- #1
Rijad Hadzic
- 321
- 20
Homework Statement
A skater of mass m standing on ice throws stone of mass M with speed v in a horizontal direction. Find the distance over which the skater will move in the opposite direction if the coeffecient of kinetic friction between the ice and the skater is mk
Homework Equations
[itex]V_{0x} + a_xt = V_x [/itex]
[itex]\Delta x = (1/2)(V_{0x} + V_x)t[/itex]
[itex] \Delta x = V_{0x}t + (1/2)a_xt^2[/itex]
[itex] \Delta x = V_xt - (1/2)a_xt^2 [/itex]
[itex] V_x^2 = V_{0x}^2 + 2a\Delta x [/itex]
[itex] F= ma [/itex]
[itex]P_{initial} = P_{final} [/itex]
The Attempt at a Solution
[/B]
Ok, so using the last eq, P_i = P_f.
The initial P = 0, so P_f = 0 as well.
Well momentum for the stone = [itex] Mv [/itex], thus [itex] mv_{skater} = Mv [/itex]
to make momentum = 0 again.
So now I know
[itex] v_{skater} = (Mv)/m [/itex]
Using
[itex]\Delta x = (1/2)(V_{0x} + V_x)t[/itex]
I get
[itex] \Delta x = (1/2)(Mv/m)t [/itex]
using
[itex] \Delta x = V_{0x}t + (1/2)a_xt^2[/itex]
I get
[itex] \Delta x = (1/2)a_xt^2 [/itex]
Setting the two [itex] \Delta x [/itex] = to each other I get
[itex](1/2)a_xt^2 = (1/2)(Mv/m)t [/itex]
[itex] a_xt = (Mv/m) [/itex]
[itex] t = (Mv/a_xm)[/itex]
Now I plug t back into: [itex] (Mv/2m)t [/itex]
[itex] (M^2v^2)/(2m^2a_x) = \Delta x [/itex]
but my books answer is
[itex] (M^2v^2)/(2m^2 (mewk) g ) [/itex]
does anyone know where I went wrong?
Looking at my answer a_x must = (mewk) g but I'm not sure how to figure this out?