91Ó°ÊÓ

The given data can be listed below as-

• The mass of the object A is${\mathrm{m}}_{\mathrm{A}}=8\mathrm{kg}$ .

• The initial momentum of the object A is ${\stackrel{\mathrm{s}}{\mathrm{p}}}_{\mathrm{A},\mathrm{i}}=(20,−5,0)\mathrm{kg}×\mathrm{m}/\mathrm{s}$.

• The mass of the object B is ${\mathrm{m}}_{\mathrm{B}}=11\mathrm{kg}$.

• The initial momentum of the object B is ${\stackrel{s}{p}}_{B,i}=(5,6,0)\mathrm{kg}×\mathrm{m}/\mathrm{s}$.

This law illustrates that the total momentum of a body before and after collision remains equal if no external force gets involved.

The equation of the momentum gives the total initial momentum, impulse applied, final momentum and the momentum of object B.

a) From the law of conservation of energy, the equation of the total momentum of the system is expressed as-

${\mathrm{p}}_{\mathrm{i}}={\mathrm{p}}_1+{\mathrm{p}}_2$

Here,${\mathrm{p}}_{\mathrm{i}}$ is the total initial momentum of the system, ${\mathrm{p}}_1$and ${\mathrm{p}}_2$are the initial momentum of the object A and object B respectively.

Substituting the values in the above equation, we get-

${\mathrm{p}}_{\mathrm{i}}=(20,−5,0)\mathrm{kg}×\mathrm{m}/\mathrm{s}+(5,6,0)\mathrm{kg}×\mathrm{m}/\mathrm{s}$

${\mathrm{p}}_{\mathrm{j}}=(25,1,0)\mathrm{kg}×\mathrm{m}/\mathrm{s}$

Thus, the total initial momentum of this system just before the collision is $(25,1,0)\raisebox{1ex}{$\mathrm{kg}×\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

b) From the law of conservation of momentum, the equation of the impulse applied on the two-object system can be expressed as-

$\mathrm{I}={\mathrm{F}}_{\text{net}}\mathrm{t}$

Here,$\mathrm{I}$ is the total impulse applied on the two-object system,${\mathrm{F}}_{\mathrm{net}}$ is the net external force on the objects that is and t is the duration of the collision.

Substituting the values in the above equation, we get-

$\mathrm{I}=0\mathrm{N}×1\mathrm{s}$

$=0\mathrm{N}×\mathrm{S}$

Thus, the approximate value of the impulse applied to the two-object system due to forces exerted on the system by objects outside the system is $0\mathrm{N}×\mathrm{S}$.

c) From the law of conservation of momentum, the equation of the impulse applied on the two-object system can be expressed as-

$\mathrm{l}={\mathrm{p}}_{\mathrm{f}}−{\mathrm{p}}_{\mathrm{i}}$

Here, I is the impulse applied on the system${\mathrm{p}}_{\mathrm{i}}$ and${\mathrm{p}}_{\mathrm{f}}$ are the initial and the final momentum of the system respectively.

Substituting the values in the above equation, we get-

$0={\mathrm{P}}_{\mathrm{f}}-{\mathrm{P}}_{\mathrm{i}}$

${\mathrm{P}}_{\mathrm{f}}={\mathrm{P}}_{\mathrm{i}}$

Thus, the total final momentum of the system just after the collision is .$(25,1,0)\raisebox{1ex}{$\mathrm{kg}×\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

d) From the law of the conservation of momentum, the equation of the total momentum of the system can be expressed as-

${\mathrm{P}}_{\mathrm{f}}={\mathrm{P}}_{\mathrm{Af}}+{\mathrm{P}}_{\mathrm{Bf}}$

Here,${\mathrm{P}}_{\mathrm{f}}$ is the total momentum of the system,${\mathrm{P}}_{\mathrm{Af}}$ is the momentum of the object A that is and${\mathrm{P}}_{\mathrm{Bf}}$ is the momentum of the object B.

Substituting the values in the above equation, we get-

role="math" localid="1658149428363" $(25,1,0)\raisebox{1ex}{$\mathrm{kg}×\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.=(18,5,0)\raisebox{1ex}{$\mathrm{kg}×\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.+{\mathrm{P}}_{\mathrm{Bf}}$

${\mathrm{P}}_{\mathrm{Bf}}=(7,-4,0)\raisebox{1ex}{$\mathrm{kg}×\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

Thus, the momentum of the object B just after the collision is $(7,-4,0)\raisebox{1ex}{$\mathrm{kg}×\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.