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$\dfrac{{mMv}}{{\left( {M + m} \right)}}$.
B) KE is$\dfrac{1}{2}\left( {M{v^2}} \right)$.
C) Momentum is$\dfrac{{mv\left( {M + m} \right)}}{M}$.
D) KE is$\dfrac{{{m^2}{v^2}}}{{2\left( {M + m} \right)}}$.

Answer 1

Momentum is defined as the product of mass and velocity. The conservation of linear momentum says that if the system does not have any effect of any external force then the linear momentum will be constant.

Formula used: The formula of the linear momentum is given by,
$p = m \cdot v$
Where momentum is p, the mass is m and the velocity is v.
The formula of the kinetic energy is equal to,
$KE = \dfrac{1}{2}m \times {v^2}$
Where mass is m of the body and the velocity is v.

Complete step by step answer:
It is given that 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 we need to find the velocity of the combined system (bag + bullet):
The formula of the linear momentum is given by,
$p = m \cdot v$
Where momentum is p, the mass is m and the velocity is v.
The momentum of the bullet is equal to:
$\Rightarrow {p_1} = mv$………eq. (1)
Let the velocity of the combined system be V.
The momentum of the system is equal to,
$\Rightarrow {p_2} = \left( {M + m} \right) \cdot V$………eq. (2)
According to the linear conservation of momentum,
$\Rightarrow {p_1} = {p_2}$
Replace the value of momentum of bullet and momentum of combined bullet and bag from equation (1) and equation (2) in above equation.
$\Rightarrow {p_1} = {p_2}$
$\Rightarrow \left( {M + m} \right) \cdot V = mv$
$\Rightarrow V = \dfrac{{mv}}{{\left( {M + m} \right)}}$………eq. (3)
The formula of the kinetic energy is equal to,
$KE = \dfrac{1}{2}m \times {v^2}$
Where mass is m of the body and the velocity is v.
The kinetic energy of the system is equal to,
$\Rightarrow KE = \dfrac{1}{2}m \times {v^2}$
$\Rightarrow KE = \dfrac{1}{2}\left( {M + m} \right) \times {V^2}$………eq. (4)
Replace the value of velocity of the system from the equation (3) to equation (4).
$\Rightarrow KE = \dfrac{1}{2}\left( {M + m} \right) \times {V^2}$
$\Rightarrow KE = \dfrac{1}{2}\left( {M + m} \right) \times {\left[ {\dfrac{{mv}}{{\left( {M + m} \right)}}} \right]^2}$
$\Rightarrow KE = \dfrac{1}{2}\left( {M + m} \right) \times \dfrac{{\left( {{m^2}{v^2}} \right)}}{{{{\left( {M + m} \right)}^2}}}$
$\Rightarrow KE = \dfrac{1}{2} \cdot \dfrac{{\left( {{m^2}{v^2}} \right)}}{{\left( {M + m} \right)}}$
The kinetic energy of the system is equal to$KE = \dfrac{1}{2} \cdot \dfrac{{\left( {{m^2}{v^2}} \right)}}{{\left( {M + m} \right)}}$.

The correct answer for this problem is option D.

Note: It is advisable to remember and understand the concept of conservation of linear momentum. The kinetic energy is a scalar quantity. The momentum is a scalar quantity and if there is any change in the path then the conservation of the linear momentum is applied in that direction only.