Answer:
0.624 kg
Explanation:
We are given that
Mass of one block,[tex]m_1=1.3 kg[/tex]
[tex]v_1=1.3 m/s[/tex]
[tex]v_2=5 m/s[/tex]
[tex]V=2.5 m/s[/tex]
We have to find the mass of second block.
Mass of second block,[tex]m_2=\frac{m_1V-m_1v_1}{v_2-V}[/tex]
Substitute the values
[tex]m_2=\frac{1.3\times 2.5-1.3\times 1.3}{5-2.5}[/tex]
[tex]m_2=0.624 kg[/tex]
Hence, the mass of second block=0.624Kg
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 :
[tex]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[/tex]
So, the mass of the second block is 0.624 kg.
