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

Answer:

0.624 kg

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,$m_2=\frac{m_1V-m_1v_1}{v_2-V}$

Substitute the values

$m_2=\frac{1.3\times 2.5-1.3\times 1.3}{5-2.5}$

$m_2=0.624 kg$

Hence, the mass of second block=0.624Kg

Answer 2

Answer:

The mass of the second block is 0.624 kg.              

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 :

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

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