# 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 $$V_F = 30 m/s$$

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