
A heavy truck and bike are moving with the same kinetic energy. If the mass of the truck is four times that of the bike, then calculate the ratio of their momenta.
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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