# Why is final momentum found by two formulae in case of increasing force not equal?

So, let us say a body moves with 10 m/s, mass = 10 kg. At time t=0 a force of 30 N is acting on it. Gradually (linearly) the force increases and finally it becomes 130 N after a period of 2 second .

Now, we can say that change in force every second = $$30+50t.$$

Now, $$dp = dt (30+50t)$$

Now, integrating and putting limits as t = 0 to t= 2 seconds. We get final momentum = $$160\ kg/m s^2$$. Now, we cannot say that F=160/2 since value of instantaneous force is not constant.

Part 1: Finding final velocity and momentum:

If I say $$130N=10*a$$, then a=$$13\ m/s^2.$$

Final velocity = u+at = 10 + (13)(2) = 36m/s.

Now, Final momentum becomes $$m*v =10*36$$ = $$360\ kgm/s^2.$$

Q1 Now, why is 160 not equal to 360 as the final momentum?

Q2 are final velocity and acceleration correct?

• Could you clarify the following statement? "So, let us say a body moves with 10 m/s, mass = 10 kg having a force on 30 N on it. Then, let us say it’s final force acting on it becomes 130 N after 2 second ." Is the initial velocity 10[m/s] and then a force of 30N is applied? or is the 30N required in order to move at 10[m/s] – NMech Apr 25 at 18:56
• I think your equation F = 30+50t was an incorrect start. Instead, you should write F = 30+m*a=130 and solving for a, then find V after 2s. The integration shall be performed on p = int (mv). – r13 Apr 26 at 0:10

If I understand the problem the you have:

• a mass moving at 10[m/s]
• at time t=0, a force of 30 [N] is applied, and by time t=2 it increases to 130 [N]

The question is what is the final momentum.

The final momentum (at time $$t_2=2[sec]$$) will be equal to the initial momentum plus the impulse of the force (see below):

$$m\cdot v_2 = m\cdot v_0 + \int_{t_0}^{t_2} F(t)dt$$

As you have gathered the force wrt time is equal to $$F(t) = 30 +50 t$$

Therefore $$m\cdot v_2 = m\cdot v_0 + \int_{t_0}^{t_2} (30 +50 t)dt$$ $$m\cdot v_2 = 10 [kg] \cdot 10 \left[\frac{m}{s}\right] + \left[30t +50 t^2\right]_{t_0}^{t_2}$$ $$m\cdot v_2 = 10 \cdot 10 \left[kg \frac{m}{s}\right] + \left(30\cdot 2 +25\cdot 2^2\right) \left[N\cdot s\right]$$ $$m\cdot v_2 = \left(100 + 60 +100\right) \left[N\cdot s\right]$$ $$m\cdot v_2 = 260\left[N\cdot s\right]$$

The problem with the final velocity is the following:

The acceleration can be calculated by:

$$F=m\cdot a$$

Therefore: $$a = \frac{F(t)}{m} = 3+5t$$

So the velocity is:

$$v_2 = v_0 + \int_{t_0}^{t_2} a(t) dt)$$ $$v_2 = v_0 + \int_{t_0}^{t_2} (3 +5 t)dt)$$ $$v_2 = 10 + \left[3t +2.5 t^2\right]_{0s}^{2s}$$ $$v_2 = 10 + \left(6 +10\right)$$ $$v_2 = 26 \;\frac{m}{s}$$

So finally the momemtum will be $$\cdot v_2 =260 [Ns]$$.

• Thanks a lot . A very nice method to find answer for final momentum. Actually , the question also includes how to find final velocity and total distance covered by the body ? Where in final velocity , I was getting wrong. I got final velocity , but we also know that acceleration keeps on changing. So , an equation for total distance covered is what I didn’t get – Srijan M.T Apr 26 at 4:14
• I have updated it. I was having trouble understanding your question what exactly the setup was so I had left it incomplete (and incorrect thanks again phil Sweet) yesterday. Anyway, I've woken up now so, and it seems I was on the right track, so I assume this is your answer. – NMech Apr 26 at 4:35
• @PhilSweet you are right. Thanks for spotting it. I was half asleep when I was posting that, and I had completed it this morning. – NMech Apr 26 at 4:43
• @NMech Ohh. I just couldn’t think to find acceleration that way. I’ll keep it in mind. Thanks a lot – Srijan M.T Apr 26 at 5:31
• When you wrote m.v2 = mv1 + integration of 30+50t. Then , what happened to the constant of integration ? – Srijan M.T Jun 23 at 6:34

I see this problem differently.

The 10kg mass was moving at a constant speed at 10 m/s, so $$V_0 = 10 m/s$$. 2 seconds After a 30N force was applied, the force increased to 130N due to acceleration, this can be expressed as $$a = (F_2 - F_1)/m = (130 - 30)/10 = 10 m/s^2$$. Now we can find the final velocity, $$V_F = V_0 + a*t = 10 + 10*2 = 30 m/s^2$$.

For $$V_0 = 10 m/s$$ and $$V_F = 30 m/s$$

$$p = m*(V_0 + V_F) = 10*(10 + 30) = 400 kg*m/s$$

Edit: Adding figure below. • Isn’t it that by writing 10m/s^2 as your acc. You have assumed that acc is 10 throughout the journey of 2seconds whereas it is (Given in the Q)3m/s^2 at t=0 sec and then increased 13m/s^2. – Srijan M.T Apr 26 at 14:09
• No, it moves at a constant speed with a = 0, then a sudden blow of 30N (gulf swing) forced it to speed up that increases the velocity, thus acceleration a = (Vf - V0)/t. That's my argument, the mass does not have initial acceleration, the 30 N force is externally applied, not due to moving, because prior to that moment F = ma = 10*0 =0. . – r13 Apr 26 at 14:22
• @Srijan I added a figure. I think you need to set up the reference timeline, and realize that with constant speed prior to time 0, both the acceleration and force equal to zero, because the object can't accelerate itself, so F = ma = 0 at t = 0- until the external force is applied.at t = 0+. that triggers the acceleration, thus at t = 2s, F = 30 + ma = 130N. Hope you got the picture. – r13 Apr 26 at 17:56
• Your point is also making sense. Nice one. – Srijan M.T Apr 27 at 14:12
• @Srijan The argument really depends on the initial force whether it causes immediate acceleration or not. I am trying to get to the bottom of it too. – r13 Apr 27 at 15:55