Question

Block 1 has a mass of 12 kg is moving to the right on a levelsurface at a speed of 2 m/s. Block 2 has a mass of 2.5 kg andis at rest on the surface. Block 1 collides with block 2, causingblock 2 to move to the right with a speed of 4 m/s. How fast,and in what direction, is block 1 moving after the collision?

Answer 1

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.

Step-by-step 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,

`m_1u=m_1v_1+m_2v_2`

Where v₁ is the velocity of block 1 after the collision.

On rearranging the above equation,

`v_1=(m_1u-m_2v_2)/(m_1)`

On substituting the known values,

`\begin{gathered} v_1=(12*2-2.5*4)/(12) \\ =1.17\text{ m/s} \end{gathered}`

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.