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# A weird physics question...pls help !! watch

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1. yes i am man..
why?

hey...help me ans q3 plsss
2. (Original post by MalaysianDude)
yes i am man..
why?

hey...help me ans q3 plsss
im doing it on the 14th too
i just helped u already
3. 1) A truck with mass of 10000kg moves with 0.5ms^-1
sand from above the truck is passed onto the truck with rates of 40kgs^-1
how much additional horizontal force is needed for the truck to move at constant speeds of 0.50ms^-1 when weight is added to the truck constantly throughout the journey

Well we know that you want the truck to move constantly, so this means the extra force we are looking for is actually the force required to accelerate 40kg of sand per second. But simple PCLM will not work here, because after the first second, the mass of the truck increases. So we can write the mass of the truck is equal to: 10000 + t(40), as every second this increases by 40kg. So we can write:

(10000 + t(40))0.5 = (10000 + t(40) + 40)v

As the collision is inelastic, we group the two masses together (the mass on the LHS is not shown because the sand and the train collide when the sand's horizontal velocity is 0).

To find the velocity, in terms of time, if the train is left alone:

v = (5000 + 20t)/(10040 + 40t)

Which is reducing as long at the time is positive, which you'd expect. Now find the decelleration this causes by... you guessed it... differentiation! Using the quotient rule:

dv/dt = [(5000 + 20t)D(10040 + 40t) - D(5000 + 20t)(10040 + 40t)]/(10040 + 40t)²
dv/dt = [(5000 + 20t)(40) - (20)(10040 + 40t)]/(10040 + 40t)²
dv/dt = (200000 + 800t) - 200800 + 800t)/(10040 + 40t)²
dv/dt = -800/(10040 + 40t)² = a

Which is what we want - decelleration. Now to find the force: F = ma, so:

F = -800(10000 + 40t)/(10040 + 40t)²

That's my answer anyway. The force needed therfore decreases, I think.
Is this right? I'm sure I made some errors, half of this stuff I've picked up from P3 and M2 topics. Lol.

2) A parachutist falls from an airplane travelling at 67ms^-1.After traveling for 80m away from the plane,it opens up the parachute and its velocity at this time is approximately 40ms^-1.
Determine the magnitude and direction of the resultant velocity of the parachuutist at this point

Taking horizontal first, and ignoring air resistance, he is still travelling at 67 m s^-1 in the direction of the plane. Now taking vertical components:

u = 0
v = ?
x = 80
a = 9.81

[v² = u² + 2ax]
v² = 0 + 2+80+9.81 = 1569.6
v = 39.62 m s^-1

Since we assume the plane is travelling horizontal, these two vectors are at right angles and their resultant we know is 40 (or 39.62, as i'm guessing the question meant it's vertical velocity).

R = root[(39.62)^2 + (67)^2 ] = 77.84 m s ^-2
Can't be bothered with angle, sry!

3) How do you connect newton`s third law pair.Give 2 examples.

They act in opposite directions, on seperate bodies, with equal magnitudes.
Eg. The force I feel towards the Earth, and the force the Earth feels towards me. And the contact force I feel from the ground, and the contact force the ground feels from me.
4. Someone tell me if my working is right in the first opne please?
I surprised myself with that, didn't think dv/dt would come up so easily.
5. (Original post by mik1a)
Someone tell me if my working is right in the first opne please?
I surprised myself with that, didn't think dv/dt would come up so easily.
first one is wrong i dont know why though it seems like it shud work.
there is a more simple way look on previous pages i posted how to do it
6. (Original post by mik1a)
1) A truck with mass of 10000kg moves with 0.5ms^-1
sand from above the truck is passed onto the truck with rates of 40kgs^-1
how much additional horizontal force is needed for the truck to move at constant speeds of 0.50ms^-1 when weight is added to the truck constantly throughout the journey

Well we know that you want the truck to move constantly, so this means the extra force we are looking for is actually the force required to accelerate 40kg of sand per second. But simple PCLM will not work here, because after the first second, the mass of the truck increases. So we can write the mass of the truck is equal to: 10000 + t(40), as every second this increases by 40kg. So we can write:

(10000 + t(40))0.5 = (10000 + t(40) + 40)v

As the collision is inelastic, we group the two masses together (the mass on the LHS is not shown because the sand and the train collide when the sand's horizontal velocity is 0).

To find the velocity, in terms of time, if the train is left alone:

v = (5000 + 20t)/(10040 + 40t)

Which is reducing as long at the time is positive, which you'd expect. Now find the decelleration this causes by... you guessed it... differentiation! Using the quotient rule:

dv/dt = [(5000 + 20t)D(10040 + 40t) - D(5000 + 20t)(10040 + 40t)]/(10040 + 40t)²
dv/dt = [(5000 + 20t)(40) - (20)(10040 + 40t)]/(10040 + 40t)²
dv/dt = (200000 + 800t) - 200800 + 800t)/(10040 + 40t)²
dv/dt = -800/(10040 + 40t)² = a

Which is what we want - decelleration. Now to find the force: F = ma, so:

F = -800(10000 + 40t)/(10040 + 40t)²

That's my answer anyway. The force needed therfore decreases, I think.
Is this right? I'm sure I made some errors, half of this stuff I've picked up from P3 and M2 topics. Lol.

2) A parachutist falls from an airplane travelling at 67ms^-1.After traveling for 80m away from the plane,it opens up the parachute and its velocity at this time is approximately 40ms^-1.
Determine the magnitude and direction of the resultant velocity of the parachuutist at this point

Taking horizontal first, and ignoring air resistance, he is still travelling at 67 m s^-1 in the direction of the plane. Now taking vertical components:

u = 0
v = ?
x = 80
a = 9.81

[v² = u² + 2ax]
v² = 0 + 2+80+9.81 = 1569.6
v = 39.62 m s^-1

Since we assume the plane is travelling horizontal, these two vectors are at right angles and their resultant we know is 40 (or 39.62, as i'm guessing the question meant it's vertical velocity).

R = root[(39.62)^2 + (67)^2 ] = 77.84 m s ^-2
Can't be bothered with angle, sry!

3) How do you connect newton`s third law pair.Give 2 examples.

They act in opposite directions, on seperate bodies, with equal magnitudes.
Eg. The force I feel towards the Earth, and the force the Earth feels towards me. And the contact force I feel from the ground, and the contact force the ground feels from me.
sorry mik1a your first answer is wrong man..
it looks complicated but its actually really simple and direct..
but the second one takes a little thinking and it seems to me you really worked to get the answer

anyways the first answer is 20N
7. 1. F = dp/dt = (10040*.5- 10000*.5)/1 = 20N

2. v = root(67^2 + 40^2) = 78m/s
angle = arctan(40/67)= 30.8 degrees below horizontal

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