Question

A 20 0 kg body is moving through space in the positive direction of an x axis with a speed of 200 m/swhen, due to an internal explosion, it breaks into three parts. One part, with a mass of 10.0 kg, moves away from the point of explosion with a speed of 100 m/sin the positive y direction. A second part, with a mass of 4.00 kg, moves in the negative x direction with a speed of 500 m/s. (a) In unit-vector notation, what is the velocity of the third part? (b) How much energy is released in the explosion? Ignore effects due to the gravitational force.

Short answer

1. The velocity of third part in unit vector notation is,${\overrightarrow{\mathrm{v}}}_3=\left(1.00\hat{\mathrm{i}}-0.167\hat{\mathrm{j}}\right)\mathrm{km}/\mathrm{s}$
2. The energy released in the explosion is,$∆\mathrm{K}=3.23\times {10}^6\mathrm{J}$

Step-by-step solution

Step 1: Given Data

The mass of the original body is, M = 20.0 kg .

Initial velocity of the body is,${\overrightarrow{v}}_0=200\hat{i}$ .

The mass of the fragment is,$m_1=10.0kg$ .

The velocity of the fragment is, ${\overrightarrow{v}}_1=100\hat{j}$.

The mass of the second fragment is,${\mathrm{m}}_2=4.0\mathrm{kg}$ .

The velocity of the second fragment is,${\overrightarrow{v}}_2=-500\hat{i}$ .

Determining the concept

By using the conservation of linear momentum, find the velocityof the third part in unit vector notation. Also, by using equation for energy released, find the energy released in the explosion.

Formulae are as follow:

Conservation of linear momentum is,$M{\overrightarrow{v}}_0=m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2+m_3{\overrightarrow{v}}_3........\left(1\right)$

The energy released is,$∆\mathrm{K}=K_f-K_i,\dots \left(2\right)$

where,$M,m_1,m_2,m_3$ are masses, ${\overrightarrow{v}}_0,{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3$are velocities and $∆\mathrm{K},K_f,K_i$are kinetic energies.

(a) Determining the velocity of the third part in unit vector notation

The mass of the third fragment is,

$\begin{array}{l}M=m_1+m_2+m_3\\ m_3=M-m_1-m_2\\ m_3=20.0-10.0-4.0\\ m_3=6.00\mathrm{kg}\end{array}$

According to conservation of linear momentum,

$M{\overrightarrow{v}}_0=m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2+m_3{\overrightarrow{v}}_3$

The energy released in explosion equal to, the change in kinetic energy.

Using the above momentum conservation equation,

$$\begin{gathered} {\overrightarrow{v}}_3=\frac{M{\overrightarrow{v}}_0-m_1{\overrightarrow{v}}_1-m_2{\overrightarrow{v}}_2}{m_3} \\ {\overrightarrow{v}}_3=\frac{20.0\times 200\hat{i}-10.0\times 100\hat{j}-4.0\times \left(-500\right)\hat{i}}{6.00} \\ {\overrightarrow{v}}_3=\left(1.00\times {10}^3\right)\hat{i}-\left(0.167\times {10}^3\right)\hat{j}\mathrm{m}/\mathrm{s} \\ {\overrightarrow{\mathrm{v}}}_3=\left(1.00-0.167\hat{\mathrm{j}}\right)\mathrm{km}/\mathrm{s} \end{gathered}$$

Thus, the magnitude of$v_3$ is,

$$\begin{gathered} v_3=\sqrt{{\left(1.00\times {10}^3\right)}^2+{\left(-0.167\times {10}^3\right)}^2} \\ {v}_3=1.0138\times {10}^3\mathrm{m}/\mathrm{s} \end{gathered}$$

.It points at,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{-167}{1000}\right) \\ \theta =-9.48° \end{gathered}$$

That is, at $9.5°$measured clockwise from the +x axis.

Hence, the velocity of the third part is${\overrightarrow{v}}_3=\left(1.00\hat{i}-0.167\hat{j}\right)km/s$

(b) Determining the energy released in the explosion

The energy released is $∆K$:

$$\begin{gathered} ∆\mathrm{K}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}} \\ ∆\mathrm{K}=\left(\frac{1}{2}{\mathrm{m}}_1{\mathrm{v}}_1^2+\frac{1}{2}{\mathrm{m}}_2{\mathrm{v}}_2^2+\frac{1}{2}{\mathrm{m}}_3{\mathrm{v}}_3^2\right)-\frac{1}{2}{\mathrm{Mv}}_0^2 \\ ∆\mathrm{K}=\left(\frac{1}{2}\times 10.0\times {100}^2+\frac{1}{2}\times 4.00\times {\left(-500\right)}^2+\frac{1}{2}\times 6.00\times {\left(1.0138\times {10}^3\right)}^2\right)-\frac{1}{2}\times 20.0\times {200}^2 \\ ∆\mathrm{K}=3.23\times {10}^6\mathrm{J} \end{gathered}$$

Hence,the energy released in the explosionis$3.23\times {10}^6\mathrm{J}$

Therefore, by using the conservation of linear momentum and law of conservation of energy, the velocity and energy released in the collision can be found.