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Original post by Some one
Alright so I understand that speed at X decreases as mass increases, but how tf does speed at Y remain the same?!?! Isnt the mass decreasing?
(If you haven't guessed already the ans is B)
Alright so I understand that speed at X decreases as mass increases, but how tf does speed at Y remain the same?!?! Isnt the mass decreasing?
(If you haven't guessed already the ans is B)
I believe you are wondering why the speed doesn't increase at Y.
At X, the sand has no initial horizontal momentum so the trolley has to share its momentum with the sand and its own momentum (and consequently velocity) decreases.
At Y, when the sand leaves it has a horizontal momentum which it takes away with it so the trolley's momentum remains unchanged since the sand takes it momentum away.
Hope that helps
Original post by InkRed
I believe you are wondering why the speed doesn't increase at Y.
At X, the sand has no initial horizontal momentum so the trolley has to share its momentum with the sand and its own momentum (and consequently velocity) decreases.
At Y, when the sand leaves it has a horizontal momentum which it takes away with it so the trolley's momentum remains unchanged since the sand takes it momentum away.
Hope that helps
At X, the sand has no initial horizontal momentum so the trolley has to share its momentum with the sand and its own momentum (and consequently velocity) decreases.
At Y, when the sand leaves it has a horizontal momentum which it takes away with it so the trolley's momentum remains unchanged since the sand takes it momentum away.
Hope that helps
Oooh. It does make some sense. But why does the sand not have horizontal momentum at X?
Original post by Some one
Oooh. It does make some sense. But why does the sand not have horizontal momentum at X?
Momentum P is the product of mass and velocity.
The sand falls on the trolley vertically. The horizontal component of its velocity is 0 as it is moving downwards not rightwards. Hence horizontal mom. Is also 0.
Original post by InkRed
Momentum P is the product of mass and velocity.
The sand falls on the trolley vertically. The horizontal component of its velocity is 0 as it is moving downwards not rightwards. Hence horizontal mom. Is also 0.
The sand falls on the trolley vertically. The horizontal component of its velocity is 0 as it is moving downwards not rightwards. Hence horizontal mom. Is also 0.
Thanks a bunch mate!
Do you mind if I ask another?
Original post by Some one
Thanks a bunch mate!
Do you mind if I ask another?
Do you mind if I ask another?
Not in the least! Do ask...
Original post by InkRed
Not in the least! Do ask...
Ok so how does this represent displacement? if the ball is falling it's displacement should decrease rather than increase. So how do we conclude that this graph represents displacement?
Original post by Some one
Ok so how does this represent displacement? if the ball is falling it's displacement should decrease rather than increase. So how do we conclude that this graph represents displacement?
Ok so how does this represent displacement? if the ball is falling it's displacement should decrease rather than increase. So how do we conclude that this graph represents displacement?
The displacement of the ball will increase regardless of the fact it is thrown downwards or upwards.
This is the difference b/w disp. and height. Height of an object is measured from the ground so an object travelling upwards would have an increasing height while one travelling downwards would have a decreasing height.
Howevver this isn't so with disp. Disp. is measured from the objects starting position. if the object moves away (in any direction) from its starting position, its displacement will increase.
When it touches the ground for the first time, its disp will be maximum as thats the furthest it can get from its starting position (say 5m above the ground). Upon touching the ground the ball rebounds and moves up and gets closer to where it was released. If it rebounds say 2m up, its disp will be 3m only as thats the distance b/w its current position and starting position so you can see the dip in the graph.
The next time the ball rebounds to say 1m. Its disp. will decrease and the graph will dip again but less than the previous rebound
The disp. will always increase, The only other way you can draw this graph is to make it entirely under the x- axis. It will be negative but still showing inc. disp.
Original post by Some one
Ok so how does this represent displacement? if the ball is falling it's displacement should decrease rather than increase. So how do we conclude that this graph represents displacement?
Ok so how does this represent displacement? if the ball is falling it's displacement should decrease rather than increase. So how do we conclude that this graph represents displacement?
Have a look at these sample graphs for this situation
Original post by Some one
OOh Ok thanks a bunch!
Just one last question pls si
I know that acceleration is negative but why does acceleration increase rather than decrease when it hits the ceiling?
Just one last question pls si
I know that acceleration is negative but why does acceleration increase rather than decrease when it hits the ceiling?
The sharp peak is formed when the ball touches the ceiling. Since there is no air resistance, the ball touches the ceiling when exactly half the time has passed. This eliminates option A and C.
Since vertically upwards is the +ve direction and acceleration is downwards it will be in the negative axis. So not option B.
This only leaves option D.
As to why accel increases....say the ball hits the ceiling with +5m/s and rebounds with the same speed but in opposite direction it will be -5m/s
This happens in very short time say 0.001s
a= v-u/t
a= 5-(-5)/0.001
a=10/0.001
a=10,000 so yeah large increase
you can ask more but quote me...i nearly missed this
Original post by InkRed
The sharp peak is formed when the ball touches the ceiling. Since there is no air resistance, the ball touches the ceiling when exactly half the time has passed. This eliminates option A and C.
Since vertically upwards is the +ve direction and acceleration is downwards it will be in the negative axis. So not option B.
This only leaves option D.
As to why accel increases....say the ball hits the ceiling with +5m/s and rebounds with the same speed but in opposite direction it will be -5m/s
This happens in very short time say 0.001s
a= v-u/t
a= 5-(-5)/0.001
a=10/0.001
a=10,000 so yeah large increase
you can ask more but quote me...i nearly missed this
Since vertically upwards is the +ve direction and acceleration is downwards it will be in the negative axis. So not option B.
This only leaves option D.
As to why accel increases....say the ball hits the ceiling with +5m/s and rebounds with the same speed but in opposite direction it will be -5m/s
This happens in very short time say 0.001s
a= v-u/t
a= 5-(-5)/0.001
a=10/0.001
a=10,000 so yeah large increase
you can ask more but quote me...i nearly missed this
Wow You're a life saver
Soooo this question has been a nightmare. Mind explaining why D is correct and rest of them are wrong?
Original post by InkRed
The sharp peak is formed when the ball touches the ceiling. Since there is no air resistance, the ball touches the ceiling when exactly half the time has passed. This eliminates option A and C.
Since vertically upwards is the +ve direction and acceleration is downwards it will be in the negative axis. So not option B.
This only leaves option D.
As to why accel increases....say the ball hits the ceiling with +5m/s and rebounds with the same speed but in opposite direction it will be -5m/s
This happens in very short time say 0.001s
a= v-u/t
a= 5-(-5)/0.001
a=10/0.001
a=10,000 so yeah large increase
you can ask more but quote me...i nearly missed this
Since vertically upwards is the +ve direction and acceleration is downwards it will be in the negative axis. So not option B.
This only leaves option D.
As to why accel increases....say the ball hits the ceiling with +5m/s and rebounds with the same speed but in opposite direction it will be -5m/s
This happens in very short time say 0.001s
a= v-u/t
a= 5-(-5)/0.001
a=10/0.001
a=10,000 so yeah large increase
you can ask more but quote me...i nearly missed this
Original post by Some one
Soooo this question has been a nightmare. Mind explaining why D is correct and rest of them are wrong?
Soooo this question has been a nightmare. Mind explaining why D is correct and rest of them are wrong?
Yo!
Sorry i didn't see this]
Original post by InkRed
Yo!
Sorry i didn't see this]
Sorry i didn't see this]
Np. But now you have
for the first question you can put in some numbers...
suppose the truck without sand is traveling a 6 m/s and weighs 10 kg... its momentum is now 60 units
2 kg of sand is added... the momentum of the full truck is still 60 units but the mass is 12 kg so the speed is now 5 m/s
as the 2kg of sand drops out it is moving at 5 m/s so takes 10 units of momentum with it... so 50 units are left for the truck... 10 x ? = 50.... ? = 5 m/s so the truck does not slow down or speed up.
answer = B
suppose the truck without sand is traveling a 6 m/s and weighs 10 kg... its momentum is now 60 units
2 kg of sand is added... the momentum of the full truck is still 60 units but the mass is 12 kg so the speed is now 5 m/s
as the 2kg of sand drops out it is moving at 5 m/s so takes 10 units of momentum with it... so 50 units are left for the truck... 10 x ? = 50.... ? = 5 m/s so the truck does not slow down or speed up.
answer = B
Original post by the bear
for the first question you can put in some numbers...
suppose the truck without sand is traveling a 6 m/s and weighs 10 kg... its momentum is now 60 units
2 kg of sand is added... the momentum of the full truck is still 60 units but the mass is 12 kg so the speed is now 5 m/s
as the 2kg of sand drops out it is moving at 5 m/s so takes 10 units of momentum with it... so 50 units are left for the truck... 10 x ? = 50.... ? = 5 m/s so the truck does not slow down or speed up.
answer = B
suppose the truck without sand is traveling a 6 m/s and weighs 10 kg... its momentum is now 60 units
2 kg of sand is added... the momentum of the full truck is still 60 units but the mass is 12 kg so the speed is now 5 m/s
as the 2kg of sand drops out it is moving at 5 m/s so takes 10 units of momentum with it... so 50 units are left for the truck... 10 x ? = 50.... ? = 5 m/s so the truck does not slow down or speed up.
answer = B
Can u do the 3rd one (stationary waves) . The one thats unsolved
Original post by Some one
Can u do the 3rd one (stationary waves) . The one thats unsolved
http://physics.stackexchange.com/questions/260798/which-arrangement-of-pipe-and-sources-can-create-a-standing-wave
Original post by Some one
Np. But now you have
You have yet to give your exam? I comprehend. I thought you were done with it and hence didnt care for the the question any more. You can always ask me any other mcq
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