Trajectory and conservation of linear momentum

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving trajectory and conservation of linear momentum. The scenario describes a person on frictionless ice tossing a rock and seeks to determine the angle of the toss and the person's resulting speed. Key equations related to projectile motion and momentum conservation are referenced.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the angle of the toss and the distance the rock travels, questioning the initial assumptions about the distance and the effects of the person's sliding motion.
  • Some participants attempt to derive equations for time of flight and horizontal distance, while others express confusion over the setup and the implications of the given values.
  • There is a discussion about the significance of the rock's mass and its impact on the calculations, with some suggesting that the distance the person slides may be negligible.

Discussion Status

The conversation is ongoing, with participants sharing various interpretations and calculations. Some have identified potential inconsistencies in the problem setup, while others are working through the implications of their findings. There is no clear consensus, but several participants have provided insights and alternative approaches to the problem.

Contextual Notes

Participants note discrepancies in the mass of the rock (4.5 g vs. 4.5 kg) and discuss how this affects the solvability of the problem. There is also mention of the need to consider the distance the person slides during the toss, which complicates the calculations.

bakin
Messages
55
Reaction score
0

Homework Statement



While standing on frictionless ice, you (mass 65.0 kg ) toss a 4.50g rock with initial speed 12.0m/s . The rock is 15.2 m from you when it lands. Take the axis on the ice surface in the horizontal direction of the motion of the rock. Ignore the initial height of the toss.
At what angle did you toss it?
How fast are you moving? (cm/s)

Homework Equations



trajectory equation r=(vo²/g)sin2theta

m1v1+m2v2=0

The Attempt at a Solution



I solved for theta, and got

theta=(1/2) sin-1(gR/v²) but that's giving me sin-1(1.03)

obviously something is wrong. please help!

And for the 2nd part, would you just do m1v1+m2v2cos(theta from part a)=0 ??
 
Physics news on Phys.org
Ok, so I was looking at it wrong. The distance 15.2m is the distance from where the rock lands to where the guy slid during the time it was in the air. How do I find that distance without knowing the angle theta?
 
Amazing how they can come up with new trajectory questions!
I had to dig out my trig identity book to do this one . . . and it didn't work. Same problem you had with your range equation. There appears to be an inconsistency in the information given.
I drew 12 at an angle of A and found the initial velocities to be 12cos(A) horizontal and 12*sin(A) vertical. Considering t to be the time to maximum height,
HORIZONTAL: d = vt -> 15.2/2 = 12*cos(A)*t -> t = 0.633/cos(A)
VERTICAL: V = Vi + at -> 0 = 12*sin(A) - 9.81*t -> t = 1.22*sin(A)

Equating the two expressions for t we have 1.22*sin(A) = 0.633/cos(A)
so sin(A)*cos(A) = 0.5188
The trig book says sin(A)*cos(A) = 0.5*sin(2A) so
sin(2A) = 2*0.5188
which has no solution.
 
How come you did 15.2/2 in the horizontal equation?
 
I was thinking the horiz distance to the point of maximum height is half the range.
But I didn't have your insight about the distance including the person's slide, so back to the drawing board!

Perhaps you could begin with the range being 15.2 - x, where x is the slide distance.
Maybe write x as Vs*t (time of flight) and use conservation of momentum to get another equation with a Vs in it, related to the 12*cos(A).

Good luck!
 
Already did :wink: and...I actually came up with the same answer :mad:

Spent like 15-20 min working through it, too.

edit: conceptually, how far would a person slide throwing something that weighs .0045kg? Not very far at all, I'm imagining. Especially when it's not completely horizontal.
 
Yes, I see the problem. The slide speed is 0.0008308*cos(A) and that is too small to matter. So, the problem as stated cannot be solved. Perhaps your marker would be okay with you changing the 15.2 to 10, which would certainly make it solvable.
You could begin with the momentum bit:
mv = mv
65Vs = .0045*12*cos(A)
which shows that you can ignore the distance of the slide (insignificant in 3 digit accuracy).
 
When you do find the angle, how would you do the second part?

m1v1=m2v2, or do you need to multiply it by the angle at all?

edit: even though my problem says 4.5g, and the other thread has 4.5KG, i solved it the same way and put in the same answer. It turns out the actual answer was 37.7 degrees. Oh well.
 
  • #10
Oh, the 4.5 kg instead of grams would make a huge difference! It would make the problem solvable!

Back to including the slide. The range equation is
Eq1: 15.2 - Vs*t = 12^2/g*sin(2A) (t is the time of flight in all this)
Conservation of momentum gives
Eq2: 65Vs = 4.5*v*cos(A) or Vs = 0.8308*cos(A)
Vertical velocity at the turnover point is 0 = Vi + at = 12*sin(A) - gt/2 or
Eq3: t = 2.45*sin(A)
Sub eq3 into eq1 to get
15.2 - 2.45*Vs*sin(A) = 14.7*sin(2A)
Sub eq2 in for Vs:
15.2 - 2.45*0.8308*cos(A)*sin(A) = 14.7*sin(2A)
15.2 - 2*sin(A)*cos(A) = 14.7*sin(2A) a bit of a roundoff here to get the 2
15.2 - sin(2A) = 14.7*sin(2A)
15.2 = 15.7*sin(2A)
sin(2A) = 0.97
A = 37.8 degrees

To get the speed of the person, you would just use eqn2.
 
  • #11
if you don't mind me asking, how did you get 12^2/g*sin(2A) from .5at^2 for the first equation?
 
  • #12
That is some impressive math Delphi51. To simplify things, you can find the center of the mass of system. This gives where each object was located originally. Then you can find the actual distance the rock went and solve for theta using the range equation. the rest is plug and chug
 

Similar threads

Using conservation of energy and momentum in projectile motion
  • · Replies 18 ·
Replies
18
Views
2K
Inelastic collisions -- how is momentum conserved but not energy?
  • · Replies 5 ·
Replies
5
Views
3K
Momentum conservation: block-wedge problem
  • · Replies 4 ·
Replies
4
Views
2K
Challenging problem about an impact with a smooth frictionless surface
  • · Replies 55 ·
2
Replies
55
Views
6K
Conservation of Linear momentum
  • · Replies 5 ·
Replies
5
Views
2K
Confusion with "explosive" part, Conservation of Momentum
  • · Replies 6 ·
Replies
6
Views
2K
Physics 30 question about conservation of momentum
  • · Replies 11 ·
Replies
11
Views
2K
Elastic collision of block on block
  • · Replies 4 ·
Replies
4
Views
2K
Conservation of Momentum of a snow ball
  • · Replies 4 ·
Replies
4
Views
4K
Conservation of momentum: throwing bricks
  • · Replies 1 ·
Replies
1
Views
2K