VCE Physics Question Thread!

[FarAwaySS2]

I know this question is meant to be quite easy but I can't seem to get my head around it:

A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s/s, and keeps accelerating at this rate until it has passed the bike.

How far does the police car travel before it overtakes the motorbike?


Thanks!

[dankfrank420]

I know this question is meant to be quite easy but I can't seem to get my head around it:

A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s/s, and keeps accelerating at this rate until it has passed the bike.

How far does the police car travel before it overtakes the motorbike?


Thanks!

That's the thing with physics, it appears easy on the surface but is actually a bit challenging underneath.

Is the answer 612.5m? Here's how I worked it out:

d (motorbike) = 35t
d (car) = (0)t + 1/2(4)t^2

Equate the two formulas, and t = 0 or t = 17.5

sub in 17.5 into the initial motorbike formula, and you get 612.5m.

[AirLandBus]

This thread seems so dead. We need some more activity for us physics kiddies.

[FarAwaySS2]

Yup. That's the answer! Thank youuuuu! 

That's the thing with physics, it appears easy on the surface but is actually a bit challenging underneath.

Is the answer 612.5m? Here's how I worked it out:

d (motorbike) = 35t
d (car) = (0)t + 1/2(4)t^2

Equate the two formulas, and t = 0 or t = 17.5

sub in 17.5 into the initial motorbike formula, and you get 612.5m.

[bts]

Did they first assume light was the particle model or the wave model? who?

Thank you

[Kel9901]

Did they first assume light was the particle model or the wave model? who?

Thank you

Newton proposed the particle (corpuscular) theory for light, but Young's double slit experiment's results could only be explained by the wave model, so the wave model became the 'only' model for light as there was not yet any evidence supporting the particle model only. The experiment that lead to wave-particle duality was the photoelectric effect, whose results could be explained only by the particle model

[RedCapsicum]

Hi guys can some one help me with this question.

So a 2.5kg mass is rotated in a conical pendulum where the length of the string is 0.68 metres and thr angle between the string and the vertical is 35 degrees.
Find
a. The tension in the string
b. The speed of the mass

If the pendulum is now spun faster so that it's period is now 1.2 seconds find
A. The tension in the string
b. The angel the string makes with the vertical 

[JackSonSmith]

How do I express direction in terms of degrees. eg. 53.1 degrees counterclockwise from the ground (is there a more "scientific" way to express this?)

[Kel9901]

Hi guys can some one help me with this question.

So a 2.5kg mass is rotated in a conical pendulum where the length of the string is 0.68 metres and thr angle between the string and the vertical is 35 degrees.
Find
a. The tension in the string
b. The speed of the mass

If the pendulum is now spun faster so that it's period is now 1.2 seconds find
A. The tension in the string
b. The angel the string makes with the vertical

q1:
a. net force vertically on the mass is 0 newtons, therefore downward force due to weight=upward force due to tension.
T cos(35)=mg=2.5 * 10=25 therefore T=30.5 N

b. net force horizontally is equal to T sin(35)=17.5 N.
a=F/m=7 ms^-2
r=0.68 sin(35)=0.39 m
v^2=ar=2.73 --> v=1.65 ms^-1

q2: remember mass is still 2.5 kg and length of string is still 0.68 m
a. r=0.68 sin(θ), v=2πr/T=3.56 sin(θ)
a=v^2/r=18.6 sin(θ)
F=ma=46.6 sin(θ)
horizontal component of tension force=T sin(θ) --> T=46.6 N

b. vertical component of tension force=T cos(θ)=mg
46.6 cos(θ)=25
cos(θ)=0.54
θ=57.6 degrees

Hope this helped!

edit:

How do I express direction in terms of degrees. eg. 53.1 degrees counterclockwise from the ground (is there a more "scientific" way to express this?)

53.1 degrees from the horizontal is probably best. I definitely wouldn't use bearings.

[RedCapsicum]

Thank-you so much for your help! 

[JackSonSmith]

A person pushes a lawnmower of mass 20kg at a constant speed with a force of 100N directed along the handle, which is at 35 degrees to the horizontal.

What force must the person exert on the lawnmower in order to give it a velocity of 2.0 m/s in 2.5 seconds?

[AirLandBus]

A person pushes a lawnmower of mass 20kg at a constant speed with a force of 100N directed along the handle, which is at 35 degrees to the horizontal.

What force must the person exert on the lawnmower in order to give it a velocity of 2.0 m/s in 2.5 seconds?

I have a answer, whether it is right or not is another issue. Whats the answer in the book say? Cross check with mine.

[Kel9901]

A person pushes a lawnmower of mass 20kg at a constant speed with a force of 100N directed along the handle, which is at 35 degrees to the horizontal.

What force must the person exert on the lawnmower in order to give it a velocity of 2.0 m/s in 2.5 seconds?

Lawnmover moving at constant speed therefore acc. to Newton's first law forces are balanced therefore frictional forces=100 cos(35)=81.9 N.

assimung it accelerates from rest:
u=0, v=2, t=2.5, a=?
v=u+at
2=0+2.5a
a=0.8 ms^-2
Net force acting on lawnmower=ma=16 N=driving force (in the direction of motion)-frictional forces=F cos(35)-81.9
F cos(35)=97.9 N
F=120 N

Person must exert a force of 120 N.

[AirLandBus]

Lawnmover moving at constant speed therefore acc. to Newton's first law forces are balanced therefore frictional forces=100 cos(35)=81.9 N.

assimung it accelerates from rest:
u=0, v=2, t=2.5, a=?
v=u+at
2=0+2.5a
a=0.8 ms^-2
Net force acting on lawnmower=ma=16 N=driving force (in the direction of motion)-frictional forces=F cos(35)-81.9
F cos(35)=97.9 N
F=120 N

Person must exert a force of 120 N.

Can confirm, also got 120 N.

[JackSonSmith]

A ball of mass 0.15kg hits the floor at 12 m/s and rebounds at 9 m/s. The ball was in contact with the floor for 0.12s.

a) What is the maximum height the ball will rise to on the rebound?  Ans = 4.05m

b) What is the average size of the reaction force from the floor on the ball? Ans = 26.25N

Could someone please give me a clue as to how to work this out?