Question

A 74 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?

Answer 1

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

$\textrm{total momentum before collision}=\textrm{total momentum after collision}\\m_{1}\times u+ m_{2}\times u=m_{1}\times v_{1}+m_{2}\times v_{2}$

substituting the values which are given above we get

$74\times 11 + 13\times 11 = 74\times v_{1} +13\times -6\\ 957 = 74\times v_{1} -78\\v_{1}=\frac{1035}{74} \\v_{1}=13.98\ m/s$

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