• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Physics: Post your doubts here!

usama321

usama321

Messages
1,258
Reaction score
1,397
Points
173
Thought blocker said:
View attachment 38385
Click to expand...
PE lost from P to Q is converted to KE. Thus the 50J at P is converted to KE, that is 50J. It also has initial KE of 5J, so total becomes 55J. 10J is lost in work against friction, so the rest that is left is 45J.
Thought blocker said:
Answer is B or D ?
Explain please

View attachment 38388
Click to expand...

The answer should be B

At the greatest height, the velocity of the ball would be 0. Thus this option can be ruled out.

For B, as we are neglecting air resistance, no energy is lost while doing work against friction. Thus, all the initial KE is converted to PE, and back to KE as it falls. This should be correct.

For C, the principle of conservation of momentum applies to a system, not just the ball. We would have considered this if it had stated that the momentum of the ball and the earth remained constant.

For D, we know that the ball is decelerating. Thus, it does not cover the same amount of distance every second. As PE depends on height, it therefore can not increase uniformly
 
Thought blocker

Thought blocker

Messages
8,477
Reaction score
34,837
Points
698
usama321 said:
PE lost from P to Q is converted to KE. Thus the 50J at P is converted to KE, that is 50J. It also has initial KE of 5J, so total becomes 55J. 10J is lost in work against friction, so the rest that is left is 45J.


The answer should be B

At the greatest height, the velocity of the ball would be 0. Thus this option can be ruled out.

For B, as we are neglecting air resistance, no energy is lost while doing work against friction. Thus, all the initial KE is converted to PE, and back to KE as it falls. This should be correct.

For C, the principle of conservation of momentum applies to a system, not just the ball. We would have considered this if it had stated that the momentum of the ball and the earth remained constant.

For D, we know that the ball is decelerating. Thus, it does not cover the same amount of distance every second. As PE depends on height, it therefore can not increase uniformly
Click to expand...
ty :)
 
zonaali

zonaali

Messages
4
Reaction score
4
Points
1
sagar65265 said:
The two most important things to note for this question are as follows:

i) Since the stri
sagar65265 said:
The two most important things to note for this question are as follows:

i) Since the string is "mass-less", the tension in every part of the string is the same. In other words, the string pulls each block with the same force; it pulls the 2.0 kg block to the right with the same force as it tries to pull the 1.0 kg block upwards.

ii)Since the string is assumed to be in-extensible (in other words the length of the string cannot change, it cannot stretch at all) the acceleration magnitude of the two blocks are the same; if the 1.0 kg block moves a small distance downwards, the 2.0 kg block has to move the same distance in the horizontal direction. So, since both of them have to fall the same distance in the same time interval, they have the same velocity. From here, since any change in the speed of the 1.0 kg block during a particular time interval has to result in the SAME change in speed of the 2.0 kg block during the EXACT SAME time interval, the two accelerations at any point are the same.

Let's define the directions, first: for the 1.0 kg block the positive direction is downwards and the negative direction is upwards (it doesn't travel or accelerate sideways, so we don't need to worry about that direction). Also, for the 2.0 kg block, the positive direction is to the right and the negative direction is to the left.
We've chosen the axis so that the accelerations are all positive in sign, so that we don't have to juggle around any confusing negative values.

So, taking Newton's second law for the 1.0 kg block (Net Force = Mass * Acceleration):

(1.0 kg)(9.81 ms^-2) - T = (1.0 kg)a
9.81 - T = a

Alright, so that's the first equation. Considering the 2.0 kg block:

T = (2.0 kg)a
T = 2.0a

And there's the second one. Since there is no friction acting on the 2.0 kg block, that's it!

Now, since the two accelerations and the two tension forces are the same, we can substitute the second equation into the first as:

9.81 - 2.0a = 1.0a
9.81 = 3.0a

Therefore, a = 9.81/3 = 3.27 ms^-2

Since the acceleration here is constant and does not change with time, we can use the constant acceleration kinematics equations with:
u = 0 ms^-1
a = 3.27 ms^-2
s = 0.50 m (since both blocks travel this distance before the 1.0 kg block hits the ground and stops)
v = ?

v^2 = u^2 + 2 * a * s
v^2 = 0 + 2 * 3.27 * 0.5
v^2 = 3.27

Therefore, the final speed is √(3.27) = 1.8(08) ms^-1 = 1.8 ms^-1 = A.

Hope this helped!
Good Luck for all your exams!
Click to expand...
Thanks alot sagar. Its been a great help. Thank you very much. God bless.
ng is "mass-less", the tension in every part of the string is the same. In other words, the string pulls each block with the same force; it pulls the 2.0 kg block to the right with the same force as it tries to pull the 1.0 kg block upwards.

ii)Since the string is assumed to be in-extensible (in other words the length of the string cannot change, it cannot stretch at all) the acceleration magnitude of the two blocks are the same; if the 1.0 kg block moves a small distance downwards, the 2.0 kg block has to move the same distance in the horizontal direction. So, since both of them have to fall the same distance in the same time interval, they have the same velocity. From here, since any change in the speed of the 1.0 kg block during a particular time interval has to result in the SAME change in speed of the 2.0 kg block during the EXACT SAME time interval, the two accelerations at any point are the same.

Let's define the directions, first: for the 1.0 kg block the positive direction is downwards and the negative direction is upwards (it doesn't travel or accelerate sideways, so we don't need to worry about that direction). Also, for the 2.0 kg block, the positive direction is to the right and the negative direction is to the left.
We've chosen the axis so that the accelerations are all positive in sign, so that we don't have to juggle around any confusing negative values.

So, taking Newton's second law for the 1.0 kg block (Net Force = Mass * Acceleration):

(1.0 kg)(9.81 ms^-2) - T = (1.0 kg)a
9.81 - T = a

Alright, so that's the first equation. Considering the 2.0 kg block:

T = (2.0 kg)a
T = 2.0a

And there's the second one. Since there is no friction acting on the 2.0 kg block, that's it!

Now, since the two accelerations and the two tension forces are the same, we can substitute the second equation into the first as:

9.81 - 2.0a = 1.0a
9.81 = 3.0a

Therefore, a = 9.81/3 = 3.27 ms^-2

Since the acceleration here is constant and does not change with time, we can use the constant acceleration kinematics equations with:
u = 0 ms^-1
a = 3.27 ms^-2
s = 0.50 m (since both blocks travel this distance before the 1.0 kg block hits the ground and stops)
v = ?

v^2 = u^2 + 2 * a * s
v^2 = 0 + 2 * 3.27 * 0.5
v^2 = 3.27

Therefore, the final speed is √(3.27) = 1.8(08) ms^-1 = 1.8 ms^-1 = A.

Hope this helped!
Good Luck for all your exams!
Click to expand...
 
taffycandy

taffycandy

Messages
38
Reaction score
16
Points
18
hey can u help me with this question
Q) a certain charge Q is dived into two parts q and Q-q,which are then separated by a certain distance.what must q be in terms of Q to maximize the electrostatic repulsion between the two charges?
Q)
two free particles (that is,free to move )with charges +q and +4q are at distance L apart . A third charge is placed so that the entire system is in equilibrium is unstable.
 
You must log in or register to reply here.
Top