Kinematics questions. (Linear & projectile)

[coconut62]

Pictures of the questions are attached below.

For question 2, I totally cannot get any answer from the four choices after I substituted everything in the equation.

For question 4, I used V^2= U^2 + 2AS, with U being zero, A being 9.8 and S being 2.0. I got C but the answer is D. Can someone explain it to me?

For question 5:
Firstly, why can't I calculate theta by directly using tan θ = 30/30 which makes θ=45°?
And how do I get the answer?

I have used (u sin θ)/9.8 = 1.2 but there are still two unknowns (u and θ). I can't find another equation to do simultaneous.

Big thanks to anyone who can help me solve these.

p/s: those are A-levels questions.

 

[Doc Al]

For question 2, I totally cannot get any answer from the four choices after I substituted everything in the equation.

What equations did you use? (Note that you can immediately eliminate several of the choices based on units.)

For question 4, I used V^2= U^2 + 2AS, with U being zero, A being 9.8 and S being 2.0. I got C but the answer is D. Can someone explain it to me?

The initial velocity isn't zero.

For question 5:
Firstly, why can't I calculate theta by directly using tan θ = 30/30 which makes θ=45°?

The angle formed by that 30/30 right triangle is not the same as the angle of the trajectory. (It would be if the ball just went straight, but it doesn't.)

And how do I get the answer?

I have used (u sin θ)/9.8 = 1.2 but there are still two unknowns (u and θ). I can't find another equation to do simultaneous.

That's one of the equations you'll need. Find another relating to distance.

 

[coconut62]

What equations did you use? (Note that you can immediately eliminate several of the choices based on units.)

s= ut+ 1/2 at^2.

Based on units, the choices left are A and B.

So is that equation correct?

The initial velocity isn't zero.

Why?

The helicopter goes up with a steady velocity, so it's motion is controlled, not free fall or projectile motion right? So how can I determine it's velocity when it's at a height of 2.0m?

Furthermore, when the commando jumps off it, why isn't his velocity zero? How come the commando have a velocity the moment he jumps off?

The angle formed by that 30/30 right triangle is not the same as the angle of the trajectory. (It would be if the ball just went straight, but it doesn't.)

That's one of the equations you'll need. Find another relating to distance.

I used the range equation, and I found that u^2 sin 2θ = 588.

Then I made it: 2 u^2 sin θ cosθ = 588, then divide it by u sin θ = 11.76

so u cos θ = 25, which makes u tan θ = 25/11.76.

∴ θ= 25.2. Is this the way?


(TO LENGTHEN THE REPLY TO FOUR CHARACTERS. HOHOHO.)

 

[Doc Al]

s= ut+ 1/2 at^2.

Based on units, the choices left are A and B.

So is that equation correct?

Yes. Hint: Apply it twice.

Why?

The helicopter goes up with a steady velocity, so it's motion is controlled, not free fall or projectile motion right? So how can I determine it's velocity when it's at a height of 2.0m?

You are given the velocity of the helicopter.

Furthermore, when the commando jumps off it, why isn't his velocity zero? How come the commando have a velocity the moment he jumps off?

Because he is moving along with the helicopter.

I used the range equation, and I found that u^2 sin 2θ = 588.

Then I made it: 2 u^2 sin θ cosθ = 588, then divide it by u sin θ = 11.76

so u cos θ = 25, which makes u tan θ = 25/11.76.

∴ θ= 25.2. Is this the way?

That's a fine way to do it.

You can also use an equation for the horizontal distance:
u cosθ 1.2 = 30.

(TO LENGTHEN THE REPLY TO FOUR CHARACTERS. HOHOHO.)

That's because you messed up the quotes.

 

[coconut62]

New questions

For the objective question, the answer is D.
I understand why it is negative, but I don't know why it needs a little time for the acceleration to become zero. Notice that the dotted lines are not parallel. What happened to the motion of the ball the time when it's going to touch the ground?

For the structured question, it says velocity decreased from 30m/s to 22m/s, and distance is 26m. So I used the area under graph: 1/2 x (30+22) x t = 26, which makes t=1. My understanding is that the deceleration took 1s. Then, I calculate the remaining displacement by using the remaining area under graph: 1/2 x 4 x 22 = 44m. How come the answer is 30m?

 

[Doc Al]

For the objective question, the answer is D.
I understand why it is negative, but I don't know why it needs a little time for the acceleration to become zero. Notice that the dotted lines are not parallel. What happened to the motion of the ball the time when it's going to touch the ground?

When the ball touches the ground, the ground exerts an upward force on the ball to change its velocity. The force isn't infinite, so it takes some short time for the ball to rebound off the floor.

For the structured question, it says velocity decreased from 30m/s to 22m/s, and distance is 26m. So I used the area under graph: 1/2 x (30+22) x t = 26, which makes t=1. My understanding is that the deceleration took 1s.

Looks good. So what's the acceleration?

Then, I calculate the remaining displacement by using the remaining area under graph: 1/2 x 4 x 22 = 44m. How come the answer is 30m?

Where did you get the 4 from? Figure out the time it takes to come to rest.

 

[coconut62]

When the ball touches the ground, the ground exerts an upward force on the ball to change its velocity. The force isn't infinite, so it takes some short time for the ball to rebound off the floor.

So what's with the slightly slanting part? Is it because of the friction between the ball and the ground that makes its acceleration decrease? (This idea is stupid, but)
At the exact moment when the ball touches the ground, isn't the acceleration going to become zero straightaway? And remain zero throughtout the short impact time?

Looks good. So what's the acceleration?

Ah, I got it. By using formula. But can you explain why it isn't reasonable to directly calculate the area under graph? It's the distance traveled anyway, no?

Where did you get the 4 from? Figure out the time it takes to come to rest.

Rest time 30 - time after deceleration 26.

Because he is moving along with the helicopter.

But, when he jumps off, isn't the only force acting on him G?(ignoring air resistance)
I mean, when you jump out of the plane at a height of 2.0m as stated, you should jump out exactly at a height of 2.0m right? If you consider the helicopter continues going up during his jumping time, then the height won't be 2.0m anymore. And even so, how does it affect his initial velocity? The helicopter goes vertically up and the person jumps vertically down, how come his velocity is not completely independent of the helicopter? Does he jump from a door below the helicopter or from the edge of the helicopter? And do they make any differences? (sorry )

 

[Doc Al]

Please make proper use of quotes. (Using color to distinguish quoted text is not a good idea.)

 

[Doc Al]

So what's with the slightly slanting part? Is it because of the friction between the ball and the ground that makes its acceleration decrease? (This idea is stupid, but)
At the exact moment when the ball touches the ground, isn't the acceleration going to become zero straightaway? And remain zero throughtout the short impact time?

No. The acceleration of the ball is not zero during the collision (except for when the ground force exactly balances the weight of the ball, for perhaps an instant). Realize that the ball is squishy. It gets compressed as it rebounds. As soon as it loses contact with the ground, once again gravity is the only force acting.

Ah, I got it. By using formula. But can you explain why it isn't reasonable to directly calculate the area under graph? It's the distance traveled anyway, no?

Nothing wrong with calculating the area under the graph, if you do it correctly.

Rest time 30 - time after deceleration 26.

Take another look. It comes to rest before time = 30.

But, when he jumps off, isn't the only force acting on him G?(ignoring air resistance)

After he jumps, only gravity acts.

I mean, when you jump out of the plane at a height of 2.0m as stated, you should jump out exactly at a height of 2.0m right? If you consider the helicopter continues going up during his jumping time, then the height won't be 2.0m anymore.

Assume he jumps when the height is exactly 2.0 m.

And even so, how does it affect his initial velocity? The helicopter goes vertically up and the person jumps vertically down, how come his velocity is not completely independent of the helicopter?

If step out of a car moving at 60 mph is your speed independent of the car? At the moment he jumps, his speed is exactly that of the helicopter. (Assuming he doesn't shove off and give himself extra speed.)

Does he jump from a door below the helicopter or from the edge of the helicopter? And do they make any differences?

No difference.

 

[coconut62]

Thank you very much. Really helps me a lot.
Can I just post in this thread everytime I have questions?

1. If ball X is dropped vertically off a height and ball Y is thrown horizontally from the same height at the same time, why will they reach the ground at the same time? Isn't Y's path longer?

The following two questions are attached together with my workings.

2. Value of T2 is totally unreasonable. How should I arrange the forces? Since the two angles are both 45°, I can either put tan 45 = T2/100 or tan 45= 100/T2. How could that be?

3. Is my working correct? I don't know how to continue. I cannot just substitute e=1/2 into the equation, right? Even if I do, I still couldn't get the value of Va.

(here is the 4rd picture)
http://postimage.org/image/475ly8uzh/

 

[Doc Al]

Thank you very much. Really helps me a lot.

You are welcome.

Can I just post in this thread everytime I have questions?

You'll probably get help a lot quicker if you post a new thread. We have many top notch helpers here.

1. If ball X is dropped vertically off a height and ball Y is thrown horizontally from the same height at the same time, why will they reach the ground at the same time? Isn't Y's path longer?

The path may be longer but it's also moving faster. The only thing that counts towards the time it takes to reach the ground is the vertical motion. And since both have zero initial speed in the vertical direction, they will both fall at the same rate.

Any horizontal speed only makes it go sideways as it falls. Who cares?

The following two questions are attached together with my workings.

2. Value of T2 is totally unreasonable. How should I arrange the forces? Since the two angles are both 45°, I can either put tan 45 = T2/100 or tan 45= 100/T2. How could that be?

Well, what does tan 45° equal? Does it matter?

How did you get a value of T2 = 0.01 N? Check your work.

3. Is my working correct? I don't know how to continue. I cannot just substitute e=1/2 into the equation, right? Even if I do, I still couldn't get the value of Va.

Unfortunately, I cannot view the picture you linked.

 

[coconut62]

All solved. Bless you.

Unfortunately, I cannot view the picture you linked.

Here it is.

--------------------------
Another question: (Since it's still projectile motion, I might as well continue here)

Please view the second image.

I have problems with d) and f).

For d), I have found the t above the line to be 1.53s. For the part below it, I tried two ways:

1. Vertical component of velocity: 15sin30
t = 15sin30/G = 0.76

2. Height: 50m
t= 50/(15sin30) = 6.67 (?!)

Please explain to me why it's wrong.

For f), I calculate the distance by adding the range of the projectile motion and the horizontal component of the remaining trajectory.

Range = (U^2 sin 2θ)/g = 18^2 sin 2(30)/9.81 = 28.6m
Horizontal component = vt = 18cos30(2.52) = 39.3m

It's wrong. The real answer is 72.9.

*The answer for d) is 4.05s. I minus 1.53 from it and got the 2.52.

Thanks in advance.

 

[Doc Al]

3. Is my working correct? I don't know how to continue. I cannot just substitute e=1/2 into the equation, right?

Your work is fine. Express Va in terms of u. What minimum value of e will give Va a negative direction?

Even if I do, I still couldn't get the value of Va.

You can express Va and Vb in terms of u.

 

[Doc Al]

Another question: (Since it's still projectile motion, I might as well continue here)

Please view the second image.

I have problems with d) and f).

For d), I have found the t above the line to be 1.53s. For the part below it, I tried two ways:

1. Vertical component of velocity: 15sin30
t = 15sin30/G = 0.76

2. Height: 50m
t= 50/(15sin30) = 6.67 (?!)

Please explain to me why it's wrong.

Method 1 would find the time for something to reach that speed start from zero. (That's the time it takes to rise to the top.)

Method 2 assumes a constant speed for the descent.

Neither method will work. Why not just use:
[tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]

For f), I calculate the distance by adding the range of the projectile motion and the horizontal component of the remaining trajectory.

Range = (U^2 sin 2θ)/g = 18^2 sin 2(30)/9.81 = 28.6m
Horizontal component = vt = 18cos30(2.52) = 39.3m

It's wrong. The real answer is 72.9.

The horizontal speed is not correct.

And the easy way, once you've found the total time, is just distance = vt (for the horizontal motion). No need for the range equation.

 

[coconut62]

Method 1 would find the time for something to reach that speed start from zero. (That's the time it takes to rise to the top.)

Now I use v^2=u^2 + 2as, where v is the vertical component, u = 15sin30, a=9.8, s=50.
Then I substitute the V value into a= (v-u)/t. t= 2.52s.

Is this the way, or are there any better ones?

Neither method will work. Why not just use:
[tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]

What does the o mean?
Is this derived from the s=ut + 1/2at^2 formula?

The horizontal speed is not correct.

Ooh, it's 18 m/s.
But (18 x 2.52) + 28.6 is still not the answer. Which part is wrong?

And, please see the image again. I thought I would easily solve e) after I finished d) and f) but

I used v^2 = u^2 + 2as for vertical component.
v^2 = (15sin30)^2 + 2(9.8)(50), v= 32.2 m/s

Then to find the actual velocity, I calculate v' cos (90-30)= 32.2, which makes v' 64.4m/s which is far from the answer

 

[Doc Al]

Now I use v^2=u^2 + 2as, where v is the vertical component, u = 15sin30, a=9.8, s=50.
Then I substitute the V value into a= (v-u)/t. t= 2.52s.

Is this the way, or are there any better ones?

That's one way. I would use the formula that I provided.

What does the o mean?

The 0 signifies initial position and initial velocity. (As in T = 0.)

Is this derived from the s=ut + 1/2at^2 formula?

Yes, same thing.

Ooh, it's 18 m/s.
But (18 x 2.52) + 28.6 is still not the answer. Which part is wrong?

I believe you misapplied the range equation.

But once you have the total time, finding the horizontal distance is easy without needing the range equation.

And, please see the image again. I thought I would easily solve e) after I finished d) and f) but

I used v^2 = u^2 + 2as for vertical component.
v^2 = (15sin30)^2 + 2(9.8)(50), v= 32.2 m/s

That's fine.

Then to find the actual velocity, I calculate v' cos (90-30)= 32.2, which makes v' 64.4m/s which is far from the answer

You found the vertical component of the final velocity. What's the horizontal component?

 

[coconut62]

I believe you misapplied the range equation.

R= (u^2 sin 2θ)/g

= 15^2 sin 2(30)/9.81 where is wrong?

And why is the time taken to attain max height (15sin30)/g, regardless of the real height? Wouldn't that be assuming that the ball just went straight?

The other questions are solved. Tq.

*And why is the vertical component of a point at the other side of the trajectory = 0? By other side I mean the part on the right of the symmetrical line.

 

[Doc Al]

R= (u^2 sin 2θ)/g

= 15^2 sin 2(30)/9.81 where is wrong?

Wrong speed and wrong angle. (You need the angle with respect to the ground.)

And why is the time taken to attain max height (15sin30)/g, regardless of the real height? Wouldn't that be assuming that the ball just went straight?

The height reached depends only on the vertical motion, which is why you use the initial velocity in the vertical direction.

*And why is the vertical component of a point at the other side of the trajectory = 0? By other side I mean the part on the right of the symmetrical line.

I don't understand your question. Are you asking about position (height) or velocity?

Give me a specific example of what you mean.

 

[coconut62]

I don't understand your question. Are you asking about position (height) or velocity?

Give me a specific example of what you mean.

Velocity.

Its a projectile motion diagram, an upside down U letter.

There's a point X on the left half trajectory. It's horizontal component is U cos θ, and vertical component < u sin θ. And there's a point Y on the right half of the trajectory(the path the object falls down), it's horizontal component is still ucosθ, but it's vertical component is 0.

 

[Doc Al]

Velocity.

Its a projectile motion diagram, an upside down U letter.

There's a point X on the left half trajectory. It's horizontal component is U cos θ, and vertical component < u sin θ. And there's a point Y on the right half of the trajectory(the path the object falls down), it's horizontal component is still ucosθ, but it's vertical component is 0.

How can the vertical component be zero? The only place where the vertical component of the velocity is zero is at the apex of the trajectory.

 

[coconut62]

So at any point on the falling path, the vertical component is the same as the mirrored point on the other side?

Then how about the acceleration at the apex?

 

[Doc Al]

So at any point on the falling path, the vertical component is the same as the mirrored point on the other side?

The same magnitude, but different direction.

Then how about the acceleration at the apex?

The acceleration is the same everywhere, including at the apex.

 

[coconut62]

Okay, that's all for kinematics. Thanks a lot :)