Question

A body is traveling at $\mathbf{2}\mathbf{.}\mathbf{0}\frac{\mathbf{m}}{\mathbf{s}}$ along the positive direction of an xaxis; no net force acts on the body. An internal explosion separates the body into two parts, each of 4.0 kg, and increases the total kinetic energy by 16 J. The forward part continues to move in the original direction of motion. What are the speeds of (a) the rear part and (b) the forward part?

Short answer

a) Speed for rear part is $0\frac{\mathrm{m}}{\mathrm{s}}$.

b) Speed for forward part is $4.0\frac{\mathrm{m}}{\mathrm{s}}$.

Step-by-step solution

Step 1: Given

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 $m_1=m_2=4\mathrm{kg}$.

Determine the formula for the conservation of energy.

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.

(a) Determine the speed for rear part

According to law of conservation of momentum.

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

Substitute the values and solve as:

$8\times 2=4(v_f+v_r)$

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

Now,

$\Delta K=16$

$K_f-K_i=16$

Substitute the values and solve as:

$0.5\times m_1(v_f^2+v_r^2)-0.5\times M\times v_0^2=16$

$0.5\times 4\times (v_f^2+v_r^2)-0.5\times 8\times 2^2=16$

$v_f^2+v_r^2=16$ …… (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 $v_r=0\frac{\mathrm{m}}{\mathrm{s}}$.

(b) Determine the speed for forward part

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.