Question

A 1.3 kgkg block slides along a frictionless surface at 1.3 m/sm/s . A second block, sliding at a faster 5.0 m/sm/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.5 m/sm/s . What was the mass of the second block?

Answer 1

The mass of the second block is 0.624 kg.

Step-by-step explanation:

Given that,

Mass of the block 1, m₁ = 1.3 kg

Speed of block 1, u₁ = 1.3 m/s

Speed of block 2, u₂ = 5 m/s

The final velocity of the combined blocks is 2.5 m/s, V = 2.5 m/s

It is a case of inelastic collision. Using the conservation of linear momentum as :

`m_1u_1+m_2u_2=(m_1+m_2)V\\\\1.3* 1.3+5m_2=(1.3+m_2)2.5\\\\1.69+5m_2=3.25+2.5m_2\\\\m_2=0.624\ kg`

So, the mass of the second block is 0.624 kg.

Answer 2

0.624 kg

Step-by-step explanation:

We are given that

Mass of one block,
`m_1=1.3 kg`

`v_1=1.3 m/s`

`v_2=5 m/s`

`V=2.5 m/s`

We have to find the mass of second block.

Mass of second block,
`m_2=(m_1V-m_1v_1)/(v_2-V)`

Substitute the values

`m_2=(1.3* 2.5-1.3* 1.3)/(5-2.5)`

`m_2=0.624 kg`

Hence, the mass of second block=0.624Kg