Solving Physics Homework Challenges

[Clari]

1. A ball is dropped from a height of 20m and rebounds with a velocity which is 3/4 of the velocity with which it hit the ground. What is the time interval between the first and second bounces?

I don't have any idea about dealing with this problem.

2.Explain why a projectile fired from a long-barrelled gun is subject to less acceleration than a projectile fired from a short-barrelled gun if the range is the same in both cases.

umm...I don't even know what is the difference between long-barrelled gun and short-barrelled gun.

3. A bird of mass 0.5kg hovers by beating its wing of effective area 0.3 m^2. Estimate the velocity imparted to the air, which has a densiy of 1.3 kgm^-3, by the beating of the wings.

again, I don't know how to do this question... >_<

 

[HallsofIvy]

You really don't have any ideas?

1. You are told the ball is dropped from 20 m so, since it's acceleration is -g= -9.8 m/s², its velocity after time t is -gt and it will have fallen d= -(g/2)t² meters in t seconds. How long will it take to fall 20 meters (solve -4.9t²= 20)? What is its speed after that time (plug that value of t into -9.8t).

You are told that the speed of the ball after the bounce is 3/4 its speed before. You just calculated the velocity before (do you remember the difference between speed and velocity? The velocity you just calculated should have been negative. The velocity after the bounce must be positive.) so it's easy to find the velocity of the ball after: call that "v₀". The height of the ball is now given by h= -(g/2)t²+ v₀t= -4.9t²+ v₀t.
Obviously, if t= 0 the height is 0 since the ball bounces from the floor. What will t be when the height is 0 again?

2. "I don't even know what is the difference between long-barrelled gun and short-barrelled gun."

Well, I presume the difference is that one has a longer barrel than the other!
(Why will people not believe what they are told!)

If the range of the two guns is the same then the speed of the bullets as they come out of the barrel must be the same. The bullet going up the longer barrel has further to go to that point and so takes a longer time to get out of the barrel- a longer time to reach the same speed. What does that tell you about the acceleration?

3. A bird of mass 0.5kg has a weight of 0.5g= 4.9 Newtons. In order to support its weight (i.e. "hover") it must create a downward force on the air of 4.9 Newtons.
I can't, off hand, think of any formula so I might try to get one by "dimensional analysis" We are given that the area of the birds wing(s) (surely it has more than one!) is 0.3 m². The density of air is given as 1.5 kg/m³. Multiplying those gives a figure of 0.45 kg/m. The force necessary is 4.9 N= 4.9 kg m²/sec². If we multiply 0.45 kg/m by m²/sec², which is a speed squared, we will, at least, have the correct units so I would be inclined to use 0.45 v²= 4.9 so v²= 4.9/.45= 10.89 and
v= 3.30 m/s. My estimate is that the air under the wings is moved at 3.30 m/s.

 

[Clari]

Hello HallsofIvy,
I really appreciate your help, especially the wonderful explanation and the use of dimensional analysis...lol...but for no.1, I am not sure in "it's acceleration is -g= -9.8 m/s2, its velocity after time t is -gt and it will have fallen d= -(g/2)t2 meters in t seconds"...why the value is negative...acceleration due to gravity is downward, so it should not be negative...the -ve or +ve things are really toublesome sometimes...
Thank you very much again!

 

[UrbanXrisis]

"it's acceleration is -g= -9.8 m/s2, its velocity after time t is -gt and it will have fallen d= -(g/2)t2 meters in t seconds"...why the value is negative...acceleration due to gravity is downward, so it should not be negative...

It depends on your point of reference. If I was on the ground and someone dropped a ball above my head, then I would say that the ball is accelerating at -9.8m/s^2 due to gravity. The guy dropping the ball would say that the ball is accelerating at 9.8m/s^2.

basically, if you're the origin, accelerating away is positive, accelerating towards you is negative. But acc. due to gravity is usually notated as negative

 

[quasar987]

The + or - sign depends on how you (he who is solving the problem)chose to set up the coordinate system. It is conventional to take the positive direction as upward, although in freefall problems, we may chose for convinience to make the positive direction downward.

HallsofIvy chose to treat the problem with a coordinate system pointing upward, but if you specify by a little drawing or by a note that you chose the positive direction as downward, you can take all his calulations, change the - sign for a + sign and it'll still be good. As long as your calculations are coherent with your chose of coordinates, there's no problem.

Edit: I seem to be giving a different explanation than from UrbanXrisis. Someone correct me if I'm wrong please.

 

[HallsofIvy]

Actually quasar987 and UrbanXrisis are saying essentially the same thing: it depends upon your choice of coordinate system. UrbanXrisis is saying he prefers to measure distance from him which, I think, is not always best. UrbanXrisis: If an object was moving toward you from the right and another toward you from the left, would you say that both have negative speed?

Yes, I chose my coordinate system with positive upward. I would think that is standard. I haven't seen many problems with g= +9.8.
If you do choose the coodinate system with positive downward, then g= +9.8, v= 9.8t (positive speed downward) so that t would be exactly the same as before, h= 4.5t²+ C and now you have to decide where your origin is. I chose the origin "on the floor" so that the initial height was 20 feet and got h= -4.5t²+ 20, solving -4.5t²+ 20= 0, which is the same as 4.5t²= 20, to find the time until the ball bounces. Taking positive downward you could still take the floor to be at 0 but now the initial height would have to be -20 m (which looks odd to me) so h= 4.5t²- 20 and you solve 4.5t²- 20= 0 which is the same as 4.5t²= 20 just as before.
Probably a more typical choice, with positive downward, would be to take the origin at the intial position of the ball so h(0)= C= 0 and h(t)= 4.5t². Now the floor will be at h= 20 and we need to solve 4.5t²= 20, again exactly the same equation as before.

 

[Orjuan]

How many grams of U-233 could be absorbed for ongoing radiation dose equal to 20 mSvif the half life of U-233 is 0.1592My and it decays through alpha particle of a 4.91MeV

 

[Orjuan]

How many grams of U-233 could be absorbed for ongoing radiation dose equal to 20 mSvif the half life of U-233 is 0.1592My and it decays through alpha particle of a 4.91MeV

1mSv is equal to how many Bq