Question

If a man with mass 70 kg, standing still, throws an object with mass 5 kg at

Answer 1

The asnwer to your question is A

Answer 2

Answer:

-3.6 m/s

Explanation:

Here we need to use the conservation of momentum law.

So, the momentum of the man initially is zero and so of the object. The momentum of the object after throw is 5 kg multiplied by 50 m/s which is 250 kg m/s.

So the equation is as follows:

0 + 0 = (250 + 70 x v) kg m/s

Which leads to v = (-250/70) m/s ≈ - 3.6 m/s

I am not 100 % sure though! Hope this helps! :)