# M3 past paper questionWatch

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#1
A particle P of mass m is dropeed from a height h above the surface of the liquid in a container and falls freely under gravity intill it enters the liquid. As P enters the liquid, a quarter of its kinetic energy is lost. At time t after entering the liquid P has fallen a vertical distance of x through the liquid and the resistance to motion is 3mx^2. Calculate the speed of P when it has just entered the liquid and hence show taht its speed at time t later is given by:
v^2 = 2gx - 2x^3 + (3/2)gh

any help would be great thanks.
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14 years ago
#2
(Original post by droid)
A particle P of mass m is dropeed from a height h above the surface of the liquid in a container and falls freely under gravity intill it enters the liquid. As P enters the liquid, a quarter of its kinetic energy is lost. At time t after entering the liquid P has fallen a vertical distance of x through the liquid and the resistance to motion is 3mx^2. Calculate the speed of P when it has just entered the liquid and hence show taht its speed at time t later is given by:
v^2 = 2gx - 2x^3 + (3/2)gh

any help would be great thanks.
First let us find its velocity immediately after entering the liquid:

Using the constant acceleration equations:
S=h, a=g, u=0, v=?=velocity before entering
v^2 = u^2 + 2as
v^2 = 2gh
KE before entering = 0.5m(2gh) = mgh
A quarter of KE is lost, hence is enters with (3/4)mgh of KE
KE = 0.5mV^2 = 0.75mgh
0.5V^2 = 0.75gh
V^2 = (3/2)gh

Considering motion inside the liquid:
f=ma
mg - 3mx^2 = ma
g-3x^2 = a
0.5v^2 = gx - x^3 + C
v^2 = 2gx - 2x^3 + k
x=0, v^2=(3/2)gh, k=(3/2)gh
Hence v^2 = 2gx - 2x^3 + (3/2)gh.
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