Let us assume the positive \hat{x} direction is towards the right.
During any collision, the total momentum is conserved.
That means, the total momentum before the collision is the same as the total momentum after the collision.
p before = P after
Given,
Mass 1, m_1 = 111.1kg
Mass 2, m_2 = 222.2kg
Initial Velocity of Mass 1, 1,1 3.00.î m/s
Initial Velocity of Mass 2, \overrightarrow v_{2,i} =0
Velocity of Mass 1 after collision, \overrightarrow v_{1,f} =-1.000\hat{x}\hspace{0.1cm}m/s
The total Momentum of the system before the collision is,
\overrightarrow p_{before} =m_1\overrightarrow v_{1,i} + m_2\overrightarrow v_{2,i}
=> \overrightarrow p_{before} =(111.1)(3.00\hat{x}) + 0
=> \overrightarrow p_{before} =333.3\hspace{0.1cm}\hat{x}\hspace{0.1cm}kgm/s
The total Momentum of the system after the collision is,
\overrightarrow p_{after} =m_1\overrightarrow v_{1,f} + m_2\overrightarrow v_{2,f}
=> \overrightarrow p_{after} =(111.1)(-1.000\hat{x}) + (222.2)\overrightarrow v_{2,f}
=> \overrightarrow p_{after} =(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f}
We know the total momentum before the collision is the same as the total momentum after the collision.
p before = P after
=>333.3\hat{x} =(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f}
=>(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f} = 333.3\hat{x}
=> (222.2)\overrightarrow v_{2,f} = 333.3\hat{x} + 111.1\hat{x}
=> (222.2)\overrightarrow v_{2,f} = 444.4\hat{x}
=> \overrightarrow v_{2,f} = \frac{444.4}{222.2}\hspace{0.1cm}\hat{x}\hspace{0.1cm}m/s
=> \overrightarrow v_{2,f} = 2.00\hspace{0.1cm}\hat{x}\hspace{0.1cm}m/s
Therefore the velocity of m_2 after collision is v_{2,f} = 2.00\hspace{0.1cm}m/s going to the right
Please check the attached file for a brief answer.
Learn more about a velocity at
https://brainly.com/question/25749514
#SPJ4
