- #1
Karol
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- 22
Homework Statement
Mass m is raised to height h and released. it hits, elastically, mass M which is on a surface with friction coefficient μ1. mass m, same as the first, lies on top of M. the coefficient between them is μ2.
1) What's the box's velocity just before the hit
2) What is the tension T1 in the rope immediately after releasing
3) What is the tension T2 in the rope just before the hit
4) What is M's acceleration immediately after the hit
5) What is m's acceleration (the box on top) immediately after the hit
6) How much time M travels till it stops
7) Draw on one coordinate system M and m's velocities till they halt, relative to the floor
8) What distance M travels till it halts
9) What should be the minimal length of M so that m won't fall
Homework Equations
Potential energy of height: ##EP=mgh##
Kinetic energy: ##EK=\frac{1}{2}mv^2##
Conservation of momentum: ##m_1v_1+m_2v_2=m_1v_1'+m_2v_2'##
Acceleration, distance and speed in constant acceleration: ##V^2=V_0^2+2ax##
Centripetal force: ##F_c=\frac{mv^2}{r}##
The Attempt at a Solution
Height transformed into speed:
$$mgh=\frac{1}{2}mv^2~~\rightarrow~~v_2=2gh$$
Tension before release:
$$T_1=mg\cos\alpha=mg\frac{l-h}{l}$$
Tension before hit:
$$F_c=\frac{mv^2}{r}~~\rightarrow~~T_2=m\left[ \frac{v^2}{l}+g \right]=mg\left( \frac{2h}{l}+1 \right)$$
Conservation of momentum and energy, V is M's velocity:
$$\left\{ \begin{array}{l} \frac{1}{2}mv^2=\frac{1}{2}\left[ mv_1^2+MV_0^2 \right] \\ mv=MV_0-mv_1 \end{array}\right.$$
$$\rightarrow V_0=\frac{m\sqrt{2gh}\left[ M+\sqrt{M(2M+m)} \right]}{M(M+m)}$$
M's acceleration after the hit:
$$(M+m)g\mu_1+mg\mu_2=Ma_1~~\rightarrow~~a_1=\frac{g\left[ (M+m)\mu_1+m\mu_2 \right]}{M}$$
m's acceleration after the hit:
$$mg\mu_2=ma_2~~\rightarrow~~a_2=g\mu_2$$
The time till equality of velocities is t1:
$$V_0-a_1t_1=a_2t_1:~~V_0-\frac{(M+m)\mu_1+m\mu_2}{M}gt_1=g\mu_2t_1$$
$$\rightarrow t_1=\frac{MV_0}{(M+m)(\mu_1+\mu_2)g}$$
Their common velocity, after time t1:
$$V_2=a_2t_1=g\mu_2\frac{MV_0}{(M+m)(\mu_1+\mu_2)g}=\frac{MV_0\mu_2}{(M+m)(\mu_1+\mu_2)}$$
When they move at one velocity, their common deceleration a3 is (f=friction force):
$$f=(M+m)a_3~~\rightarrow~~(M+m)g\mu_1=(M+m)a_3~~\rightarrow~~a_3=g\mu_1$$
Time till both stop t2:
$$0=V_2-a_3t_2: ~~\frac{MV_0\mu_2}{(M+m)(\mu_1+\mu_2)}=g\mu_1t_2$$
$$\rightarrow t_2=\frac{MV_0\mu_2}{(M+m)(\mu_1+\mu_2)\mu_1g}$$
Total time to full stop:
$$t_{tot}=t_1+t_2=\frac{MV_0}{(M+m)(\mu_1+\mu_2)g}\left( 1+\frac{\mu_2}{\mu_1}\right)$$
The graph:
$$a_1=\frac{g\left[ (M+m)\mu_1+m\mu_2 \right]}{M}=\left[ \frac{M+m}{M}\mu_1+\frac{m}{M}\mu_2 \right]g>a_3=\mu_1g$$
The distance M travels till equality of velocities: ##V_2^2=V_0^2-2a_1x_1##
$$\frac{\mu_2^2M^2V_0^2}{(M+m)^2(\mu_1+\mu_2)^2}=V_0^2-2\frac{g\left[ (M+m)\mu_1+m\mu_2 \right]}{M}X_1~~\rightarrow~~x_1=...$$
The distance M travels from the equality till it stops: ##0=V_2^2-2A_3x_2##
$$0=V_0^2-2g\mu_1x_3~~\rightarrow~~x_2=...$$
$$x_{tot}=x_1+x_2$$
But the expressions are complicated.
The time till M stops:
M's absolute velocity+m's relative velocity yields m's absolute velocity: $$V_{M(abs)}+v_{m(rel)}=v_{m(abs)}$$
$$\rightarrow~v_{m(rel)}=v_{m(abs)}-V_{M(abs)}=a_2t-(V_0-a_1t)=g\mu_2t-\left( V_0-\frac{(M+m)\mu_1+m\mu_2}{M}gt \right)$$
By equating ##v_{m(rel)}## to 0 i find t, the time till m stops (equal velocities):
$$v_{m(rel)}=0~~\rightarrow~~t=\frac{MV_0}{g[(M+m)(\mu_1+\mu_2)]}$$
The distance traveled by m: ##x_1=\frac{1}{2}a_{m(rel)}t^2##:
$$a_{m(rel)}=a_{m(abs)}-V_0=g\mu_2-\frac{(M+m)\mu_1+m\mu_2}{M}g=\frac{(M-m)\mu_2-(M+m)\mu_1}{M}g$$
The distance x1 involves the square of the above t, and it's only a part of the total distance M travels. the second part, x2, from ##0=V_2^2-2a_3x_2##
It's tiring, maybe there's a simpler way, for some of the stages?