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Author Topic: VCE Physics Question Thread!  (Read 603397 times)  Share 

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paper-back

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Re: VCE Physics Question Thread!
« Reply #825 on: February 08, 2015, 12:57:37 pm »
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In momentum questions which require the use of the formula m1u1+m2u2=m1v1+m2v2, do we assign directions?
E.g. Question
A cart with mass 2kg moving at 2m/s in westerly direction collides with cart moving at 3m/s in easterly direction. Upon collision, carts connect and move of at 1m/s in westerly direction
What's the mass of second cart?

Would we go;
(2)(2)-(3)m2=(2+m2)(1)?

Maths Forever

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Re: VCE Physics Question Thread!
« Reply #826 on: February 08, 2015, 01:11:56 pm »
+3
In momentum questions which require the use of the formula m1u1+m2u2=m1v1+m2v2, do we assign directions?
E.g. Question
A cart with mass 2kg moving at 2m/s in westerly direction collides with cart moving at 3m/s in easterly direction. Upon collision, carts connect and move of at 1m/s in westerly direction
What's the mass of second cart?

Would we go;
(2)(2)-(3)m2=(2+m2)(1)?

Always choose a direction to be positive! Let the easterly direction be positive.

p (Before) = m1v1 + m2v2 = 2 x -2 + m2 x 3

p (After) = (m1 + m2) v = (2 + m2) x -1

p (Before) = p (After) ......conservation of momentum

Therefore, -4 + 3 x m2 = -2 - m2

-2 = -4 m2

Hence, m2 = 0.5 kg

But your way (letting westerly direction to be positive) would also work. Hope this helps!

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Cosec

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Re: VCE Physics Question Thread!
« Reply #827 on: February 08, 2015, 01:16:04 pm »
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As far as im aware the general rule of thumb is, if its a vector, assign a direction. Which should be first nature now.

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Re: VCE Physics Question Thread!
« Reply #828 on: February 08, 2015, 01:18:27 pm »
+1
As far as im aware the general rule of thumb is, if its a vector, assign a direction. Which should be first nature now.

That's not even the rule of thumb - if you don't assign a direction, the maths breaks, lol.

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Re: VCE Physics Question Thread!
« Reply #829 on: February 08, 2015, 01:33:39 pm »
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That's not even the rule of thumb - if you don't assign a direction, the maths breaks, lol.

Hahha, gez cant get away from you. But yeah, thats what i was implying.

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Re: VCE Physics Question Thread!
« Reply #830 on: February 08, 2015, 01:35:14 pm »
+1
Hahha, gez cant get away from you. But yeah, thats what i was implying.


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Re: VCE Physics Question Thread!
« Reply #831 on: February 09, 2015, 06:30:37 pm »
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Hi,

Can someone please fully explain the following question for me? I'm confused about the way gravity and the normal force are involved for these sort of questions:

A rubber ball of mass 80 g bounces vertically on a concrete floor. The ball strikes the floor at 10 m/s and rebounds at 8.0 m/s

The time of contact between the ball and the floor during the bounce was 0.050 s.


a) Calculate the average net force acting on the ball during its contact with the floor.
b) Calculate the average force that the floor exerts on the ball.
c) Calculate the average force that the ball exerts on the floor.

thanks
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Maths Forever

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Re: VCE Physics Question Thread!
« Reply #832 on: February 09, 2015, 07:13:47 pm »
+1
Hi,

Can someone please fully explain the following question for me? I'm confused about the way gravity and the normal force are involved for these sort of questions:

A rubber ball of mass 80 g bounces vertically on a concrete floor. The ball strikes the floor at 10 m/s and rebounds at 8.0 m/s

The time of contact between the ball and the floor during the bounce was 0.050 s.


a) Calculate the average net force acting on the ball during its contact with the floor.
b) Calculate the average force that the floor exerts on the ball.
c) Calculate the average force that the ball exerts on the floor.

thanks

I = Δp = F(average) Δt = m Δv

Always assign a positive direction! Let Up be positive.

a) Sub in values: Δt = 0.050 s, m = 0.080 kg, Δv = Final - Initial = (8.0) - (- 10.0) = 18.0 m/s

F(average) x 0.050 = 0.080 x 18.0

Therefore F (average) = 1.44 / 0.050 = 28.8 N

Therefore average net force is 28.8 N in the upwards direction.

b) F (av) (floor on ball) - W (weight force) = F (av) (net)

F (av) (floor on ball) = N (normal force)

Therefore, N - mg = 28.8 (from part a)

N = 28.8 + (0.080 x 10) = 28.8 + 0.8 = 29.6 N

Therefore F (av) (floor on ball) = 29.6 N in the upwards direction.

c) F (av) (floor on ball) and F (av) (ball on floor) are a Newton's 3rd Law action reaction pair.

Hence F (av) (ball on floor) = - 29.6 N

Therefore F (av) (ball on floor) = 29.6 N in the downwards direction.

This should be correct! Hope it helps!
« Last Edit: February 09, 2015, 09:26:16 pm by Maths Forever »
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Cosec

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Re: VCE Physics Question Thread!
« Reply #833 on: February 09, 2015, 07:38:17 pm »
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This question relates to impulse.

Since I = Δp = F(average) Δt
and I = Δp = m Δv

Always assign a positive direction! Let Up be positive.

Therefore F(average) Δt = m Δv

a) Sub in values: Δt = 0.050 s, m = 0.080 kg, Δv = Final - Initial = (8.0) - (- 10.0) = 18.0 m/s

F(average) x 0.050 = 0.080 x 18.0

Therefore F (average) = 1.44 / 0.050 = 28.8 N

Therefore average net force is 28.8 N in the upwards direction.

b) I (floor on ball) = Δp (ball) = p (final) - p (initial) = m ( v (final) - v (initial) ) = 0.080 x (8.0 - (-10.0) )
= 0.080 x 18.0 = 1.44 N s (standard unit for impulse: Newton second)

Therefore I (floor on ball) = 1.44 N s in the upwards direction.

c) I (floor on ball) and I (ball on floor) are a Newton's 3rd law action reaction pair.

Hence I (ball on floor) = - 1.44 N s

Therefore I (ball on floor) = 1.44 N s in the downwards direction.

I think this is all correct! Hope it helps!

Looks good to me! Instead though i did part a with the constant acceleration motion formulae. Still got 28.8. Got a bit stumped though when i hit b and c, then i re read it and noticed "average", gets me every time. Haahha.

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Re: VCE Physics Question Thread!
« Reply #834 on: February 09, 2015, 07:47:42 pm »
0
This question relates to impulse.

Since I = Δp = F(average) Δt
and I = Δp = m Δv

Always assign a positive direction! Let Up be positive.

Therefore F(average) Δt = m Δv

a) Sub in values: Δt = 0.050 s, m = 0.080 kg, Δv = Final - Initial = (8.0) - (- 10.0) = 18.0 m/s

F(average) x 0.050 = 0.080 x 18.0

Therefore F (average) = 1.44 / 0.050 = 28.8 N

Therefore average net force is 28.8 N in the upwards direction.

b) I (floor on ball) = Δp (ball) = p (final) - p (initial) = m ( v (final) - v (initial) ) = 0.080 x (8.0 - (-10.0) )
= 0.080 x 18.0 = 1.44 N s (standard unit for impulse: Newton second)

Therefore I (floor on ball) = 1.44 N s in the upwards direction.

c) I (floor on ball) and I (ball on floor) are a Newton's 3rd law action reaction pair.

Hence I (ball on floor) = - 1.44 N s

Therefore I (ball on floor) = 1.44 N s in the downwards direction.

I think this is all correct! Hope it helps!

Thanks for the reply but I'm still slightly confused.
If the question asks for the 'average force' how can the answer be in N s? Should it not just be N?

Part a) you gave an answer in Newtons, but in part b) and c) you gave an answer in N s, even though it was asking for force? Can you please explain this to me because it's really got me confused.

Also
it might be worth mentioning that this question was taken from the god-awful Heinemann textbook - below is their version of the fully-worked solution for parts b) and c).

b)   The forces acting are gravity Fg downwards and normal reaction force FN upwards.
   Fg = mg = 0.78 N
   totalF = Fg + FN
   This gives FN = 29 N up.
c)   As described by Newton’s third law, this is equal and opposite to the force that the floor exerts on the ball, so F = 29 N down.


Rendering me totally confused

thanks and sorry for my ignorance in advance
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Re: VCE Physics Question Thread!
« Reply #835 on: February 09, 2015, 07:54:10 pm »
+3
Hi,

Can someone please fully explain the following question for me? I'm confused about the way gravity and the normal force are involved for these sort of questions:

A rubber ball of mass 80 g bounces vertically on a concrete floor. The ball strikes the floor at 10 m/s and rebounds at 8.0 m/s

The time of contact between the ball and the floor during the bounce was 0.050 s.


a) Calculate the average net force acting on the ball during its contact with the floor.
b) Calculate the average force that the floor exerts on the ball.
c) Calculate the average force that the ball exerts on the floor.

thanks

regarding nature of forces: net force=both normal reaction and weight forces, forces in parts b and c are just normal reaction forces.

a) FnetΔt=mΔv
Fnet*0.050=0.08*18
Fnet=28.8 N (UP)

b) Fnet=N-W=28.8
N-0.08*10=28.8
N=29.6 N

c) According to Newton's 3rd law, this is also 29.6 N

edit: this question looks perfectly fine imo no need to hate on heinemann
« Last Edit: February 09, 2015, 07:57:38 pm by Kel9901 »
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Maths Forever

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Re: VCE Physics Question Thread!
« Reply #836 on: February 09, 2015, 08:33:26 pm »
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Thanks for the reply but I'm still slightly confused.
If the question asks for the 'average force' how can the answer be in N s? Should it not just be N?

Part a) you gave an answer in Newtons, but in part b) and c) you gave an answer in N s, even though it was asking for force? Can you please explain this to me because it's really got me confused.

Also
it might be worth mentioning that this question was taken from the god-awful Heinemann textbook - below is their version of the fully-worked solution for parts b) and c).

b)   The forces acting are gravity Fg downwards and normal reaction force FN upwards.
   Fg = mg = 0.78 N
   totalF = Fg + FN
   This gives FN = 29 N up.
c)   As described by Newton’s third law, this is equal and opposite to the force that the floor exerts on the ball, so F = 29 N down.


Rendering me totally confused

thanks and sorry for my ignorance in advance

Sorry, I took parts b and c to mean impulse. A silly misinterpretation! Please see my edited version above, and thanks to Kel9901!
« Last Edit: February 09, 2015, 09:27:52 pm by Maths Forever »
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Re: VCE Physics Question Thread!
« Reply #837 on: February 09, 2015, 09:52:11 pm »
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Im confused now, haha. It says average, when i think of average i think of how the value changes over a period of time (im thinking calculus) and as a result thought that it would how the force changes over that time interval of impact.
Additionally, when your looking at a object that the only two forces acting on it are the weight and normal reaction forces and then you form the equation
Fw = -Fn  cause you have to take into account the different directions in which they act.
And therefore the Sum of forces = Fw + Fn.

Why above is it (Sum of forces) expressed as F (av) (floor on ball) - W (weight force) = F (av) (net)
 
Can somebody explain just to clarify for me.

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Re: VCE Physics Question Thread!
« Reply #838 on: February 09, 2015, 10:12:20 pm »
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Im confused now, haha. It says average, when i think of average i think of how the value changes over a period of time (im thinking calculus) and as a result thought that it would how the force changes over that time interval of impact.
Additionally, when your looking at a object that the only two forces acting on it are the weight and normal reaction forces and then you form the equation
Fw = -Fn  cause you have to take into account the different directions in which they act.
And therefore the Sum of forces = Fw + Fn.

Why above is it (Sum of forces) expressed as F (av) (floor on ball) - W (weight force) = F (av) (net)
 
Can somebody explain just to clarify for me.

Sorry, the notation you are confused about is only because of my lack of mathematical symbols.

The average was asked for in part (a). So therefore, we also take the average net force as the sum of all forces in parts (b) and (c).

We are calculating the average reaction force in part (b), or the average force that the floor exerts on the ball?

Does this help? Sorry for any confusion.
« Last Edit: February 10, 2015, 04:17:34 pm by Maths Forever »
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Re: VCE Physics Question Thread!
« Reply #839 on: February 10, 2015, 03:55:40 pm »
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Hey, I'm just wondering if electricity and motion the only topics that are taught in 1/2 and then expanded on in 3/4?