Question

A 4.0 kg mess kit sliding on a frictionless surface explodes into two 2.0 kgparts: 3.0 m/sdue north, and 5.0 m/s, 30$°$north of east. What is the original speed of the mess kit?

Short answer

The original speed of the mess kit is, v = 3.5 m/s

Step-by-step solution

Step 1: Given Data

Total mass of the kit is, M = 4.0 kg

After explosion the mass of each part is, m = 2.0 kg

After explosion speed in north direction is,${\mathrm{v}}_1=3.0\mathrm{m}/\mathrm{s}$

After explosion speed in north-east direction is,

Angle is,$\theta =30°$

Determining the concept

By using the formula for linear momentum, find the linear momentum of each part and adding them, total linear momentum can be found. By dividing the total mass, the original velocityof the mess kit can be found. According to conservation of linear momentum, momentum that characterizes motion never changes in an isolated collection of objects.

Formula is as follow:

The linear momentum,$\overrightarrow{P}=m\overrightarrow{v}$

where,$\overrightarrow{P}$is linear momentum, m is mass and $\overrightarrow{v}$is velocity.

Determining the original speed of the mess kit

Our + x direction is east and + y direction is north.

The linear momentum for the two m = 20 kg parts are,

$$\begin{gathered} {\overrightarrow{p}}_1=m{\overrightarrow{v}}_1=m{v}_1\hat{j} \\ \mathrm{where},{\mathrm{v}}_1=3.0\mathrm{m}/\mathrm{s},\mathrm{and} \\ {\overrightarrow{\mathrm{p}}}_2=\mathrm{m}{\overrightarrow{\mathrm{v}}}_2 \end{gathered}$$

But, puttingrole="math" localid="1661239882899" ${\overrightarrow{v}}_1=v_{2x}\hat{i}+v_{2y}\hat{j}$

$$\begin{gathered} \overrightarrow{p_2}=m(v_{2x}\hat{i}+v_{2y}\hat{j} \\ \overrightarrow{p_2}=m{v}_2\left(\cos \theta \hat{i}+\sin \theta \hat{j}\right) \end{gathered}$$

where, $v_2=5.0\mathrm{m}/\mathrm{s}\mathrm{and}\mathrm{\theta }=30°$. The combined linear momentum of both parts is,

$$\begin{gathered} \overrightarrow{P}=\overrightarrow{p_1}+\overrightarrow{p_2} \\ \overrightarrow{P}=m{v}_1\hat{j}+m{v}_2\left(\cos \theta \hat{i}\sin \theta \hat{j}\right) \\ \overrightarrow{P}=(m{v}_2\cos \theta )\hat{i}+\left(m{v}_1+m{v}_2\sin \theta \right)\hat{j} \\ \overrightarrow{P}=\left(2.0\times 5.0\times \cos 30°\right)\hat{i}+\left(2.0\times 3.0+2.0\times 5.0\times \sin 30°\right) \\ \overrightarrow{P}=\left(8.66\hat{i}+11\hat{j}\right)\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

From the conservation of linear momentum, this is also the linear momentum of the whole kit before it splits. Thus, the speed of the $4.0kg$kit is,

role="math" localid="1661240710497" $$\begin{gathered} v=\frac{P}{M} \\ v=\frac{\sqrt{P_x^2+P_y^2}}{M} \\ v=\frac{\sqrt{8.{66}^2+{11}^2}}{4.0} \\ v=3.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore,the original speed of the mess kit is, v = 3.5 m/s

Thus, by using the linear momentum formula, the velocity can be found.