contestada

Block A can slide relative to block B which, in turn, can slide on a perfectly smooth horizontal plane. If the initial velocity of A is v0, and the initial velocity of B is 0, find the final velocities of the two blocks and the distance that A slides relative to B.

Respuesta :

ajeigbeibraheem

Answer:

the final velocity of the two blocks is [tex]v = \frac{mv_o}{m+M}[/tex]

the distance that A slides relative to B is [tex]S = \frac{v_o^2M}{2 \mu g (M+m)}[/tex]  

Explanation:

From the diagram below;

acceleration of A relative to B is : [tex]a = - ( \mu g + \frac{ \mu mg}{M})[/tex]

where

v = u + at

[tex]0 = v_o + ( - \mu g - \frac{\mu m g }{M})t[/tex]

Making t the subject of the formula; we have:

[tex]t = \frac{v_o M}{(\mu g )(M+m)}[/tex]

[tex]v^2 = u^2 +2 as\\\\0^2 = v_o^2 - 2 (\mu g ) (\frac{M+m}{M})S\\\\[/tex]

[tex]S = \frac{v_o^2M}{2 \mu g (M+m)}[/tex]  which implies the distance that A slides relative to B.

The final velocities of the two blocks can be determined as follows:

v = u + at

[tex]v = v_o - \mu g \frac{v_oM}{\mu g (M+m)}\\\\v = \frac{\mu g mv_o}{m+M}\\\\[/tex]

[tex]v = \frac{mv_o}{m+M}[/tex]

Thus, the final velocity of the two blocks is [tex]v = \frac{mv_o}{m+M}[/tex]

Ver imagen ajeigbeibraheem

The final velocities of the two blocks A and B are :

[tex]$v=\frac{mv_0}{m+M}$[/tex]

And, the distance that the block A slides relative to the block B is :

[tex]$S=\frac{v_0^2 M}{2 \mu g(m+M)}$[/tex]

Finding the distance travelled

Given :

Initial velocity of block A = [tex]$v_0$[/tex]

Initial velocity of block B = [tex]0[/tex]

Now from the diagram, the acceleration can be found out as :

Acceleration of block A relative to block B :

[tex]$a=- \left( \mu g + \frac{\mu mg}{M} \right)$[/tex]

Now we know, final velocity :

[tex]$v=u+at$[/tex]

[tex]$0=v_0 - \left(\mu g + \frac{\mu mg}{M} \right) \times t$[/tex]

[tex]$t=\frac{v_0 M}{(\mu g)(m+M)}$[/tex]

Now,

[tex]$v^2=u^2+2aS$[/tex]

[tex]$0=v_0^2+2(\mu g) \left(\frac{m+M}{M} \right) S$[/tex]

Therefore,

[tex]$S=\frac{v_0^2 M}{2 \mu g (m+M)}$[/tex]

This is the distance through which block A slides relative to block B.

Final velocities of the two blocks

We know,

[tex]$v=u+at$[/tex]

[tex]$v=v_0-(\mu g)\left( \frac{v_0 M}{(\mu g)(m+M)} \right)$[/tex]

[tex]$v=\frac{m v_0}{m+M}$[/tex]

This is the final velocity of the two blocks.

Learn more about "relative velocities" here :

https://brainly.com/question/24337516

Ver imagen AbsorbingMan

Otras preguntas