91Ó°ÊÓ

i) Mass of object is $M=8\mathrm{kg}$.

ii) Speed of object is $v_0=2\frac{\mathrm{m}}{\mathrm{s}}$.

iii) Increase in kinetic energy is $∆K=16\mathrm{J}$.

iv) Mass of each part after collision is $m_1=m_2=4\mathrm{kg}$.

Using the law of conservation of energy, find the relation between velocity of front and rear parts. Then, using the formula for increase in K.E, find another equation. Solving these two equations, the velocities of the front and rear part can be found.

Formulae are as follow:

${\mathit{m}}_{\mathbf{1}}{\mathit{u}}_{\mathbf{1}}\mathbf{+}{\mathit{m}}_{\mathbf{2}}{\mathit{u}}_{\mathbf{2}}\mathbf{=}{\mathit{m}}_{\mathbf{1}}{\mathit{v}}_{\mathbf{1}}\mathbf{+}{\mathit{m}}_{\mathbf{2}}{\mathit{v}}_{\mathbf{2}}$

$KE=\frac{1}{2}m{v}^2$

Here, m₁, m₂ are masses, u₁, u₂ are initial velocities, v₁, v₂ are final velocities, m is mass and KE is kinetic energy.

According to law of conservation of momentum.

$M{v}_0=(m_1{v}_f+m_2{v}_r)$

Substitute the values and solve as:

$8×2=4(v_f+v_r)$

$v_f+v_r=4\frac{\mathrm{m}}{\mathrm{s}}$ ……. (1)

Now,

$\mathrm{Δ}K=16$

$K_f-K_i=16$

Substitute the values and solve as:

$0.5×m_1(v_f^2+v_r^2)-0.5×M×v_0^2=16$

$0.5×4×(v_f^2+v_r^2)-0.5×8×2^2=16$

$v_f^2+v_r^2=16$ …… (2)

Now from equation (1) solve as:

$v_f=4-v_r$

Plugging this in equation (2) and solve as:

$(4-v_r{)}^2+v_r^2=16$

$16-8v_r+v_r^2+v_r^2=16$

$2v_r^2-8v_r=0$

$v_r=0or4\frac{\mathrm{m}}{\mathrm{s}}$

Hence, rear part will have velocity $v_r=0\frac{\mathrm{m}}{\mathrm{s}}$.

Substitute the values in equation as follows:

$$\begin{gathered} v_f+v_r=4v_f+v_r=4 \\ {v}_f+4=4 \\ {v}_f=0\frac{m}{s} \end{gathered}$$

Or,

$$\begin{gathered} v_f+0=4 \\ {v}_f=4\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Hence, front part will continue to move in the forward direction, it will have velocity$v_f=4\frac{m}{s}$.

Therefore, the velocity of objects after explosion can be found from the initial velocity of the system and increase in its K.E can be found using the law of conservation of energy.