Cons. of momentum: person moves at angle atop free-to-move cart

AI Thread Summary
The discussion focuses on calculating the new velocity of a cart when a student walks on it at an angle. The initial momentum of the system is established using the combined mass of the student and cart, with the student walking at 1.8 m/s at 25 degrees West of North. Through conservation of momentum, the final velocity of the cart is derived, resulting in a speed of approximately 8.31 m/s at an angle of 21.4 degrees South of East. However, a correction suggests the angle should actually be 21.4 degrees South of West. The calculations confirm the new velocity and direction of the cart after the student's movement.
strangeeyes
Messages
8
Reaction score
0

Homework Statement


A 63kg student stands on 34kg cart that is free to move in any direction. The cart and student are moving together with a velocity of 3.2 m/s [W]. The student then starts to walk so that her velocity is 1.8 m/s [25 degrees W of N] relative to the floor. Calculate the new velocity of the cart.

mass of student = mg = 63kg
mass of cart = mc = 34kg
velocity of system = vs = 3.2 m/s [W]
velocity of student = vg = 1.8 m/s [25 W of N]
velocity of cart = ?

Homework Equations


p=p'

The Attempt at a Solution


[PLAIN]http://img338.imageshack.us/img338/4709/momdia.jpg

let 's' represent the system
the final momentum of the system is equal to the vector addition of the student (g) and the cart (c)
ps'= pc' +pg'
rearrange to find pc':
pc'=ps'-pg'

ps=ps'
=(mg+mc)vs
=310.4 kg * m/s [W]

-pg'=63(1.8)
=113.4 kg * m/s [25 E of S]
now using components:
[PLAIN]http://img571.imageshack.us/img571/2365/diamom.jpg

-pgx'=63(1.8)sin25
=47.3249 kg * m/s [E]
-pgy'=63(1.8)cos25
=102.78 kg * m/s

using Pythagorus
pc'^2=pgy'^2 + [ps' + (-pgx)]^2
pc'=282.439 kg * m/s

to find direction of pc':
angle =tan-1(102.78 / 262.475)
= 21.38 degrees

therefor the new velocity of the cart is:
vc'=pc'/mc
= 8.31 m/s [21.4 S of E]

am i wrong? that answer seems unlikely...
 
Last edited by a moderator:
Physics news on Phys.org
strangeeyes said:
= 8.31 m/s [21.4 S of E]

am i wrong? that answer seems unlikely...

I think the angle should be 21.4 degrees S of W.

Let's solve it a different way, using imaginary numbers. East and West are along the usual real number line, North and South lie on the + and - imaginary axis.

Mp = person mass = 63 kg
Mc = cart mass = 34 kg

The initial velocity of both person and cart is

v = -3.2 m/s

The momentum is then

p = (63kg + 34kg)*v = -310.4 kg*m/s

The person starts walking at an angle that's 25 degrees West of North. That's 90 + 25 = 115 degrees. His speed is 1.8 m/s, So,

vp = 1.8 m/s * (cos(115) + i*sin(115))

vp = (-0.761 + 1.131i) m/s

Conservation of momentum gives us

p = Mp*vp + Mc*vc

solving for vc yields

vc = (p - Mp*vp)/Mc
vc = (-7.72 -3.023i) m/s

|vc| = 8.291 m/s

arg(vc) = -158.617 deg

That's 21.383 degrees South of West
 
makes sense! thanks so much.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top