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## Homework Statement

Mass m is raised to height h and released. it hits, elastically, mass M which is on a surface with friction coefficient μ

_{1}. mass m, same as the first, lies on top of M. the coefficient between them is μ

_{2}.

1) What's the box's velocity just before the hit

2) What is the tension T

_{1}in the rope immediately after releasing

3) What is the tension T

_{2}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 t

_{1}:

$$V_0-a_1t_1=a_2t_1:~~V_0-\frac{(M+m)\mu_1+m\mu_2}{M}gt_1=g\mu_2t_1$$

$$\rightarrow t_1=\frac{MV_0}{(M+m)(\mu_1+\mu_2)g}$$

Their common velocity, after time t

_{1}:

$$V_2=a_2t_1=g\mu_2\frac{MV_0}{(M+m)(\mu_1+\mu_2)g}=\frac{MV_0\mu_2}{(M+m)(\mu_1+\mu_2)}$$

When they move at one velocity, their common deceleration a

_{3}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 t

_{2}:

$$0=V_2-a_3t_2: ~~\frac{MV_0\mu_2}{(M+m)(\mu_1+\mu_2)}=g\mu_1t_2$$

$$\rightarrow t_2=\frac{MV_0\mu_2}{(M+m)(\mu_1+\mu_2)\mu_1g}$$

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:

_{1}to a

_{3}:

$$a_1=\frac{g\left[ (M+m)\mu_1+m\mu_2 \right]}{M}=\left[ \frac{M+m}{M}\mu_1+\frac{m}{M}\mu_2 \right]g>a_3=\mu_1g$$

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 x

_{1}involves the square of the above t, and it's only a part of the total distance M travels. the second part, x

_{2}, from ##0=V_2^2-2a_3x_2##

It's tiring, maybe there's a simpler way, for some of the stages?