- #1

- 1,380

- 22

## Homework Statement

A cart of mass m is adjacent to a wall. it has a mass, also m, attached at point A to a loose spring of constant k. an identical spring is on the opposite wall. the floor of the cart is frictionless apart of the middle part CD with equivalent μ. a force F is applied to the mass and then released. at point B of the other loose spring m has relative velocity v

_{rel}.

1) What's the cart's velocity relative to the floor at the moment m hits the right spring.

2) What's the max acceleration a

_{max}of the cart from the beginning till m stops completely relative to the cart.

3) What's CD's length?

4) What is the final velocity of the cart?

5) At what distance from C m finally stops on the rough surface CD.

6) What's the impact the mass applies on the cart at the time described in question 2.

The variables are given values:

m=200[gr], μ=0.5, k=500[N/m], F=40[N], v

_{rel}=2[m/s]

## Homework Equations

Kinetic energy: ##E_k=\frac{1}{2}mv^2##

Conservation of momentum: ##m_1v_1+m_2v_2=m_1v_1'+m_2v_2'##

Reduced mass: ##\overline{m}=\frac{m_1m_2}{m_1+m_2}##

Total kinetic energy in the COM frame of reference: ##E_k=\frac{1}{2}\overline{m}V_{12}^2##

Equivalent spring of two springs in a row: ##\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}##

## The Attempt at a Solution

The velocity of m at A:

$$\left\{ \begin{array}{l} \frac{1}{2}kx^2=\frac{1}{2}mv_A^2 \\ F=kx~\rightarrow~x=\frac{F}{k} \end{array}\right.$$

$$\rightarrow~v_A^2=\frac{F^2}{mk}$$

1) v

_{m}is the boxe's absolute velocity, v

_{M}is the cart's. conservation of momentum from the beginnig+relative velocity:

$$\left\{ \begin{array}{l} mv_A=mv_m+mv_M \\ v_m=v_M+v_{rel} \end{array}\right. ~\rightarrow~v_M=\frac{v_A-v_{rel}}{2}$$

2) When m reaches B if works, from there on, on both springs, who's equivalent is ##k_{eq}=\frac{k}{2}##

In the COM's frame, after the box reached B, when the box stops the total kinetic energy which is contained in both, the box and the cart and the given relative velocity zeroes and it enters the equivalent spring:

$$E_k(tot)=\frac{1}{2}\overline{m}V_{12}^2=\frac{1}{2}k_{eq}x^2~\rightarrow~\frac{m}{2}v_{rel}^2=\frac{k}{2}x^2~\rightarrow~x^2=\frac{mv_{rel}^2}{k}$$

The cart's acceleration:

$$F=k_{eq}x=m\cdot a_{max}~\rightarrow~\frac{k}{2}v_{rel}\sqrt{\frac{m}{k}}=ma_{max}$$

$$a_{max}=\frac{v_{rel}}{2}\sqrt{\frac{k}{m}}$$

3) The difference in the total kinetic energy in the COM frame is the loss on the rough surface. since at the beginning the relative velocity is the one i found as the velocity at A:

$$\Delta E_{k(tot)}=\frac{1}{2}\frac{m}{2}\left[ v_A^2-v_{rel}^2 \right]=mg\mu\cdot x_{CD}~\rightarrow~x_{CD}=\frac{v_A^2-v_{rel}^2}{4\mu g}$$

4) Conservation of momentum between the beginning and with their common velocity when the box stops:

$$mv_A=2mv_{com}~\rightarrow~v_{com}=\frac{v_A}{2}$$

5) The energy loss in one pass over CD:

$$E_1=fx_{CD}=mg\mu\frac{v_A^2-v_{rel}^2}{4\mu g}=\frac{m(F^2-mkv_{rel}^2)}{4mk}$$

I divide the kinetic energy at the beginning by E

_{1}to get the number of passes:

$$n=\frac{\frac{1}{2}\frac{F^2}{k}}{\frac{m(F^2-mkv_{rel}^2)}{4mk}}$$

After substituting the numbers i get n=2.28, so the last pass is:

$$0.28E_1=mg\mu\cdot x~\rightarrow~x=...$$

But is there a general method without substituting actual values? in this case n was small but...

6) The impulse the box exerts on the cart is the cart's final velocity, since at the beginning it was at rest:

$$P=mv_{com}=\frac{1}{2}mv_A$$