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evansmiley
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Homework Statement
A smooth wedge of mass 2m and slope 45 degrees is placed on a smooth horizontal surface. A particle of mass m is placed on the inclined face of the wedge.
The system is released from rest.
(i) Show on separate diagrams the forces acting on the wedge and the particle.
(ii) Show that the acceleration of the wedge is g/5 m/s/s.
(iii) Find the speed of the mass relative to the wedge, when the speed of the wedge is 1 m/s.
Homework Equations
F = ma
...
The Attempt at a Solution
I hope this is all comprehensible
(i) was ok,
(ii)I solved part two as follows
K being the normal force from the mass on the wedge,
ksin45 = 2ma
Then I calculated the forces on the particle of mass m normal to the wedge's slope, using the acceleration b_y of the particle in the direction normal to the wedge's slope (y and x are my new axes, where x is in the direction of the wedge's slope and y is perpindicular):
mgcos45 - k = mb_y
as the particle contines to be on the surface of the wedge, the acceleration of the wedge in this direction is equal to the acceleration of the particle in the direction:
b_y = a_y
The component of a in the y direction = a_y = asin45
so ma/√2 = mg/√2 - k
subbing new value for k into the first equation, a works out to be g/5 m/s/s
(iii) Although this should be the easy part for some reason I'm not getting the given answer (3√2m/s/s)
I started off by getting the time taken to reach this speed - t = 5/g - but I'm not entirely sure where to go now.
I tried calculating the relative acceleration in the direction down the slope as follows : for particle m, acceleration b_x = g/√2 where direction down the slope is positive. Acceleration of wedge in this direction - a_x = -g/(5√2)
so relative acceleration is g/√2 -g/(5√2) = 4g/(5√2)
this would give it speed 4/√2 so this must be wrong. Should I be looking for the absolute relative acceleration instead?
I'm very confused so many thanks to anyone who can explain where I'm going wrong, or what stupid mistake I'm making!
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