Needs help to some Mechanics questions

[anzaar]

i solved part(ii) Q.5 also. For Q.7 use net force=ma

 

[Amy Bloom]

i solved part(ii) Q.5 also. For Q.7 use net force=ma

Sorry u didn't check the document again. U didn't find the value of k.

 

[anzaar]

Q.7.
Part(i)
2.5-T=.25a ................. (a)
T-1.5=.15a ..................(b)
solve (a) and (b)
we get 2.5-1.5=0.4a
a=2.25 m/s^2
for Part(ii)
after 2 sec B hits the ground, here we find the speed when it hits.
v=u+at=0+2.25*2=4.50 m/sec
this is the speed when A starts slang so if we consider the initial speed of A and final speed at highest point is zero.
then
v=u-gt
0=4.50-10t
t=4.50/10=0.45 sec
so total time when string becomes slack till it taut again=0.45+0.45=0.9 sec.
hope you have understood.

 

[Amy Bloom]

Thanks a lot. ^_^

 

[Amy Bloom]

anzaar Can u help me wid these please?

 

[Hephistocles]

I'm pretty sure your question one is not possible to answer without knowing how fast (backwards) the runner is travelling when t=35. For example, if v=-1, then the area of that triangle = 0.5*10*-1 = -5 so he travelled 5 metres. If v=-10, then he travelled 50 metres.

Question 2 is basic SUVAT as far as I can tell. Work out T, at the maximum height of A (when V=0), then work out the height (S) of B at that time. Part ii) use SUVAT again, and call the height of A = x. Then the height of B = x + 0.9, and you'll end up with a pair of simultaneous equations to work out.

 

[Amy Bloom]

Hello there. Can someone help me with these:
anzaar Hephistocles

 

[celoth]

Hello there. Can someone help me with these:
anzaar Hephistocles
View attachment 14494

answer for no4)
by differentiating equation of displacement,

speed = t + 0.1t^2
when t=10,
speed = 10 + 0.1(10^2)
= 20m

equation of acceleration:

a = 1+0.2t
when t=0, a=1m/s^2 , this one is initial acceleration

2x1 = 1 + 0.2t
t=5s

 

[Hephistocles]

7.i)
Using SUVAT again, except because you're on a slope you first need to split up the 60deg velocity into horizontal and vertical components. Horizontal component will be the same all the way up, but the vertical component will change, with a=-9.8 and s=2.5sin(60). Once you have the horizontal and vertical velocities, recombine to find the final speed.

ii)
Firstly, calculate the Kinetic Energy of the particle at B (KE = 0.5mv^2). Then imagine that was all converted to Gravitational Potential Energy (GPE = mgh). Equate KE = GPE and find h (I believe it will be 3.2)

iii)
Because by this stage, any GPE gained by the particle since it went past B will have been converted back into KE (assuming a perfect system)

iv)
Again, because you're on a slope, you need to split things into vertical and horizontal components. There are three things you need to be aware of - the inital speed, the frictional force and the acceleration due to gravity. You could either split the first two into horizontal and vertical components, or it would probably be quicker to instead resolve the acceleration due to gravity to work out the acceleration going down the slope. Up to you, whatever you find easier. Once you've resolved everything, it's pretty much just F=ma to work out the deceleration from friction, and finally work out speed through SUVAT.

 

[Amy Bloom]

Can anybody help, its quite urgent.

7.i)
Using SUVAT again, except because you're on a slope you first need to split up the 60deg velocity into horizontal and vertical components. Horizontal component will be the same all the way up, but the vertical component will change, with a=-9.8 and s=2.5sin(60). Once you have the horizontal and vertical velocities, recombine to find the final speed.

ii)
Firstly, calculate the Kinetic Energy of the particle at B (KE = 0.5mv^2). Then imagine that was all converted to Gravitational Potential Energy (GPE = mgh). Equate KE = GPE and find h (I believe it will be 3.2)

iii)
Because by this stage, any GPE gained by the particle since it went past B will have been converted back into KE (assuming a perfect system)

iv)
Again, because you're on a slope, you need to split things into vertical and horizontal components. There are three things you need to be aware of - the inital speed, the frictional force and the acceleration due to gravity. You could either split the first two into horizontal and vertical components, or it would probably be quicker to instead resolve the acceleration due to gravity to work out the acceleration going down the slope. Up to you, whatever you find easier. Once you've resolved everything, it's pretty much just F=ma to work out the deceleration from friction, and finally work out speed through SUVAT.

Hey thanks for the help, i'll try it again.

 

[Amy Bloom]

Hey anybody can help me with this, its quite urgent please.

 

[celoth]

Hey anybody can help me with this, its quite urgent please.
View attachment 14578

1st hafta find the acceleration:

70N - T = 7a
T - 20N = 2a
therefore, a=50/9

then u=0, s=1m (because 2kg block went up 1m, 7kg block went down 1m, so totally 2m apart.)

v^2 = 0 + 2(50/9)(1)
v = 3.33

 

[Amy Bloom]

1st hafta find the acceleration:

70N - T = 7a
T - 20N = 2a
therefore, a=50/9

then u=0, s=1m (because 2kg block went up 1m, 7kg block went down 1m, so totally 2m apart.)

v^2 = 0 + 2(50/9)(1)
v = 3.33

Thanks loads

 

[Amy Bloom]

Hello. Can anybody answer this for me please.
smzimran i need your help, please.

 

[Hephistocles]

There are three types of energy in the question: Kinetic, Gravitational Potential, and Work Done by the sea's resistance. To begin with, the missile has only kinetic energy (KE = 0.5 * mv^2). As it goes through the water, it loses energy due to work done (WD = force * distance), and also loses some kinetic energy converted into gravitational potential (GPE = mgh). The KE as it leaves the water is calculated by

Final KE = Initial KE - WD - GPE at 600m.
= 0.5*50*200*200 - 160*600 - 50*9.81*600
= 1000kJ - 96kJ - 294kJ
= 610 kJ

Once you have the KE at sea level, then you just need to use GPE = mgh again, in reverse to get the maximum height (since all the KE will have been converted into GPE).

height = 610000 / (50 * 9.81)
= 1244 m

The differences in my values to yours are probably due to using a different value for g.

 

[Amy Bloom]

Hephistocles
thanks loads mate. yeah u're right for (ii)

 

[Amy Bloom]

Can anybody out here help me with these:
(1) A particle of mass 100g moves in a straight line across a horizontal surface against resistances of constant magnitude. The particle passes through a point A with a speed of 7ms-1 and then through a point B with a speed of 3ms-1, B being 2 metres from A. Find the magnitude of the resistances.
For this my answer was 1000N but the book says its 1N, can you explain

(2) A slope is inclined at an angle theeta to the horizontal where tan theeta is 4/3. A particle Q is projected from the foot of the slope a line of greatest slope wit h a speed V ms-1 and comes instantaneously at rest after travelling 6 metres. if the coefficient of friction between Q and the slope is 0.5, calculate:
(a) the value of V [11.5 ms-1] --- i'm okay with this
(b) the speed of Q when it returns to its starting point [7.76 ms-1] ---- my biggest trouble!

 

[celoth]

Can anybody out here help me with these:
(1) A particle of mass 100g moves in a straight line across a horizontal surface against resistances of constant magnitude. The particle passes through a point A with a speed of 7ms-1 and then through a point B with a speed of 3ms-1, B being 2 metres from A. Find the magnitude of the resistances.
For this my answer was 1000N but the book says its 1N, can you explain

1st, find a, the acceleration,
since v=3, u=7, s=2 then a=-10,

then,
0-R=ma
R=-(0.1)(-10) = 1N

 

[Amy Bloom]

1st, find a, the acceleration,
since v=3, u=7, s=2 then a=-10,

then,
0-R=ma
R=-(0.1)(-10) = 1N

Hey thanks. Can u help me in (2)

 

[Amy Bloom]

Help! Just give me an idea, then i'll work it out.
celoth & Hephistocles nightrider1993