# How Does Impulse Relate to Momentum Change in Physics Problems?

• wgajtdb311
wgajtdb311
Homework Statement
Students are running an experiment to determine the mass of a block without using a scale. The students set up a block on a horizontal surface of negligible friction. They then attach a force sensor to the block that is at rest. The sensor is then used to exert a force on the block, moving it along a straight line. The force exerted as a function of time is shown in the top graph below. At time t = 12 s, the block is moving at 1.5 m/s.

(a)
i. Explain how the students could use the graph to estimate the impulse exerted on the block.
ii. Calculate the estimated impulse exerted on the block for the time interval 0 to 12 seconds.

(b) Calculate the mass of the block.
When t = 12s, the force sensor loses contact with the block, which continues to move along the horizontal surface. When t = 20s, the block encounters a rough area and comes to rest in 25 cm.

(c) On the (bottom) graph, sketch a graph of the speed of the block as a function of time.

d) Calculate the coefficient of kinetic friction between the block and the rough surface.
Relevant Equations
p = mv
J = Ft
F = ma

(bottom graph relates only to c)

(a)
i. The students can calculate the area under the graph to find the impulse exerted on the block. This is because the area under a force vs. time graph is the change in momentum or the impulse.
ii. Knowing that the graph is linear and begins at around 3 N when t = 0s, the function F(t) = 3 + t/4 can be used. When taking the integral from 0 to 12 of this function, an impulse of 54 kg * m/s is given.

(b)
I = mv
54 = (1.5)m
m = 34 kg

(c) (unsure of this one)

From 0-20s, there is a linearly increasing force on the block, meaning that the acceleration of the block is increasing linearly, so the speed is increasing exponentially. However, after 20 s, the block goes over a rough surface, causing it to slow down and eventually stop over a much shorter time interval.

(d) (unsure of this one as well)
Assuming that the function described in aii still holds true after 12 s but before 20 s, then the force at 20 s can be found with F(20). This means that at 20s, there is a force of 8 N being applied to the box. The box slows down to a stop after 20 s, so:
Fnet = Ff (friction force)
8 = mgμ
8 = (34)(9.8)μ
μ = 0.024 ???

If anyone could help with these, that would be greatly appreciated!

In part a you have done this integral to find the mass ##m##:

$$m v - \cancel{mv_o}^{0} = \int_{0}^{12} \left( 3[\text{N}] + \frac{1}{4}t \left[ \frac{\text{N}}{\text{s}}\right] \right) ~dt$$

Which the functions linearity is supposed to hold until ##t = 20 [\text{s}]## (if I understand correctly)?

What do you get for the velocity (##v##) if you then evaluate the impulse at the upper limit of ##20 [\text{s}]## ?

The box slows down to a stop after 20 s, so:

For d) the box doesn't come to rest 20 after hitting the rough patch. The box comes to rest over a distance of ##25 [\rm{cm}]##. So it has its velocity at 20 , and just 25 cm away from that position it comes to rest. Think Work-Energy.

Last edited:
wgajtdb311
Thanks for the response. Looking at part a again with the formula provided, the mass would be 36 kg (not 34 kg, which was a calculation error )

mv = J
m(1.5) = 54
m = 36 kg

Knowing this, we could then find the velocity at t = 20 s by evaluating the impulse at 20 s.

mv = ∫0 to 20 of F(t) dt
(36)v = 110
v = 3.06 m/s (at t = 20 s)

With this, I could then adjust the graph to be more accurate by marking a point at t = 12 s where the speed is 1.5 m/s, and another at t = 20 s where the speed is 3.06 s. The speed would still be exponentially increasing (concave up) from t = 0 s to t = 20 s.

However, this still leaves the friction portion of the graph along with calculating the coefficient of kinetic energy. From what I can tell, we would need one of the kinematic equations to calculate the acceleration during the third portion of the graph, but what we would do after that I'm not sure. Using the acceleration as the net force and setting that equal to the force of friction gives an unrealistically high coefficient of kinetic friction:
vF^2 = vI^2 + 2a(xF - xI) (sorry, I'm not familiar with subscripts and superscripts on this site)
0^2 = 3.06^2 + 2a(0.25)
-9.3636 = 0.5a
a = -18.7 m/s^2

Fnet = Ff
ma = mgμ
(36)(18.7) = (36)(9.8)μ
μ = 1.91

Could the acceleration be used for another equation? Or am I doing this part wrong?

wgajtdb311 said:
Thanks for the response. Looking at part a again with the formula provided, the mass would be 36 kg (not 34 kg, which was a calculation error )

mv = J
m(1.5) = 54
m = 36 kg

Knowing this, we could then find the velocity at t = 20 s by evaluating the impulse at 20 s.

mv = ∫0 to 20 of F(t) dt
(36)v = 110
v = 3.06 m/s (at t = 20 s)

With this, I could then adjust the graph to be more accurate by marking a point at t = 12 s where the speed is 1.5 m/s, and another at t = 20 s where the speed is 3.06 s. The speed would still be exponentially increasing (concave up) from t = 0 s to t = 20 s.
It increasing like with ##t^2## not really what we call "exponential increase".

wgajtdb311 said:
Fnet = Ff
ma = mgμ
(36)(18.7) = (36)(9.8)μ
μ = 1.91

Could the acceleration be used for another equation? Or am I doing this part wrong?
ok, but you cant just ignore the signs. If you have thought of ##a## to the right as positive, and you find ##a## is negative, then unless you changed convention half way through( ok, but unwise), ##a## is negative in the calulation....what is also negative?

I don't think its unrealistically high coefficient of friction. Its a 36 kg block, moving at 3 m/s coming to rest in a 1/4 of a meter. It pretty much hit rubber or glue...

wgajtdb311
erobz said:
It increasing like with t2 not really what we call "exponential increase".
I'm not sure what you mean by this, would the line still be curved concave up from t = 0 s to t = 20 s? My rationale for this is that there is an increasing force being applied to the object so it would experience an increasing acceleration and therefore a velocity increasing over time.

erobz said:
ok, but you cant just ignore the signs. If you have thought of a to the right as positive, and you find a is negative, then unless you changed convention half way through( ok, but unwise), a is negative in the calulation....what is also negative?
Would the work-energy equation be better in this situation? If W = Fd, and W = 1/2mvF^2 - 1/2mvI^2, then:

W = 1/2(36)(0)^2 - 1/2(36)(3.06)^2
W = 168.5 J
W = Fd
168.5 = F(0.25)
F = 674 N

Ff = F
mgμ = 674
(36)(9.8)μ = 674
μ = 1.91

This gives the same answer, which would make sense, but I'm not sure if I've done it correctly.

wgajtdb311 said:
I'm not sure what you mean by this, would the line still be curved concave up from t = 0 s to t = 20 s? My rationale for this is that there is an increasing force being applied to the object so it would experience an increasing acceleration and therefore a velocity increasing over time.
I mean I believe the wording “exponential increase” is not accurate. Maybe I’m wrong but I believe it’s reserved for thing like ##2^x## not curves like ##x^2##.
wgajtdb311 said:
Would the work-energy equation be better in this situation? If W = Fd, and W = 1/2mvF^2 - 1/2mvI^2, then:

W = 1/2(36)(0)^2 - 1/2(36)(3.06)^2
W = 168.5 J
W = Fd
168.5 = F(0.25)
F = 674 N

Ff = F
mgμ = 674
(36)(9.8)μ = 674
μ = 1.91

This gives the same answer, which would make sense, but I'm not sure if I've done it correctly.
Yeah, either or works.

wgajtdb311
wgajtdb311 said:
The sensor is then used to exert a force on the block….
When t = 12s, the force sensor loses contact with the block,
That means there is nothing accelerating the block after t=12s.

erobz
haruspex said:
That means there is nothing accelerating the block after t=12s.
Oh geez. I missed that bit…calling @wgajtdb311 for a redo.

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