Answer:
After throwing the object the, the velocity of the man is 13.98 m/s
Explanation:
Given:
Let,
mass of man, m1 = 74 kg
mass of box, m2 = 13 kg
Initial velocity u = 11 m/s (initially both are together hence initial velocity will be same for both)
Final velocity of man = v1
Final velocity of box = v2 = -6 m/s (the velocity is recoil velocity therefore it is negative)
To Find:
Final velocity of man,after throwing the object = v1 = ?
Solution:
Recoil velocity:
It is the backward velocity experienced.
Here recoil velocity is the backward velocity experience while throwing the box behind.Hence the velocity of the box is negative 6 m/s.
The recoil velocity is the result of conservation of linear momentum of the system. Therefore we will follow the law of conservation of momentum.
Law of conservation of momentum :
Total momentum of an isolated system before collision is always equal to total momentum after collision
[tex]\textrm{total momentum before collision}=\textrm{total momentum after collision}\\m_{1}\times u+ m_{2}\times u=m_{1}\times v_{1}+m_{2}\times v_{2}[/tex]
substituting the values which are given above we get
[tex]74\times 11 + 13\times 11 = 74\times v_{1} +13\times -6\\ 957 = 74\times v_{1} -78\\v_{1}=\frac{1035}{74} \\v_{1}=13.98\ m/s[/tex]
Therefore, After throwing the object the, the velocity of the man is 13.98 m/s
