Answer
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Hint: Use the formula of kinetic energy ${\text{K}}{\text{.E}}{\text{. = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{m}}{{\text{v}}^{\text{2}}}$and then using this formula derive the formula for momentum by multiplying and dividing by m. As momentum p = mv. From here, we can deduce the momentum of the truck. Then similarly deduce the momentum of the bike. Finally, evaluate the ratio of their momentum.
Complete step by step solution:
Given: Kinetic energy of the truck = Kinetic energy of the bike
Mathematically ${\text{K}}{\text{.E}}{\text{.(truck) = K}}{\text{.E}}{\text{.(bike)}}$
Kinetic energy of truck is $\dfrac{{\text{1}}}{{\text{2}}}{{\text{m}}_{\text{1}}}{{\text{v}}_{\text{1}}}^{\text{2}}$
Multiplying and dividing by m1, we get
$\dfrac{{\text{1}}}{{\text{2}}}\dfrac{{{{\text{m}}_{\text{1}}}^{\text{2}}{{\text{v}}_{\text{1}}}^{\text{2}}}}{{{{\text{m}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{p}}_1}^{\text{2}}}}{{{\text{2}}{{\text{m}}_{\text{1}}}}}$
Kinetic energy of bike is $\dfrac{{\text{1}}}{{\text{2}}}{{\text{m}}_2}{{\text{v}}_2}^{\text{2}}$
Multiplying and dividing by m2, we get
$\dfrac{{\text{1}}}{{\text{2}}}\dfrac{{{{\text{m}}_2}^{\text{2}}{{\text{v}}_2}^{\text{2}}}}{{{{\text{m}}_2}}}{\text{ = }}\dfrac{{{{\text{p}}_2}^{\text{2}}}}{{{\text{2}}{{\text{m}}_2}}}$
$\therefore {\text{ }}\dfrac{{{{\text{p}}_{\text{1}}}^{\text{2}}}}{{{\text{2}}{{\text{m}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{p}}_{\text{2}}}^{\text{2}}}}{{{\text{2}}{{\text{m}}_{\text{2}}}}}...{\text{(i)}}$
Where $m_1$ = Mass of the truck
$v_1$ = Velocity of the truck
$m_2$ = Mass of the bike
$v_2$ = Velocity of the bike
$p_1$ = momenta of truck
$p_2$ = momenta of bike
Mass of the truck is four times that of the bike i.e.,
${\text{4}}{{\text{m}}_{\text{1}}}{\text{ = }}{{\text{m}}_{\text{2}}}...{\text{(ii)}}$
On substituting the value of m1 from (ii) to (i), we get
\[
\dfrac{{{{\text{p}}_{\text{1}}}^{\text{2}}}}{{{\text{2}}{{\text{m}}_2}}}{\text{ = }}\dfrac{{{{\text{p}}_{\text{2}}}^{\text{2}}}}{{{\text{8}}{{\text{m}}_{\text{2}}}}} \\
\Rightarrow \dfrac{{{{\text{p}}_{\text{1}}}^{\text{2}}}}{1}{\text{ = }}\dfrac{{{{\text{p}}_{\text{2}}}^{\text{2}}}}{4} \\
\therefore \dfrac{{{{\text{p}}_{\text{1}}}}}{{{{\text{p}}_{\text{2}}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}} \\
\]
Therefore, the ratio of momenta of truck and bike is 1:2.
Note: Kinetic energy is the energy possessed by the particle due to its motion. The formula of kinetic energy is ${\text{K}}{\text{.E}}{\text{. = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{m}}{{\text{v}}^{\text{2}}}$ where m = mass of an object and v = velocity of an object. SI unit of the mass is $Kg$ and the SI unit of the velocity is $m/ s^2$. Thus, SI unit of kinetic energy is Joule and $1 Joule$ = $1 kg m/ s^2$. Momentum of an object is equal to the mass of the object times the velocity of the object. The formula for momentum is P = mv where m = mass of an object and v = velocity. SI unit of momentum is $kg m/s$.
Complete step by step solution:
Given: Kinetic energy of the truck = Kinetic energy of the bike
Mathematically ${\text{K}}{\text{.E}}{\text{.(truck) = K}}{\text{.E}}{\text{.(bike)}}$
Kinetic energy of truck is $\dfrac{{\text{1}}}{{\text{2}}}{{\text{m}}_{\text{1}}}{{\text{v}}_{\text{1}}}^{\text{2}}$
Multiplying and dividing by m1, we get
$\dfrac{{\text{1}}}{{\text{2}}}\dfrac{{{{\text{m}}_{\text{1}}}^{\text{2}}{{\text{v}}_{\text{1}}}^{\text{2}}}}{{{{\text{m}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{p}}_1}^{\text{2}}}}{{{\text{2}}{{\text{m}}_{\text{1}}}}}$
Kinetic energy of bike is $\dfrac{{\text{1}}}{{\text{2}}}{{\text{m}}_2}{{\text{v}}_2}^{\text{2}}$
Multiplying and dividing by m2, we get
$\dfrac{{\text{1}}}{{\text{2}}}\dfrac{{{{\text{m}}_2}^{\text{2}}{{\text{v}}_2}^{\text{2}}}}{{{{\text{m}}_2}}}{\text{ = }}\dfrac{{{{\text{p}}_2}^{\text{2}}}}{{{\text{2}}{{\text{m}}_2}}}$
$\therefore {\text{ }}\dfrac{{{{\text{p}}_{\text{1}}}^{\text{2}}}}{{{\text{2}}{{\text{m}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{p}}_{\text{2}}}^{\text{2}}}}{{{\text{2}}{{\text{m}}_{\text{2}}}}}...{\text{(i)}}$
Where $m_1$ = Mass of the truck
$v_1$ = Velocity of the truck
$m_2$ = Mass of the bike
$v_2$ = Velocity of the bike
$p_1$ = momenta of truck
$p_2$ = momenta of bike
Mass of the truck is four times that of the bike i.e.,
${\text{4}}{{\text{m}}_{\text{1}}}{\text{ = }}{{\text{m}}_{\text{2}}}...{\text{(ii)}}$
On substituting the value of m1 from (ii) to (i), we get
\[
\dfrac{{{{\text{p}}_{\text{1}}}^{\text{2}}}}{{{\text{2}}{{\text{m}}_2}}}{\text{ = }}\dfrac{{{{\text{p}}_{\text{2}}}^{\text{2}}}}{{{\text{8}}{{\text{m}}_{\text{2}}}}} \\
\Rightarrow \dfrac{{{{\text{p}}_{\text{1}}}^{\text{2}}}}{1}{\text{ = }}\dfrac{{{{\text{p}}_{\text{2}}}^{\text{2}}}}{4} \\
\therefore \dfrac{{{{\text{p}}_{\text{1}}}}}{{{{\text{p}}_{\text{2}}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}} \\
\]
Therefore, the ratio of momenta of truck and bike is 1:2.
Note: Kinetic energy is the energy possessed by the particle due to its motion. The formula of kinetic energy is ${\text{K}}{\text{.E}}{\text{. = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{m}}{{\text{v}}^{\text{2}}}$ where m = mass of an object and v = velocity of an object. SI unit of the mass is $Kg$ and the SI unit of the velocity is $m/ s^2$. Thus, SI unit of kinetic energy is Joule and $1 Joule$ = $1 kg m/ s^2$. Momentum of an object is equal to the mass of the object times the velocity of the object. The formula for momentum is P = mv where m = mass of an object and v = velocity. SI unit of momentum is $kg m/s$.
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