Answer:
The speed of metal block B is 5 m/s after the collision
Explanation:
Law Of Conservation Of Linear Momentum
The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is
P=mv.
If we have a system of bodies, then the total momentum is the sum of them all
[tex]P=m_1v_1+m_2v_2+...+m_nv_n[/tex]
If some collision occurs, the velocities change to v' and the final momentum is:
[tex]P'=m_1v'_1+m_2v'_2+...+m_nv'_n[/tex]
In a system of two masses, we have:
[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]
The metal block A has a mass of m1=3.2 Kg and moves at v1=4 m/s. Metal block b has a mass of m2=1.6 Kg and is initially at rest v2=0.
After the collision occurs, block A moves at v1'=1.5 m/s. We need to calculate the speed of the metal block B. Solving for v2':
[tex]\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2}[/tex]
Substituting the given values:
[tex]\displaystyle v'_2=\frac{3.2*4+1.6*0-3.2*1.5}{1.6}[/tex]
[tex]\displaystyle v'_2=\frac{8}{1.6}[/tex]
[tex]\displaystyle v'_2=5\ m/s[/tex]
The speed of metal block B is 5 m/s after the collision
