Question

A bag of mass M hangs by a long thread and a bullet (mass m) comes horizontally with velocity v and gets caught in the bag. Then for the combined system (bag + bullet)
A) Momentum is $\dfrac{mvM}{m+M}$
B) Kinetic is $\dfrac{m{{V}^{2}}}{2}$
C) Momentum is $\dfrac{mv\left( m+M \right)}{M}$
D) Kinetic Energy is $\dfrac{{{\left( mv \right)}^{2}}}{2\left( m+M \right)}$

Answer 1

The bullet to the bag forms conservation of momentum between the bag and the bullet received, meaning, the force of the bullet received by the bag is equal to the force of the bag perceived by the bullet. The force of the bag contains the mass of bullet and bag and is written as:
$\text{Momentum}{{\text{m}}_{\text{bag}}}=(m+M)V$
And the momentum of the bullet is written as:
$\text{Momentum}{{\text{m}}_{\text{bullet}}}=mV$
Similarly, the conservation is also further followed by the formula of Kinetic Energy as well:
$\text{Kinetic energy}{{\text{y}}_{\text{bag}}}=\dfrac{1}{2}(m+M){{V}_{c}}$
where ${{V}_{c}}$ is the combined velocity, $V$ is the velocity of the bullet, m is the mass of the bullet, M is the mass of the bag.

Complete step by step solution:
We first need to find the velocity that is perceived by the bag when the bullet hits the bag, in terms of mass of both the bullet and the bag:
The momentum exerted by the bullet is:
$\text{Momentum}{{\text{m}}_{\text{bullet}}}=mV$
Placing the value of the velocity of the bullet as $v$, we get the momentum of the bullet:
$\text{Momentum}{{\text{m}}_{\text{bullet}}}=mv$
Now to find the momentum exerted on the bag by the bullet is:
$\text{Momentum}{{\text{m}}_{\text{bag}}}=(m+M){{V}_{c}}$
To get the velocity ${{V}_{c}}$, we equate the values of the momentum of bag and bullet as:
$(m+M){{V}_{c}}=mv$
${{V}_{c}}=\dfrac{m}{(m+M)}v$
Placing the value of the velocity of the bullet as ${{V}_{c}}$, we get the kinetic energy of the bullet plus bag system or bag alone as:
$\text{Kinetic Energy}{{\text{y}}_{\text{bag}}}=\dfrac{1}{2}(m+M){{V}_{c}}^{2}$
Simplifying the kinetic energy formula further by placing ${{V}_{c}}=\dfrac{m}{(m+M)}v$, we get:
$\text{Kinetic Energy}{{\text{y}}_{\text{bag}}}=\dfrac{1}{2}(m+M){{\left( \dfrac{m}{(m+M)}v \right)}^{2}}$
$=\dfrac{1}{2}\dfrac{{{\left( mv \right)}^{2}}}{(m+M)}$
Therefore, the value of the kinetic energy of the bullet plus bag system is $\text{Kinetic Energy}{{\text{y}}_{\text{bag}}}=\dfrac{1}{2}\dfrac{{{\left( mv \right)}^{2}}}{(m+M)}$.

Note:
Another method to solve the question is by checking the options, option A and C are wrong as the formula for the momentum is the product of mass and velocity in singular form whereas in numerator both have square of mass. Option B, is wrong as the velocity is capital V, while no such unit is given in the question as the velocity is v only. Therefore, the correct answer left is (D).