Answer:
The second person will have a velocity of 0.85 m/s in the opposite direction of the first person.
Explanation:
Law Of Conservation Of Linear Momentum
The total momentum of a system of bodies is conserved unless an external force is applied. 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 a 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, the law of conservation of linear momentum is:
[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]
According to the conditions of the problem, two persons standing on a frictionless surface are initially at rest (v1=0, v2=0). Their masses are m1=50 Kg and m2=65 Kg. After the push, one person (say m1) moves backward at v1'=-1.1 m/s. We can calculate the speed of the other person by solving for v2':
[tex]\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2)[/tex]
Substituting:
[tex]\displaystyle v'_2=\frac{50*0+65*0-50(-1.1)}{65)[/tex]
[tex]\displaystyle v'_2=\frac{55}{65}[/tex]
[tex]v'_2=0.85 \ m/s[/tex]
The second person will have a velocity of 0.85 m/s in the opposite direction of the first person.
