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

In summary, a 63kg student standing on a 34kg cart with a velocity of 3.2 m/s [W] starts walking at a velocity of 1.8 m/s [25 degrees W of N] relative to the floor. The new velocity of the cart is 8.31 m/s [21.4 S of W].
  • #1
strangeeyes
8
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...
 
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  • #2
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
 
  • #3
makes sense! thanks so much.
 

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

1. What is conservation of momentum?

Conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant, regardless of any internal changes or external forces acting on the system.

2. How is momentum conserved in a person moving at an angle atop a free-to-move cart?

In this scenario, the person and the cart form a closed system. The person has a certain amount of momentum in the direction of their movement, and the cart has an equal amount of momentum in the opposite direction. When the person jumps at an angle, their momentum changes, but the total momentum of the system remains the same due to the equal and opposite momentum of the cart.

3. Is the momentum of the person and the cart the same before and after the person jumps?

Yes, according to the principle of conservation of momentum, the total momentum of the system must remain constant. So, the momentum of the person and the cart before the person jumps must be equal to the momentum of the person and the cart after the person jumps.

4. Does the mass of the person and the cart affect the conservation of momentum?

Yes, according to the equation for momentum (p=mv), the mass of an object does affect its momentum. However, in a closed system like the person and the cart, the total momentum remains constant regardless of the individual masses of the objects.

5. Can the direction of the person's jump affect the conservation of momentum?

Yes, the direction of the person's jump will affect the overall momentum of the system. However, the total momentum will still remain the same as long as the cart moves in the opposite direction with an equal and opposite momentum.

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