathrm{K}.\mathrm{E}.=\frac{1}{2}\left(\left(0.102\mathrm{kg}\right)\times {\left(300\mathrm{m}/\mathrm{s}\right)}^2+\left(3.5\mathrm{kg}\right)\times {\left(0\mathrm{m}/\mathrm{s}\right)}^2\right) \\ =\frac{1}{2}\left(0.102\mathrm{kg}\right)\left(90000{\mathrm{m}}^2/{\mathrm{s}}^2\right) \\ =4590\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =4590\mathrm{J} \end{gathered}$$.

Thus, the kinetic energy of the bullet plus the block before the collision is $4590\mathrm{J}$.

Determination of the kinetic energy of the system after collision

(c)

The expression for the kinetic energy of the bullet plus block after the collision is expressed as,

$K.E.=\frac{1}{2}\left(m_1+m_2\right)v^2$

For $$\begin{gathered} m_1=\left(0.102\mathrm{kg}\right),m_2=\left(3.5\mathrm{kg}\right)\mathrm{and}\mathrm{v}=\left(8.495\mathrm{m}/\mathrm{s}\right) \\ \mathrm{K}.\mathrm{E}.=\frac{1}{2}\left(\left(0.102\mathrm{kg}\right)+\left(3.5\mathrm{kg}\right)\right)\times {\left(8.495\mathrm{m}/\mathrm{s}\right)}^2 \\ =\frac{1}{2}\left(3.602\mathrm{kg}\right)\left(72.165{\mathrm{m}}^2/{\mathrm{s}}^2\right) \\ =129.969\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{J}}{1{\mathrm{kgm}}^2/{\mathrm{s}}^2} \\ =129.969\mathrm{J} \end{gathered}$$.

Thus, the kinetic energy of the bullet plus the block after the collision is $129.969\mathrm{J}$.

Identification of the type of the collision

(d)

In an elastic collision, the total momentum and the total kinetic energy of a system is conserved and in inelastic collision, the kinetic energy is not conserved. However, it has been observed that, the kinetic energy before and after the collision for the bullet and the block is not same. Hence, it can be concluded that this collision is inelastic.

Thus, the collision is inelastic.

Determination of the rise in the internal energy

(e)

From the first law of thermodynamics, the equation of the change in the internal energy can be expressed as,

$∆I=Q-W$

Here, Q and W are the heat added and the work done by the system.

As this system is an isolated system, hence the heat added to the system is 0 and according to the work-energy theorem, the work done is the change in the kinetic energy of the system.

Hence, the above equation can be expressed as,

$∆I=0-\left(K.E{.}_{final}-K.E{.}_{initial}\right)$

For$$\begin{gathered} K.E{.}_{final}=129.969\mathrm{J}\mathrm{and}K.E{.}_{initial}=4590\mathrm{J} \\ ∆I=0-\left(129.969\mathrm{J}-4590\mathrm{J}\right) \\ =4460.031\mathrm{J} \end{gathered}$$

Thus, the rise in the internal energy of the bullet plus block as a result of the collision is $4460.031\mathrm{J}$.