91Ó°ÊÓ

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

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

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

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

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

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

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,$M{\stackrel{→}{v}}_0=m_1{\stackrel{→}{v}}_1+m_2{\stackrel{→}{v}}_2+m_3{\stackrel{→}{v}}_3........\left(1\right)$

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

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

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,

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

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

Using the above momentum conservation equation,

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

Thus, the magnitude of$v_3$ is,

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

.It points at,

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

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

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

The energy released is $∆K$:

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

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

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.