Try this problem, useful for Physics Olympiad based on Maximum Height of Object over a Pulley.
The Problem: Maximum Height of Object over a Pulley
Two objects with masses 5Kg and 2Kg hang 0.6m above the floor from the ends of a cord 6m long passing over a frictionless pulley. Both objects start from rest. Find the maximum height reached by the 2.00Kg object.
Diagram of the pulley
Discussion:
Set up : After the (5Kg) object reaches the floor, the (2Kg) object is in free fall, with downward acceleration (g).
Execution: The (2Kg) will accelerate upward at$$ \frac{5-2}{5+2}g=3g/7$$ and the (5Kg) object will accelerate downward at (3g/7).
Let the initial height above the ground be (h_0).
When the large object hits the ground, the small object will be at a height (2h_0) and moving upward with a speed given by $$ v_0^2=2ah_0=6gh_0/7$$. The small object will rise to a distance (v_0^2/2g=3h_0/g) and so the maximum height reached will be $$ 2h_0+3h_0/7=17h_0/7=1.46m$$ above the floor, which is (0.860m) above its initial height.
Two Blocks on an Inclined Plane
Try this beautiful problem, useful for Physics Olympiad based on Two Blocks on an Inclined Plane.
The Problem:
Two blocks of masses (4kg) and (8kg) are connected by a string and slide down a (30^\circ) inclined plane. The coefficient of kinetic friction between the (4Kg) block and the plane is (0.25); that between the (8Kg) block and the plane is (0.35).
(a) Calculate the acceleration of each block.
(b) Calculate the tension in the string. Discussion:
Since the larger block has the larger coefficient of friction it will need to be pulled down the plane.
For the small block, $$ 4(sin30^\circ-(0.25)cos30^\circ)-T=4a$$
$$\Rightarrow 4a=11.11N-t$$
For larger block, $$ 15.44+T=8a$$
Now, adding the two relations $$ 26.55N=12a$$
$$ a=2.21m/s^2$$
We can find the tension (T=2.27N)
A Pulley System
Let's discuss a problem, useful for Physics Olympiad based on A Pulley System. Read the problem carefully, find it yourself, and then read the solution.
The Problem: A Pulley System
One end of a string is attached to a rigid wall at point O, passes over a smooth pulley, and carries a hanger S of mass M at its other end. One end of a string is attached to a rigid wall at point O, passes over a smooth pulley, and carries a hanger S of mass M at its other end. Another object P of mass M is suspended from a light ring that can slide without friction, along the string, as is shown in the figure. OA is horizontal. Find the additional mass to be attached to the hanger S so as to raise the object P by 10cm.
Solution:
Diagram of the system
Let us denote the tension in each string as T. $$2Tcos\theta=Mg$$$$2(Mg)cos\theta=Mg$$$$cos\theta=\frac{1}{2}$$$$ \theta=60^\circ$$$$ tan60=\frac{\frac{40\sqrt{3}}{2}}{PQ}$$$$ tan60^\circ=\sqrt{3}$$Hence,$$ PQ=20cm$$Now, when an additional mass m is hung from the pulley, the length of PQ changes to P'Q'.
\(P'Q'=PQ-10=20-10=10\).
$$ Q'S'=\sqrt{P'Q'^2+P'S^2}=\sqrt{1300}$$Now, again considering the force equation$$2Tcos\theta=Mg$$$$2(M+m)g\times\frac{10}{\sqrt{1300}}=Mg$$$$2(M+m)\times\frac{1}{\sqrt{13}}=M$$$$ 2(M+m)=\sqrt{13}M$$$$2m=M(\sqrt{13}-2)$$$$m=\frac{M\times(\sqrt{13}-2)}{2}=0.9M$$
