Question

A large, massive man and a small boy of small mass are held far apart on an icy surface where friction is negligible. The man and the boy hold opposite ends of a bungee cord that is stretched.
When the boy and the man are released, they are each pulled by the bungee cord so that they approach one another.
When they get close to each other, how will their momenta, velocity, and kinetic energies compare?
a) Magnitude of Momenta – man’s greater than boy’s, Magnitude of velocity – boy’s greater than man’s, Kinetic Energy – man’s greater than boys
b) Magnitude of Momenta – same for both, Magnitude of velocity – boy’s greater than man’s, Kinetic Energy – same for both
c) Magnitude of Momenta – same for both, Magnitude of velocity – same for both, Kinetic Energy – same for both
d) Magnitude of Momenta – boy’s greater than man’s, Magnitude of velocity – boy’s greater than man’s, Kinetic Energy – same for both
e) Magnitude of Momenta – same for both, Magnitude of velocity – boy’s greater than man’s, Kinetic Energy – boy’s greater than man’s

Answer 1

The momentum of an object can be defined as the product of the mass of the object and its velocity $\text{ }p=m\times v$
The magnitude of the velocity is the product of the momentum of the object by the mass of the object.
Kinetic energy in terms of momentum can be calculated as,
$\text{K}\text{.E}=\dfrac{{{p}^{2}}}{2m}$ ; Where $p$ is momentum and $m$ is mass

Complete step by step answer:
Consider, momentum of boy be ${{p}_{boy}}$and momentum of man be ${{p}_{man}}$
By using the law of conservation of momentum in this case we can conclude that the initial and final momentum of the boy and man are equal.

That is, Initial momentum is equal to zero and final momentum is also equal to zero.

Therefore,

& \text{ }{{p}_{boy}}+{{p}_{man}}=0 \\\ & \Rightarrow {{p}_{boy}}={{p}_{man}} \\\ & \Rightarrow \left| {{p}_{boy}} \right|=\left| {{p}_{man}} \right| \\\ \end{aligned}$$ Hence, the Magnitude of momentum is the same. Since, $$\begin{aligned} & \Rightarrow p=m\times v \\\ & \Rightarrow v=\dfrac{p}{m}......(1) \\\ \end{aligned}$$ Where $$p$$ is momentum, $$m$$ is mass and $$v$$ is velocity Thus, from equation (i) we can understand that, Velocity (v) is inversely proportional to the mass of the body, that is $$v\propto \dfrac{1}{m}$$ Therefore, as the boy is lighter than the man, the magnitude of the velocity of the boy is greater than that of the man. $${{v}_{boy}}>{{v}_{man}}$$ Now as we know, Kinetic energy is,$$\text{K}\text{.E}=\dfrac{{{p}^{2}}}{2m}$$ Again, Kinetic energy is inversely proportional to mass of the body, that is $$\text{K}\text{.E}\propto \dfrac{1}{m}$$ Therefore, as the boy is lighter than the man, kinetic energy of the boy is greater than that of the man. $$\text{K}\text{.}{{\text{E}}_{boy}}>\text{K}\text{.}{{\text{E}}_{man}}$$ **So, the correct answer is “Option E”.** **Note:** The momentum of an object is proportional to the object's velocity - if you double its velocity, you double its momentum. The kinetic energy of an object is proportional to the square of the object's velocity - if you double its velocity, you quadruple its kinetic energy.