# Can the forces change with frame of reference?

Consider a ball kept on man's head (mass $$M$$) on the Earth. Now supposing I throw the ball from height $$h$$ of tall building then why does he gets more hurt? Isn't the force still mg? I would like to know what happens in ideal case (no air resistance) and then in real case (with air resistance)

• FWIW, fictitious forces change with frame of reference. – Qmechanic Dec 1 '19 at 15:27
• Like? Which forces? – Abhishek Dec 1 '19 at 16:19

Now supposing I throw the ball from height ℎ of tall building then why does he gets more hurt? Isn't the force still mg?

The impact force of the ball falling on the man's head is not the same as the weight of the ball on the persons head. This is because it takes a force to perform work in order to absorb the kinetic energy of the ball at impact. The work needed to stop the ball is given by the work-energy theorem which states that the net work done on an object equals its change in kinetic energy, or

$$Fd=\frac{mv_{f}^2}{2}-\frac{mv_{i}^2}{2}$$

Where $$F$$ is the average impact force on the mans head, $$d$$ is the stopping distance, $$f$$ and $$i$$ indicate the final and initial velocities, respectively. If the ball is brought to a stop by the head, the $$v_f$$ = 0, and

$$Fd=-\frac{mv_{i}^2}{2}$$

The work done is negative (meaning energy is taken away from the ball) because the force acts upwards opposite the direction of the displacement $$d$$. The equation assumes the displacement $$d$$ is small so that the change in gravitational potential energy over the stopping distance can be ignored.

I would like to know what happens in ideal case (no air resistance) and then in real case (with air resistance)

With air resistance you have an upward force on the ball opposing the downward force of gravity. Therefore, for a given height the ball will impact the head with a lower velocity with air resistance than without. That, in turn, will reduce the average impact force on the man's head. So no air resistance is not the ideal case in terms of potential injury to the man. Air resistance reduces the impact force.

Hope this helps.

• Thank you! Wonderful answer! – Abhishek Dec 1 '19 at 18:51
• @Abhishek Does that mean you "Accept" my answer? I see you are a new contributor. Many new contributors don't realize they can accept an answer by pressing the "Accept" button. No pressure intended here. If you are not sure and want to wait for more answers to consider, or if you are unsure which of the other answers best suits you, that's perfectly OK. – Bob D Dec 1 '19 at 22:14
• Got it, Answer accepted! – Abhishek Dec 2 '19 at 18:30
• @Abhishek Thank you. I should also tell you that you are able to un accept an answer as well. For example, if someone else gives you an answer that works for you better than mine, you can un accept mine and accept the other, with no hard feelings on my part. Hope you continue to participate. – Bob D Dec 2 '19 at 23:03
• Yes I have tried that feature..I m getting to know things around here ...thanks anyway..it's for people like you who throw light on toddlers like to make it possible to put forth our point and get the answer! – Abhishek Dec 3 '19 at 13:38

mg is the force that the earth does on the ball, it is not the force between the head and the ball. This last force will be a function of the contact speed when they collide

The force is no longer $$mg$$. In the second scenarion the ball has a speed $$v(h)$$ just before it hits the person's head. If the speed of the ball is slowed down from $$v(h)$$ to $$0$$ by the head in time $$\tau$$, then the total force felt by the head is: $$F_{head}= M\frac{v(h)}{\tau}.$$

In the first scenario there is no such stopping force. This is true for both the case with air drag and without air drag, it's just the function $$v(h)$$ that changes between the two cases - the principle remains.

• Thank you,but in which case does he experience more force in the ideal case or the non ideal case? – Abhishek Dec 1 '19 at 16:17
• Also in the ideal case F(head)=M((v-0)/t)=Mg...is it correct? – Abhishek Dec 1 '19 at 16:19

The force is still $$mg$$, but note - it is the force that is applied on the ball, not on the man's head!

As the ball falls from above, it picks up speed due to its constant acceleration $$g$$, which respectively increases its momentum, $$p = mv$$. Bigger momentum means a bigger force, so that's why the poor person's head suffers more when you drop the ball from a certain height, than when you drop it from a height that's substantially smaller.

P.S - this is true for a scenario with air resistance, and for a scenario without air resistance.