Respuesta :
Given:
The mass of block 1, m₁=2 kg
The mass of the block 2, m₂=5 kg
The initial velocity of the block 1, u₁=3 m/s
The initial velocity of the block 2, u₂=0 m/s
The velocity of the block 2 after the collision, v₂=1.5 m/s
To find:
The magnitude and direction of the velocity of block 2 after the collision.
Explanation:
From the law of conservation of momentum, the total momentum of blocks before the collision must be equal to the total momentum of the blocks after the collision.
Thus,
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]Where v₁ is the velocity of block 1 after the collision.
On rearranging the above equation,
[tex]\begin{gathered} m_1v_1=m_1u_1+m_2u_2-m_2v_2 \\ \implies v_1=\frac{m_1u_1+m_2u_2-m_2v_2}{m_1} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} v_1=\frac{2\times3+5\times0-5\times1.5}{2} \\ =-0.75\text{ m/s} \end{gathered}[/tex]The negative sign indicates that block 1 will be sliding to the left after the collision.
Final answer:
Thus the magnitude of the velocity of block 1 after the collision is 0.75 m/s.
And the direction of block 1 after the collision is to the left.
