# Conservation of momentum and Newton's second law

I'm not sure if my approach to this question is correct. I also don't understand why I'm getting inconsistent results from my own working.

# Question # Working

## Part 1:

The time that the block moves is the moment where the push force overpowers the force due to friction

Then we have

$$F_{push} = F_{friction}$$

Our values are

$$F_p = 3t, F_f = \mu_s m g = \mu_s g = 10(0.6) = 6$$

From this we have that the block will move at the moment

$$3t = 6$$

Solving for $t$ gives

$$t = \frac{6}{3} = 2$$

Then for the first part we have that the block starts to move at time $t = 2$ seconds.

## Part 2 (using Newtons Second):

We want to know the speed of the block at $t = 5$ seconds.

From part 1 we know that the block doesn't start to move until time $t = 2$ seconds.

From newtons second law we have

$$F = ma$$

Where $F$ is the net force. This net force will be $F_n = F_p - F_f$ (push - friction).

$$F_p = 3t$$

$$F_f = \mu_k m g = \mu_k g = 0.55(10) = 5.5$$

Then we have

$$F_n = 3t - 5.5 = ma = a$$

So this is our acceleration function, integration of this gives us velocity as

$$v = \frac{3}{2}t^2 - 5.5t$$

The constant of integration is dropped as initial velocity is zero.

Therefore at time $t = 5$ seconds we have speed

$$v = \frac{3}{2}(5^2) - 5.5(5) = 10$$

this shows that the speed is 10 m/s at time $t = 5$ seconds

## Part 2 (using conservation of momentum):

Conservation of momentum states that

$$p_{init} = p_{final}$$

We also have Impulse as

$$I = F \Delta t$$

Here $\Delta t = 3$ as we're moving from $t = 2$ to $t = 5$ seconds. We also have the net force as before, which is $F = ma = a = 3t - 5.5$

Which puts Impulse as

$$I = (3t - 5.5) \times 3 = 9t - 16.5$$

Using $I = \Delta p$ where $\Delta p = m(v_f - v_0)$, note that as $m = 1$ we have $\Delta p = (v_f - v_0)$, and here $v_0 = 0$, so we just have $\Delta p = v_f$.

Using this gives

$$I = 9t - 16.5 = v_f$$

Inputting $t = 5$ gives

$$v_f = 45 - 16.5 = 28.5$$

# So

So for part 1 I got $t = 2$ seconds

For Part 2 (using Newtons) I got $v = 10$

For Part 2 (using momentum) I got $v = 28.5$

So clearly something here is wrong as I have inconsistent results, I'm not sure what though.

• in part 2 $F_p$ is not $3t$, is $3(t+2)$ because you start calculating at $t=2$
– user126422
Apr 23, 2017 at 18:45
• @WillyBillyWilliams thanks - I thought that though I start calculating at $t = 2$ considering what occurred between 1 and 2 seconds wasn't necessary as there was no movement there?
– baxx
Apr 23, 2017 at 18:48
• But the acceleration you calculated is only valid for the interval [2,5]. You can start at t=0 but then when you integrate you have to consider $a=0$ for t=[0,2]. You are actually using a negative force for t=0
– user126422
Apr 23, 2017 at 18:58
• @WillyBillyWilliams I'm not sure that I follow still - you're saying that I should use $F_p = 3(t + 2)$, is there anywhere else that I should consider $t + 2$ in the problem? And should I evaluate at $t = 5$ still?
– baxx
Apr 23, 2017 at 18:59
• it depend how you wanna do it, but the simplest is to shift the time, so the force starts as F=3(t+2) for t=0 (which is the original t=2), and instead of 5 you use 3
– user126422
Apr 23, 2017 at 19:01

For the Newton's law part, your answer is incorrect because the block doesn't start moving until t = 2 s. Thus the velocity must be 0 at t = 2 s, meaning that the constant of integration is actually 5 m/s (the formula does not even apply for t = 0 s). Then the velocity at t = 5 s is $\frac{3}{2} (5^2) - 5.5 (5) + 5 = 15 \: m/s$.

For the second part, the impulse is actually $\displaystyle \int_{t=2}^{5} Fdt = \int_{t=2}^{5} (3t-5.5) dt = 1.5t^2-5.5t |_2^5 = 15$ because the F in the formula for the impulse is the average value of the force from t=2 to t=5.

• thankyou - I don't understand why the constant of integration is 5 for the first part of your working. re Impulse: for $I = F \Delta t$, this is actually just multiplication for the case of force being a constant, where as in this problem force is a function of time meaning that we should use integration (Is this reasoning sound?).
– baxx
Apr 23, 2017 at 19:14
• The formula for acceleration applies for t= 2 s onward. Therefore you can't find the constant of integration by setting the velocity equal to 0 at t = 0. Instead, you must set v=0 at t= 2 s. For this to be true, the constant of integration must be 5 because $\frac{3}{2} (2) - 5.5(2) = -5$. Apr 23, 2017 at 19:17
• Your reasoning is correct. The impulse-momentum relation comes from Newton's second law $F = \frac{dp}{dt}$, which can be rearranged to read $\int Fdt = \int dp = \Delta p$. For a constant force, this becomes $F \Delta t = \Delta p$, but for a time-dependent force you must integrate. Apr 23, 2017 at 19:21
• I'm not sure that I follow your reasoning about why the constant must be 5? Also $1.5 * 2 - 5.5 * 2 = -8$ . So I'm not sure what I'm missing edit you've simply written $2$ instead of $2^2$ , I didn't consider that sos
– baxx
Apr 23, 2017 at 19:26
• Sorry, that was a typo. Since your formula applies for t=2 s and beyond, I just set the velocity equal to 0 at t=2 s (there's no acceleration before that) to find the constant of integration. Apr 23, 2017 at 19:35