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Gelneren [198K]
3 years ago
12

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.

Physics
1 answer:
Anna11 [10]3 years ago
4 0

Answer:

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

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

Explanation:

From the diagram below;

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

where

v = u + at

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

Making t the subject of the formula; we have:

t = \frac{v_o M}{(\mu g )(M+m)}

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

S = \frac{v_o^2M}{2 \mu g (M+m)}  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

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

v = \frac{mv_o}{m+M}

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

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