Help needed for physics question!
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A 600 kg toy car moving at 3 m/s collides and hooks up with a 900 kg toy car at rest and they move off together. What is their final velocity?
i know its by using conservation of linear momentum err i still cant work it out? -
600kg toy car? transformers? -
If use conservation of momentum,Originally posted by WoBuPaNi:A 600 kg toy car moving at 3 m/s collides and hooks up with a 900 kg toy car at rest and they move off together. What is their final velocity?
i know its by using conservation of linear momentum err i still cant work it out?
m1v1 = m2v2
m1 = 600kg
v1 = 3m/s
m2 = 600kg + 900kg = 1500kg
v2 = ?
hence, 1800 = 1500v2
v2 = 1.2 m/s -
WoW! thanks!Originally posted by eagle:If use conservation of momentum,
m1v1 = m2v2
m1 = 600kg
v1 = 3m/s
m2 = 600kg + 900kg = 1500kg
v2 = ?
hence, 1800 = 1500v2
v2 = 1.2 m/s -
need help on another question
An engine is used to pull a train of two cars out of a mine. The floor slopes upward at an angle of 22 degrees. Each car has a mass of 1.2x104 Kg and normally travels without friction on the tracks. The engine can exert a maximum force of 0.9x105 N on car A.
If the engineer again throttles back so that the force exerted by the engine on car A decreases at the constant rate of 3.3 N per second, how long before the train stoprs moving up the track?
Assume the original speed was 2.8 m/s -
No problem. I was once very powerful in physics... 6 yrs ago... when I was doing my A levels...Originally posted by WoBuPaNi:WoW! thanks! -
eagle kor kor help me 1 more timeOriginally posted by eagle:No problem. I was once very powerful in physics... 6 yrs ago... when I was doing my A levels... -
Wait ah... I work on it...Originally posted by WoBuPaNi:eagle kor kor help me 1 more time -
I'm not so sure of the answer...
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some careless mistake...
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Wondering... If u left out any other details. -
I'm not sure if it's correct this time... A bit out of touch
Assume u mean that
mass of car A = 1.2 x 10^4 kg = mass of car B (2 cars) = Mcar (mass of car)
max force = 0.9 x 10^5 kg
original speed = 2.8m/s
dF/dt = -3.3
total downward force along slope due to gravity = 2 x Mcar x g x sin 22
= 8.82x 10^4 N
So now we consider the 2 cars as one unit, which means there is a total mass of 2Mcar
If we assume constant speed intially,
initial upward force = 8.82x 10^4 N
Starting from dF/dt = -3.3
gives us F = -3.3t + F1
When t = 0, F = 8.82x 10^4 N, hence F1 = 8.82x 10^4 N
since F = ma,
ma = -3.3t + 8.82x 10^4
gives us by integration
mv = -3.3t^2 + 8.82x 10^4 t + v1
When t = 0, v = 2.8
Hence v1 = 2.8m, where m = 2Mcar
Hence, V1 = 6.72 x 10^4
hence mv = -3.3t^2 + 8.82x 10^4 t + 6.72 x 10^4
When v = 0, -3.3t^2 + 8.82x 10^4 t + 6.72 x 10^4 = 0
Solving the above quadratic equation, we get
t = -0.76 (reject) or 26728s
Hence, t = 26728s approximately = 445 mins
Someone else pls help me confirm -
Originally posted by xalkyrie: 600kg toy car? transformers?
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Why no one tell me whether I'm wrong
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someone help to confirm answer for eagle ??
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Your working method seems to be correct but not sure about calculations though.
Somehow I think that we can get the answer through another method but I will try that later. -
kaos..wad topic...
i haven learn sia...
i m jc1 btw -
lol.... just as my nick implies i also dun know i am right a not but i will give some of my reasoning:
To Eagle:
I think i found some mistake in your working, please pardon me if i identiy them wrongly i also sort of lost touch liao.
1) Your working: If we assume constant speed intially,
initial upward force = 8.82x 10^4 N
if we assume constant speed, then acceleration = 0, then force = 0 right?
2) I think u forget to resolve the forces. The force exerted by the two cars opposes the force exerted by the engine. Hence, we need to find the Resultant force.
3) I think the "Assume the original speed was 2.8 m/s" serves a bait to confuse the problem solver.
Ok, now i will write my working:
Total masses of the 2 cars: 1.2x10^4 x 2 = 2.4 x 10^4 kg
Force exerted by the cars along the slope = 2.4 x 10^4 x 9.81 x sin22
= 88197.376 N
Resultant Upward Force = 0.9 x 10^5 - 88197.376 = 1802.624N
Retardation Force by engine due to throttle back = 3.3 N/s
i.e. Time needed for the force to reach 0 = 1802.624 / 3.3 = 546 s
OK... this is the time need for the train to come to a halt from an inital velocity of 2.8m/s. After which, the train will begin to slip back down again if the engine continue to throttle back. -
help anyone to confirm the answers?
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Originally posted by Uncertain:
I thought of that, but 0.9 x 10^5 is the MAX force of the engine, not the initial driving force.
lol.... just as my nick implies i also dun know i am right a not but i will give some of my reasoning:
To Eagle:
I think i found some mistake in your working, please pardon me if i identiy them wrongly i also sort of lost touch liao.
1) Your working: If we assume [b]constant speed intially,
initial upward force = 8.82x 10^4 N
if we assume constant speed, then acceleration = 0, then force = 0 right?
2) I think u forget to resolve the forces. The force exerted by the two cars opposes the force exerted by the engine. Hence, we need to find the Resultant force.
3) I think the "Assume the original speed was 2.8 m/s" serves a bait to confuse the problem solver.
Ok, now i will write my working:
Total masses of the 2 cars: 1.2x10^4 x 2 = 2.4 x 10^4 kg
Force exerted by the cars along the slope = 2.4 x 10^4 x 9.81 x sin22
= 88197.376 N
Resultant Upward Force = 0.9 x 10^5 - 88197.376 = 1802.624N
Retardation Force by engine due to throttle back = 3.3 N/s
i.e. Time needed for the force to reach 0 = 1802.624 / 3.3 = 546 s
OK... this is the time need for the train to come to a halt from an inital velocity of 2.8m/s. After which, the train will begin to slip back down again if the engine continue to throttle back.[/b]
Which is why I wonder if any details are left out -
Furthermore, you cannot take the force to divide by 3.3N simply because the velocity is not zero when the resultant force is zero. There is inertia to be taken care of
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erm no details left out
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I have assumed initial speed to be constant...
Unless the max force is the initial driving force, in which case, u can switch the variable F1 in my working to the max driving force.
Btw, do u hv any answer for this question?
My working doesn't seem to be for normal JC physics, only physics S. Unless I am already outdated... -
yea.. t = 40s
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shld be in schoolwork forum mah
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ok... found another careless mistake in my force calculations.... I try again...
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Is that the full question? Or is it just part of a question -
Is this A lvl or O lvl std?