Answer:
New speed of the 22-kg block is 1.57 m/s
Explanation:
Mass of block [tex]m_{1}=22 kg[/tex]
Mass of another block [tex]m_{2}=29 kg[/tex]
Initial speed of the block [tex]u_{1}=5 m/s[/tex]
Initial speed of the another block [tex]u_{2}=4.6 m/s[/tex]
Initial speed of the another block [tex]v_{2}=7.2 m/s[/tex]
For conservation of momentum, we have
[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}[/tex]
Substitute all the values and solving for final speed of the 22kg block is
[tex]22\times 5 +29 \times 4.6 =22\times v_{1} +29 \times 7.2\\\\v_{1}=\frac{243.2-208.8}{22} \\\\v_{1}=1.57 m/s[/tex]
new speed of the 22-kg block is 1.57 m/s
