Solving Kinematics and Conservation of Energy Problems | Physics Homework

[TheRedDevil18]

Homework Statement

Homework Equations

The Attempt at a Solution

Question 1

1.1) 5 m.s
1.2) Acceleration
1.3) 10 m.s^-2
1.4) Displacement equals area under graph = 1/2*b*h = 1/2*2*10 = 10 m
1.5) vf = vi+at
0 = 10 + 10t
t = 1's + 2's = 3's

1.6) 1/2*1*10 = 5m
Position of ball = 10-5 = 5m

Question 2

2.1) Total mechanical energy in an isolated system is conserved

2.2) Total Mech Eergy at A = Total Mechanical Energy At B
Mech at A = mgh
= 1.5*9.8*0.6
= 8.82 J

Emech = mgh+Ek
8.82 = 1.5*9.8*0.2 +ek
ek = 5.88 J

2.3) Velocity a B:

Ek = 0.5mv^2
5.88 = 0.5*1.5*v^2
v = 2.8 m.s


Wnet = Work done by water
Wnet = change in ek
= 1/2m(vf^2-vi^2)
= 0.5*1.5(0-2.8^2)
= 5.88 J

 

[Redbelly98]

The Attempt at a Solution

Question 1

1.1) 5 m.s
1.2) Acceleration
1.3) 10 m.s^-2
1.4) Displacement equals area under graph = 1/2*b*h = 1/2*2*10 = 10 m

Looks good so far.

1.5) vf = vi+at
0 = 10 + 10t
t = 1's + 2's = 3's

It's unclear what you are doing here. You write an equation whose solution is t=-1. Then you ignore that equation, and it appears you are simply reading the time values from the graph -- which is actually the straightforward way of solving this problem.

I'm unclear why the question is asking you to use "an equation of motion ... from the time that ball was thrown." You can just read the time value directly off of the graph. And if you try to solve it using equations, I think you need two equations of motion, one describing the motion before the ball bounces and another for after the ball bounces.

1.6) 1/2*1*10 = 5m
Position of ball = 10-5 = 5m

Since the positive direction is upward, your answer says the ball ends up above the boys hand, the level at which it was initially thrown. So I disagree with your answer. Your method is on the right track, but you need to pay attention to all quantities used in the equation, and whether each one should have a positive or a negative value.

Question 2

2.1) Total mechanical energy in an isolated system is conserved

2.2) Total Mech Eergy at A = Total Mechanical Energy At B
Mech at A = mgh
= 1.5*9.8*0.6
= 8.82 J

Emech = mgh+Ek
8.82 = 1.5*9.8*0.2 +ek
ek = 5.88 J

2.3) Velocity a B:

Ek = 0.5mv^2
5.88 = 0.5*1.5*v^2
v = 2.8 m.s

I agree so far.

Wnet = Work done by water

That statement isn't right.

W_net = Sum of the work done by all forces acting on the ball.

You need to account for the work done by gravity. However -- the problem statement does not provide us with the necessary information, which would be the height of point D above point C. So this question appears to be unanswerable -- unless the question author meant to ask for the net work done on the ball, which would be the value you calculated.

Wnet = change in ek
= 1/2m(vf^2-vi^2)
= 0.5*1.5(0-2.8^2)
= 5.88 J

 

[collinsmark]

Hi TheRedDevil18,

In the future, just post one question at a time per thread. Trust me, it can get really confusing working on multiple problems at once if it turns out that there are several different posts in a thread. But anyway, ...

Homework Statement

Homework Equations

The Attempt at a Solution

Question 1

1.1) 5 m.s

You forgot a "-1" exponent in your units. It's either m/s or m·s^-1 or equivalent, but not meters times seconds.

1.2) Acceleration
1.3) 10 m.s^-2

So far so good.

1.4) Displacement equals area under graph

Yes, it is true that displacement is the area under the curve.

= 1/2*b*h = 1/2*2*10 = 10 m

I see a sort-of a shortcut was taken in your calculation, but it works!

1.5) vf = vi+at
0 = 10 + 10t
t = 1's + 2's = 3's

That 1.1.5 problem statement is strange. I understand that. But I don't think its asking for the approach you are taking.

In any case you'll have to break it up into two parts: the time before the ball touches the ground, and the time after the ball touches the ground. That's because the initial velocity and initial position are different for each part.

1.6) 1/2*1*10 = 5m
Position of ball = 10-5 = 5m

Okay, that looks good. But just for clarity, is that 5 m above or below the boy's hand? You might want to specify that.

Question 2

2.1) Total mechanical energy in an isolated system is conserved

2.2) Total Mech Eergy at A = Total Mechanical Energy At B
Mech at A = mgh
= 1.5*9.8*0.6
= 8.82 J

Emech = mgh+Ek
8.82 = 1.5*9.8*0.2 +ek
ek = 5.88 J

So far so good.

2.3) Velocity a B:

Ek = 0.5mv^2
5.88 = 0.5*1.5*v^2
v = 2.8 m.s

I don't think finding the velocity of the ball at point B will help with this problem, given how it is worded.

There's not enough information in the problem for you to give a numerical answer. To get a numerical answer, the problem would need to specify some more information about the location of point D, such as its height above point C. Without this information, there is no way to calculate a numerical answer. The best you can do is define a variable, such as the height of point D relative to point C (call it something like h_DC), and express your final answer in terms of that.

Wnet = Work done by water
Wnet = change in ek
= 1/2m(vf^2-vi^2)
= 0.5*1.5(0-2.8^2)
= 5.88 J

Your answer would be correct if the height of point D is the same as the height at point B. But looking at the figure though, this doesn't seem to be the case. (Again, it's not possible to numerically calculate unless some more information is given about the height of point D.)

[Edit: Redbelly98 beat me to the reply]

 

[TheRedDevil18]

Hi TheRedDevil18,


That 1.1.5 problem statement is strange. I understand that. But I don't think its asking for the approach you are taking.

In any case you'll have to break it up into two parts: the time before the ball touches the ground, and the time after the ball touches the ground. That's because the initial velocity and initial position are different for each part.

[Edit: Redbelly98 beat me to the reply]

Okay, I am a bit confused because on the graph the initial velocity is clearly 10m.s^-1 and the final velocity would be zero(max height) so it takes 1's to hit the ground plus the 2 seconds from the time the boy threw the ball up

 

[collinsmark]

Okay, I am a bit confused because on the graph the initial velocity is clearly 10m.s^-1 and the final velocity would be zero(max height) so it takes 1's to hit the ground plus the 2 seconds from the time the boy threw the ball up

In a way, I'm confused too, but only because the question (as originally described in the problem statement, section 1.1.5) is so odd and out of place. I mean, you can just look at the horizontal axis on the graph and conclude "3 seconds right there. It's on the graph's horizontal axis. 3. It's right there."

So I'm thinking the problem is asking you to describe either the ball's velocity as a function of time, or position as a function of time. And in either case, describe it in equation form. This is where you will need to break it up into two equations. One before 2 seconds, and one after.

[Edit: Break up the problem into an equation before the ball's velocity reaches -15 m/s, and another equation for after. Or break up the problem into an equation before the ball's position reaches 10 m below the boy's hand, and another equation for after. Since you're ultimately solving for time, it's best not to use time when breaking up the problem into its parts.]

Then solve each equation to determine each part's applicable time interval (using initial and final velocities, or initial and final positions).

That's my guess as to what the problem is asking for. [Edit: I mean, you know that the answer is 3 sec. It's just asking for a different way to get to that answer.]

 

[TheRedDevil18]

In a way, I'm confused too, but only because the question (as originally described in the problem statement, section 1.1.5) is so odd and out of place. I mean, you can just look at the horizontal axis on the graph and conclude "3 seconds right there. It's on the graph's horizontal axis. 3. It's right there."

[Edit: I mean, you know that the answer is 3 sec. It's just asking for a different way to get to that answer.]

I don't see where you are reading 3 seconds from as it is clearly not on the graph?, You may want to check the graph again

 

[collinsmark]

I don't see where you are reading 3 seconds from as it is clearly not on the graph?, You may want to check the graph again

Well, yes. You have to do a little calculation to get "3 sec." But one can (or at least almost can) do that in one's head.

When I hear the term "equations of motion," I interpret it as detailed equations describing position, velocity, perhaps momentum, etc., as a function of time.

It's very easy to eyeball the graph and determine that 3 seconds pass from the time the ball was released until the time the ball reaches its maximum height after the bounce.

But the problem statement, in section 1.1.5, asks you to find that based on an "equation of motion." I interpret that to mean that first you must represent the curve on the graph in the form of an equation. Then solve for Δt.

 

[Simon Bridge]

The context of the problem in introductory kinematics.
The person setting the question wants the student to work out the acceleration from the graph and then use that figure to work out the time after the bounce using a kinematic equation. It is not unusual to combine equation and graph work in one question - especially for an assessment - that it is arguably easier to do it in several ways is besides the point.

The main problem is in interpreting the last question - the person setting the question clearly believes there is enough there for a numerical answer. The approach would be to work out an equation that would allow you to arrive at a numerical answer from the broader context of the lesson - stating assumptions.

i.e.
- that the ball does not displace much liquid suggests that it is very small - model as a point.

- this is part of a kinematics section, so perhaps assume that the ball slows with constant acceleration.
(A very bad assumption in general but context is everything.)

- what other assumptions are needed that may be appropriate at the level needed?
Can we use what we know about point C to figure a slope for a v-t graph?

But I have the feeling the person setting the question is using faulty reasoning.

 

[Redbelly98]

I don't see where you are reading 3 seconds from as it is clearly not on the graph?, You may want to check the graph again

Ah, you're right! The 3s is not marked on the graph.

Now it makes more sense. So we're allowed to use the 2 s as the moment of bounce, as it is clearly indicated on the graph. Then we must solve an equation of motion to get the additional time beyond the 2 s.

Looks like you did things more-or-less correctly, but as I said before the solution to your equation, 0=10+10t, was t=-1.

 

[Simon Bridge]

What did I miss:
Looking at the last question.

Redraw the diagram to put origin of coord system at point C. y-axis points upwards.
Label the points by their letters.

Right away we note that the work wanted is the energy deficit getting from $(x_A,y_A)$ to $(x_D,y_D)$ ... i.e.

$W=mg(y_A-y_D)$ ... (1)
... so we need $y_D$.

So let's list what we know.
At point B, we have

$u_x^2+u_y^2=2g(y_A-y_B)$ ...(2)

If I start v-t graphs for the horizontal and vertical components of the velocity as the bob travels through the liquid (t=0 at point B) then:

The vertical v-t graph starts at $-u_y$, goes to $0$ at $t=T_C$, and goes to $+v_y$ at $t=T_D=T$
(I'm dropping subscripts where I think I can get away with it).

The horizontal v-t graph starts at $+u_x$, goes to $0$ at $t=T$, at $t=T_C$ the velocity is $+v_x$

Plot these graphs one above the other so the time-axes line up.
Assume the graphs are straight lines with the same-magnitude slope.
(i.e. the horizontal and vertical components have the same deceleration rate.)

Off the horizontal graph we get:

$aT=u_x$ ...(3)

$2(x_D-x_B)=u_x T$ ...(4)

$2x_D=(T-T_C)v_x$ ...(5)

$2x_B = -(u_x+v_x)T_C$ ... (6)

Off the vertical graph we get:

$a=(u_y+v_y)/T = u_y/T_C = v_y/(T-T_C)$ ... (7,8,9)

$2y_B=u_yT_C$ ...(10)

$2y_D=v_y(T-T_C)$ ...(11)

we know yA and yB, and m. Don't know W, yD, xD, xB, ux, uy, vx, vy, T, TC but need to find W ... so that's 11 equations and 10 unknowns? Though I may have put a couple of dependent equations in there or misread or misunderstood something.

It looks like a lot of work for 6 marks though - compared to the other 6-mark question - there may be a shortcut through the kinematic equations.