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## Homework Statement

_{0}and collides plastically but without clinging to mass A which is at rest.

The spring, initially, is loose. assume that the collision time is very short.

What's the boxes's velocity immediately after the collision?

At what distance from the wall box A stops initially?

What's the force resultant on box A immediately after it's stop at the first time?

At what distance from the wall will box B detach from A?

Where will B finally stop?

A continues to oscillate until it stops at x

_{f}. find points x

_{1}and x

_{2}at which A has momentary maximum. will x

_{f}be:

x

_{f}>x

_{2}, x

_{1}≤x

_{f}≤x

_{2}, x

_{f}<x

_{1}

The problem states quantities for everything: masses etc. but i don't give them, i solve in principle.

## Homework Equations

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

Potential energy: ##E=\frac{1}{2}kx^2##

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

## The Attempt at a Solution

Conservation of momentum:

$$m_Bv_0=(m_A+m_B)v_1\;\rightarrow\; v_1=\frac{m_B}{m_A+m_B}v_0$$

The kinetic energy is transformed into potential+friction loss:

$$\frac{1}{2}(m_A+m_B)v_1^2=(m_A+m_B)\mu g+\frac{1}{2}kx_a^2\;\rightarrow\;x_a=...$$

Immediately after both stop for the first time, the force on A is ##F=kx_a-(m_A+m_B)\mu g##

B will detach from A at the relaxed length of the spring, at x=l

_{0}, since even though the force the spring applies, kx, is smaller than the friction of any or both together, at some point, it is still acting on both masses. in order for B to detach A must be drawn backwards, it's not enough that there will be 0 net force on it.

I feel this isn't right but i have no other solution.

The potential energy when both stop at x

_{a}is transformed into kinetic energy at x=0 plus friction loss:

$$\frac{1}{2}kx_a^2=\frac{1}{2}(m_A+m_B)v_3^2+(m_A+m_B)\mu g x_a\;\rightarrow\;v_3=...$$

The points on both sides of x=0: ##kx=m_Ag\mu\;\rightarrow\; x=\frac{m_Ag\mu}{k}## are the ones with temporary maximum velocity.

m

_{A}won't stop outside x

_{1}or x

_{2}since if the mass stops, while oscillating, temporarily there, the spring's force will overcome friction.

If m

_{A}stops temporarily just outside one of these x'ses, the friction loss, in order to reach the other x, is greater than the potential energy:

$$\frac{1}{2}kx_a^2-2\cdot m_Ag\mu x_a=...<0$$

So m

_{A}will finally halt between them.

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