Mechanics 3 Watch

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Fermat
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#21
Report 13 years ago
#21
(Original post by geekypoo)
Yay i've done the last part, correct me if im wrong Fermat but it uses the conservation of energy doesnt it, so:
...
The last part uses conservsation of energy, but so also does the part before, part (iii)

part (iii)
At max height, the particle still has horizontal velocity, which adds to its energy.

The energy eqn is,

\frac{1}{2}mv_P^2 = mgh + \frac{1}{2}mv_{P_H}^2

where v_{P_H} is the horizontal velocity of the particle
and v_P is the actual velocity of the particle
and
v_{P_H} = v_Psin(\alpha)
and
v_P = 1.14 m/s

½(1.14)² = 9.8h + ½(1.14sin41.8)²
0.6498 = 9.8h + 0.2887
9.8h = 0.3611
h = 0.3611/9.8
h = 0.037 m
h = 3.7 cm
========

Alternatively, you can use the vertical velocity with suvat eqns

v² - u² = 2as
0 - (1.14cos41.8)² = -2gh
h = (1.14cos41.8)²/2g
h = 0.0368
h = 3.7 cm
========

let's hope I haven't made any more stupid mistakes!

part (iv) now becomes (using conservation of energy),

h' = 0.2 - 22/135
h' = 1/27 m
========

\frac{1}{2}m(2.8)^2 = \frac{1}{2}v_Q^2 + mgh'
v_Q^2 = (2.8)^2 - 2*9.8*(1/27)
v_Q^2 = 7.84 - 0.7259
v_Q^2 = 7.1141
v_Q = 2.67 m/s
===========

same as yourself
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