To solve this problem, let's start from the initial conditions and proceed step-by-step.
1. Initial Setup:
- A body has a mass [tex]\( m \)[/tex] and is moving with a velocity [tex]\( v \)[/tex].
- The mass [tex]\( m \)[/tex] splits into two parts, where [tex]\( \frac{m}{4} \)[/tex] comes to rest instantly.
2. Calculate Initial Kinetic Energy:
- The total initial kinetic energy (before the split) is given by:
[tex]\[
KE_{\text{initial}} = \frac{1}{2} m v^2
\][/tex]
3. After the Split:
- One part of the mass, [tex]\( m_1 = \frac{m}{4} \)[/tex], comes to rest.
- The remaining part is [tex]\( m_2 = m - m_1 \)[/tex].
Therefore, [tex]\( m_2 = m - \frac{m}{4} = \frac{3m}{4} \)[/tex].
4. Kinetic Energy of the Mass that Comes to Rest:
- The kinetic energy of the mass [tex]\( m_1 \)[/tex] that comes to rest is zero because its velocity is zero.
5. Remaining Kinetic Energy:
- Since the mass [tex]\( m_1 \)[/tex] comes to rest, the initial kinetic energy [tex]\( KE_{\text{initial}} \)[/tex] is now entirely with the remaining mass [tex]\( m_2 \)[/tex].
6. Final Kinetic Energy of the Remaining Mass:
- As the system is closed and there are no external forces acting on it, the remaining kinetic energy should be equal to the initial kinetic energy minus the kinetic energy of the mass that came to rest.
Given that the kinetic energy of the mass that came to rest is zero, the remaining kinetic energy is simply:
[tex]\[
KE_{\text{remaining}} = KE_{\text{initial}}
\][/tex]
- So, the kinetic energy of the remaining mass [tex]\( m_2 \)[/tex] is:
[tex]\[
KE_2 = \frac{1}{2} m v^2
\][/tex]
Hence, the kinetic energy of the other mass (i.e., [tex]\( m_2 \)[/tex]) is:
[tex]\[
\boxed{\frac{1}{2} m v^2}
\][/tex]
The correct answer is:
a. [tex]\( \frac{1}{2} m v^2 \)[/tex]
