5 replies to this topic

[Bryan]

So yea, I've gotten the dreaded take home test today for Physics. It's 10 problems, anyways, I've only had time to do one at the moment, but here they are.

1) How long would a day be (in hours) if the earth rotated so fast that objects at the equator were weightless? I guesstimated about 1.2

2) A block (mass(2)=3kg) rests on top of another block (mass(1)=8kg). A horizontal string connects to mass(2) block. The coefficient of kinetic friction between the two blocks is .2 and the coefficient of kinetic friction between the horizontal surface and mass(1) is .3. A 50-Newton force pulls mass(1) to the right.

A) What is acceleration of mass(1)?
B) What is tension in the string?

3) A 2100kg truck traveling east at 20m/s collides with a 1300kg sports car traveling at an unknown velocity. After the collision, the vehicles become entangled and skid to a stop. The skid marks are measured to be 48 meters at an angle of 32(degrees) relative to the east. If the coefficient of kinetic friction is measured to be .37, what was velocity of sports car? (Assume level ground)

4) A sprinter has a top speed of 11 m/s. If he starts from rest and accelerates at a constant rate, he can reach his top speed in a distance of 12meters. He then maintains that top speed for the remainder of the 100m race.

A) What will his time be for the race
B) To improve his time, he tries to decrease his distance in reaching his top speed, what must this distance be if he is to run the race in 10.0 seconds?

5) A carbon nucleus (mass(1)= 12 u) moving at 7.4 x 10^5 m/s strikes a stationary oxygen nucleus (mass(2)= 16 u). After this elastic collision, the oxygen nucleus is observed to move off at a 16(degree) angle above the horizontal. Determine the final speeds of both nuclei and the angle for the carbon nucleus.

I got 3.28 x 10^5 m/s, 3.424 x 10^5 m/s, -52.39(degrees)

6)An elevator without a ceiling is moving upward with a constant speed of 10 m/s. A boy on the elevator throws a ball directly upward from a height of 2m above the floor of the elevator at the instant the elevator floor is 28 meters above the ground. If the initial speed of the ball is 20 m/s (relative to the elevator)

A) How much time passes before the ball strikes the floor of the elevator?
B) What is the maximum height above the ground the ball rises?

7) Two identical objects moving with the same initial speed collide inelastically. After the collision, the move off with one-fourth of their initial speed. What is the angle between their initial directions?


I have the other 3 answers, I think.

Just need some help on how to set these up, and to make sure I'm doing them correctly.

Thanks =)

[TerrixBerri]

Hehe, I'm a drop out, the first question addled my damn brain into an ozzing blob good luck though!!

[ShadowLink64]

4a)

First, figure out the time for when d=0 to d=12 using [((initial velocity + final velocity)/2) * t = d]. You will have a different velocity starting than your finish velocity which is why this one requires a seperate equation. Starting velocity will be zero, and final velocity will be 11 metres per second.

Secondly, figure out the time for when d=12 to d=100 using [v=(d/t)]. This one has constant speed since he is running his max. speed from when he reaches 11 metres all up to 100 metres.

Add the two times.

--------------------
4b) Our master equation is going to be (Time 1) + (Time 2) = 10.0 seconds

Time 1 will represent the time it takes to accelerate to top speed. Time 2 will represent the time it takes to finish the remainder of the race. Both of these are dependent on the initial distance in time 1 (since to find the distance of the remainder of the race you have to go (100 - (distance of first part of race)) to find t he remaining distance.

So now... part 1. Find Time 1
The only information we are given is: Initial Velocity and Final Velocity. Therefore, form a time equation where you have "t" on one side of the equation, and a derived equation containing "d" on the other side. This derived equation will represent Time 1 on our master equation.

Part 2. Find Time 2 Do the same thing as above, but remember that the new distance is going to equal 100 minus the old distance which we are using "d" for. The only info you know is initial and final velocity once again.

Part 3 Substitute Time 1 and Time 2 in that master equation and solve algebraically for "d".

I have the answer if need be. (hope its correct) If I did it right, Cody is not quite correct in his observation that it takes less than a second to reach top speed. It takes a bit more time to reach final velocity.

[Bryan]

Sorry if I wasn't responding, my parents decided to take me out of town and I've been without internet =/ Anyways, I looked at four again and realized how stupid I was. That really seems like the simplest problem just based on the fact that it's velocity and acceleration.

I've done 1-3,5,7,9,10.

Six is really confusing me, and I can't figure out if the answer I got for 8 was correct. Potential energy confuses me =(

Thanks for the help Shadow, I really appreciate it, I'll get to solving that right now. =]

[nox]

because I'd like to help him out... that's why.

i think he means going to class

[chalow]

well since he already helped you then im of no use i actually gotta do my AP Physics take home assignment soon but im so beat right now im going to bed