A74 kg man holding a 13 kg box rides on a skateboard at a speed of 11m/s. he throws the box behind him, giving it velocity of 6 m/s with respect to the ground . how many m/s his velocity after throwing rhe object?

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

substituting the values which are given above we get

Therefore, After throwing the object the, the velocity of the man is 13.98 m/s

double the distance = divide strength by 4