Question

On a muddy football field, $110kg$ linebacker tackles an $85kg$ half back. Immediately before collision, the line backer is slipping with a velocity of $8.8m{{s}^{-1}}$ North and the half back is sliding with a velocity of $7.2m{{s}^{-1}}$ east. What is the velocity at which two players move immediately after collision?
$A.5.9m{{s}^{-1}}$
$B.6.8m{{s}^{-1}}$
$C.8.0m{{s}^{-1}}$
$D.7.6m{{s}^{-1}}$

Answer 1

We will apply the concept of vectors in terms of directions. Formula of resultant vectors will also be used in order to solve the required problem. Principle of conservation of momentum will be used. Total momentum before collision is always equal to total momentum after collision.
Formula used: We are using the following formula to get the correct solution:-
$({{m}_{1}}+{{m}_{2}})\overrightarrow{v}=({{m}_{1}}{{v}_{1}})\widehat{i}+\left( {{m}_{2}}{{v}_{2}} \right)\widehat{j}$.

Complete step by step solution:
From the problem given above we have the following parameters with us:-
Mass of linebacker before, ${{m}_{1}}=110kg$
Mass of half back, ${{m}_{2}}=85kg$
Velocity of linebacker before collision, ${{v}_{1}}=8.8m{{s}^{-1}}$.
Velocity of half back before collision, ${{v}_{2}}=7.2m{{s}^{-1}}$.
Now, let the velocity of two players just after collision be $\overrightarrow{v}$which is to be calculated.
Principle of conservation of momentum states that the total momentum before collision is equal to the total momentum after collision.
Using the principle of conservation of momentum we have
$({{m}_{1}}+{{m}_{2}})\overrightarrow{v}=({{m}_{1}}{{v}_{1}})\widehat{i}+\left( {{m}_{2}}{{v}_{2}} \right)\widehat{j}$……………. $(i)$
We consider the direction of half back towards east in positive x-direction and that of linebacker towards north in positive y-direction.
Putting values of different given parameters in equation $(i)$ we get
$\left( 110+85 \right)\overrightarrow{v}=(85\times 7.2)\widehat{i}+(110\times 8.8)\widehat{j}$
$\Rightarrow 195\overrightarrow{v}=968\widehat{i}+612\overset\frown{j}$
Magnitude of resultant of velocity can be calculated by the relation, $\overrightarrow{\left| v \right|}=\sqrt{{{a}^{2}}+{{b}^{2}}}$. This relation is used to find the magnitude of any vector in the x-y plane as they are perpendicular to each other.
Now, we have to find the magnitude of the velocity, $\overrightarrow{v}$ as follows:-
$195\left| \overrightarrow{v} \right|=\sqrt{{{968}^{2}}+{{612}^{2}}}$
$\Rightarrow \left| \overrightarrow{v} \right|=\dfrac{\sqrt{937024+374544}}{195}$
$\Rightarrow \left| \overrightarrow{v} \right|=\dfrac{\sqrt{1311568}}{195}$
Solving further we get
$\left| \overrightarrow{v} \right|=\dfrac{1145.2}{195}$……………. $(ii)$
We can write the equation $(ii)$as follows by the use of approximation:-
$\left| \overrightarrow{v} \right|=\dfrac{1145}{195}$
$\Rightarrow \left| \overrightarrow{v} \right|=5.9m{{s}^{-1}}$
This is the required magnitude of the velocity at which two players will move immediately after collision.

Therefore, we get the correct answer that is option, $(A)$.

Note:
We should apply the principle of conservation of momentum in terms of vectors. Formulas of finding the magnitude of a vector should be applied carefully. We should also take care of our calculations as this problem involves some tough calculations.