An object is moving with constant momentum $\left\{10,-12,-8\right\}\mathbf{Kg}\mathbf{\times }\mathbf{m}\mathbf{/}\mathbf{s}$. What is the rate of change of momentum$\mathbf{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}\mathbf{/}\mathbf{dt}$? What is the net force acting on the object? The object is acted on by three objects in the surroundings. Two of the forces are $\left\{100,50,-60N\right\}$and $\left\{-130,40,-70\right\}\mathbf{N}$. What is the third force?

Given that an object with a constant momentum. Also, it experiences three forces

The momentum of a system in uniform motion is constant so it does not change with time and the derivative of the momentum to the time equals zero.

$\frac{\mathrm{d}\overrightarrow{\mathrm{p}}}{\mathrm{dt}}=0$

The form of the momentum principle, the next force exerted on the object must equals the change in the momentum and as the change in the momentum is zero, therefore, the net force exerted on the object is zero.

${\overrightarrow{\mathrm{F}}}_{\mathrm{net}}=\frac{\mathrm{d}\overrightarrow{\mathrm{p}}}{\mathrm{dt}}=0$

The result of the net force will help us to get the third force on the object. The system here is the object and the surroundings are ${\mathrm{F}}_1,{\mathrm{F}}_2\mathrm{and}{\mathrm{F}}_3$The net force on the object equals the summation of the three forces and it is given by,

${\mathrm{F}}_{\mathrm{net}}={\mathrm{F}}_1+{\mathrm{F}}_2+{\mathrm{F}}_3$

Isolate $F_3$form the above equation,

${\mathrm{F}}_3={\mathrm{F}}_{\mathrm{net}}-{\mathrm{F}}_1-{\mathrm{F}}_2$

Substitute ${\mathrm{F}}_{\mathrm{net}}=0$and the given vectors role="math" localid="1656677597132" into the obtained equation,

Therefore, the rate of change of momentum is $\frac{d\overrightarrow{p}}{dt}=0$, the net force acting on the object is role="math" localid="1656677706924" $F_{net}=0$and the third force is