Given:
The mass of block 1, m₁=12 kg
The mass of block 2, m₂=2.5 kg
The velocity of block 1 before the collision, u=2 m/s
The velocity of block 2 after the collision, v₂=4 m/s
To find:
The velocity of block 1 after the collision.
Explanation:
From the law of conservation of momentum, the total momentum of the blocks before the collision will be equal to the total momentum of the blocks after the collision.
Thus,
[tex]m_1u=m_1v_1+m_2v_2[/tex]Where v₁ is the velocity of block 1 after the collision.
On rearranging the above equation,
[tex]v_1=\frac{m_1u-m_2v_2}{m_1}[/tex]On substituting the known values,
[tex]\begin{gathered} v_1=\frac{12\times2-2.5\times4}{12} \\ =1.17\text{ m/s} \end{gathered}[/tex]The positive sign of the velocity indicates that block 1 will continue to move to the right.
Final answer:
The velocity of block 1 after the collision will be 1.17 m/s and its direction is to the right.
