Question

Block 1 (2 kg) is sliding to the right on a level surface at aspeed of 3 m/s. Block 2 (5 kg) is initially at rest, and block 1collides with it. After the collision, block 2 is moving to theright with a speed of 1.5 m/s. Calculate the magnitude anddirection of the velocity of block 1 after the collision.

Answer 1

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.

Step-by-step 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,

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

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

On rearranging the above equation,

`\begin{gathered} m_1v_1=m_1u_1+m_2u_2-m_2v_2 \\ \implies v_1=(m_1u_1+m_2u_2-m_2v_2)/(m_1) \end{gathered}`

On substituting the known values,

`\begin{gathered} v_1=(2*3+5*0-5*1.5)/(2) \\ =-0.75\text{ m/s} \end{gathered}`

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.