I was trying to answer the question of the flying ball here on the basis of Newton's third law and momentum conservation. Here is what I have tried.

Lets take a ball mass of $m_1$ (index 1 is the ball) hits the man with mass $m_2$ (index 2 is the man). The velocity of the ball is $u_1$. Then the force on the man is $$ m_1(v_1-u_1)/t = -m_2(v_2-u_2)/t$$

Now the second case, man hitting ball at rest at the same speed of that of the ball in case 1. i.e, $u_1$. The force on the ball is $$ m_2(v_2-u_2)/t = -m_1(v_1-u_1)/t$$

Since $u_1 = u_2$, the equations do not give equal forces on an object.

So that the pain should be different, shouldn't it ?


Case A) Moving ball hits stationary man: $F_{1A}=\frac{m_{1}(v_{1A}-u_{1A})}{t}=-\frac{m_{2}v_{2A}}{t}$

Case B) Moving man hits stationary ball: $F_{2B}=\frac{m_{2}(v_{2B}-u_{2B})}{t}=-\frac{m_{1}v_{1B}}{t}$

You point out that $u_{1A}=u_{2B}$, let's just call it $u$. I don't see what is wrong with that. That forces in both cases must be equal just means that:

$$\text{First terms:} \frac{m_{1}(v_{1A}-u)}{t}=\frac{m_{2}(v_{2B}-u)}{t}\Rightarrow m_{1}(v_{1A}-u)=m_{2}(v_{2B}-u)$$ $$\text{Second terms:} -\frac{m_{2}v_{2A}}{t}=-\frac{m_{1}v_{1B}}{t}\Rightarrow m_{2}v_{2A}=m_{1}v_{1B}$$

Nothing here is impossible, and equal forces in both cases is not proven wrong.

If you want to find expressions for the new speeds, you can continue. Since all these terms must be equal $m_{1}(v_{1A}-u)=m_{2}(v_{2B}-u)= m_{2}v_{2A}=m_{1}v_{1B}$, we have:

$$m_{1}(v_{1A}-u)=m_{1}v_{1B}\Rightarrow v_{1A}-u=v_{1B}\\ m_{2}(v_{2B}-u)= m_{2}v_{2A}\Rightarrow v_{2B}-u= v_{2A}$$

So, the man will not reach the same final velocity in both cases $v_{1A} \neq v_{1B}$, and same for the ball $v_{2A} \neq v_{2B}$. But he will experience the same acceleration:


Same for the ball.

  • $\begingroup$ isnt momentum transfer different in both cases when ball and man take equal velocity ? $\endgroup$ – Vinayak Dec 20 '14 at 15:45
  • $\begingroup$ @Vinayak momentum is different. But (total) change in momentum is not. $\endgroup$ – Steeven Dec 20 '14 at 15:46
  • $\begingroup$ how can we find that change in momentum ? $\endgroup$ – Vinayak Dec 20 '14 at 15:48
  • $\begingroup$ For an elastic collision, total momentum change is always 0. And, you used that in your question. $\Delta p=p_{after} - p_{before} =m_1 v_1+m_2 v_2-m_1 u_1 - m_2 u_2=m_1(v_1-u_1)+m_2(v_2-u_2)=0$ $\endgroup$ – Steeven Dec 20 '14 at 15:58
  • $\begingroup$ i think my question isnt clear to u. My problem is that the pain (force ) experienced by the man is not the same in both cases when I use Newtons third law. $\endgroup$ – Vinayak Dec 20 '14 at 16:02

Isn't it just a change in the reference frame and therefore the forces are equal?

Update: Suppose your initial conditions: the ball ($m_1$) hits the man ($m_2$). Its velocity is equal to $u_1$. As you've said: $$m_1(v_1-u_1)/t=-m_2(v_2-u_2)/t$$ Now consider that an observer is moving at the velocity $u_1$. The ball looks static and the man looks like he's moving (velocity of $u_1$). You now obtain your second equation, stating, therefore, that the forces are equal, and also that the momentum transfer is the same in both.

  • $\begingroup$ Hello justguest, and welcome to Physics.SE! As it stands right now, your answer, while not incorrect, is not particularly helpful to the OP. It would be much appreciated if you elaborated on your answer and otherwise explain the cause of the OP's confusion. $\endgroup$ – Dave Coffman Apr 13 '15 at 21:15

Look at both of your equations. The second equation is just the first multiplied by $-1$ so the problem is not the equations, but your interpretation of signs. It appears that you are making different assumptions for each form of the same equation, which would naturally lead to confusion.


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