Another One Dimensional Motion Problem

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Homework Help Overview

The problem involves two objects, a rock and a ball, thrown vertically upward with different initial speeds and at different times. The objective is to determine the time and height at which they collide, as well as to explore the scenario where their order of throwing is reversed. The subject area is one-dimensional motion, specifically focusing on kinematics and the equations of motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the use of kinematic equations. Some express uncertainty about how to start, while others attempt to manipulate the equations to find the time of collision. Questions arise regarding the signs of velocity and acceleration, and whether they should be treated as positive or negative based on the direction of motion.

Discussion Status

There is ongoing exploration of the problem with various attempts to set up equations for the positions of the two objects. Participants are questioning the assumptions about the signs of acceleration and velocity, and some have provided partial calculations while seeking confirmation or guidance on their approaches. No consensus has been reached, and multiple interpretations of the problem are being considered.

Contextual Notes

Participants note the importance of correctly assigning signs to the velocities and accelerations based on the chosen coordinate system. There is also mention of the need for clarity in calculations, as some expressions are difficult to read. The discussion reflects the constraints of the homework context, where complete solutions cannot be provided.

pointintime
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Homework Statement




A rock is thrown vertically upward with a speed of 12.0 s^-1 m. Exactly 1.00 s later, a ball is thrown up vertically along the same path with a speed of 20.0 s^-1 m.

(a) At what time will they strike each other?

(b) At what height will the collision occur?

(c) Answer (a) and (b) assuming that the rder is reversed: the ball is trown 1.00 s before thr ock.

Homework Equations



X = Xo + volt + 2^-1 a t^2

The Attempt at a Solution


I have no idea how to even start this problem
were the 1 subscripts are the first object that went off and the twos are the second object
t1 = t2 - 1.00s

plugged in

Xo + Vo1 t1 + 2^-1 a t1^2

Xo + Vo1 (t2 - 1.00 s) + 2^-1 a (t2 - 1.00)^2

solved for t2 found 1 second...
 
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if i consider there velocities to be positive do i have to make there accelerations negative?
 
do they collide at 2.152 s + 1.00 s ?

not sure if that's right...
 
Hi pointintime! :smile:

(try using the X2 and X2 tags just above the Reply box :wink:)
pointintime said:
if i consider there velocities to be positive do i have to make there accelerations negative?

If you consider distance upward to be positive, then yes, you have to make the accelerations negative. :smile:

(and if you want us to check your result, do show the whole calculation! :wink:)
 
how do you recommend i solve it
 
pointintime said:
how do you recommend i solve it

The method you're using looks fine :smile:

what's worrying you about it? :confused:
 
X1 = X2
sense they collide

t1 = t2
point at which the collide

t1 = t2 - 1.00 s

X1 = Xo + Vo t + 2^-1 a t^2

X1 = (12.0 s^-1 m)t1 + 2^-1 (9.80 s^-2 m)t1^2

sense
t1 = t2 - 1.00 s

X1 = (12.0 s^-1 m)(t2 - 1.00 s) + 2^-1 (9.80 s^-2 m)(t2 - 1.00 s)^2
X1 = (12. 0 s^-1 m)t2 - 12.0 m + 2^-1 (9.80 s^-2 m)(t2^2 + 1.00 s^2 - 2 t2 1.00s)
X1 = (12. 0 s^-1 m)t2 - 12.0 m + (4.90 s^-2 m)(t2^2 + 1.00 s^2 - t2 2.00s)
X1 = (12. 0 s^-1 m)t2 - 12.0 m + (4.90 s^-2 m)t2^2 + 4.90 m - (9.80 s^-1 m)t2
X1 = (4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m

sense X1 = X2

(4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m = Xo + Vo t2 + 2^-1 a t2^2
(4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m = (20.0 s^-1 m) t2 + 2^-1 (9.80 s^-2 m) t2^2
(4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m - (20.0 s^-1 m) t2 -(4.90 s^-2 m) t2^2
(17.8 s^-1 m)t2 - 7.1 m

what now?
let me guess I did it wrong

(17.8 s^-1 m)t2 - 7.1 m = 0
(17.8 s^-1 m)t2 = 7.1 m
t2 = .3989 s

what does that mean?
 
can someone check my work i probably did it wrong and if i didn't i don't know what to do from there
 
pointintime said:
can someone check my work i probably did it wrong and if i didn't i don't know what to do from there

Well, it's bit difficult to read (please use the X2 tag in future), but the method looks ok,

except (i thought you agreed on this) the acceleration needs to be negative. :rolleyes:
 
  • #10
It does have to be negative?
If I consider it to be positive does it miss anything up
 
  • #11
pointintime said:
It does have to be negative?
If I consider it to be positive does it miss anything up

Yeees! …

if your 12 and 20 are positive, then your 9.8 must be negative (or vice versa). :wink:

what's up for velocity must be up for acceleration!​
 

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