Question

1.
A block of mass 1 kg moving with a speed of 4 ms-1, collides with
block of mass 2 kg which is at rest. The lighter block comes
collision. The speed of the heavy body after collision is
est. The lighter block comes to rest after
1) 2 ms-1
2) 1 ms-1
3) 1.5 ms-1
4) 0.5 ms-1​

Answer 1

Explanation:

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Answer 2

The speed of the heavier ball after the collision is 2 m/s.

Explanation:

Mass of block 1, $m_1=1\ kg$

Initial speed of block 1, $u_1=4\ m/s$  

Mass of block 2, $m_2=2\ kg$

Initial speed of block 2, $u_2=0$ (at rest)

After collision, final speed of block 1, $v_1=0$ (it comes to rest)

Let $v_2$ is the final speed of the heavier ball. It can be calculated using conservation of linear momentum as:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$m_1u_1=m_2v_2$

$v_2=\dfrac{m_1u_1}{m_2}$

$v_2=\dfrac{1\times 4}{2}$

$v_2=2\ m/s$

So, the speed of the heavier ball after the collision is 2 m/s. Hence, this is the required solution.

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