Let's discuss the beautiful snowball problem useful for Physics Olympiad.
The Snowball Problem:
(a) A snowball rolls off a barn roof that slopes downward at an angle of \(40^\circ\). The edge of the roof is 14.0m above the ground and the snowball has a speed of 7.00m/s as it rolls off the roof. How far from the edge of the barn does the snowball strike the ground if it doesn't strike anything else while falling?
(b) A man 1.9m tall is standing 4.0m from the edge of the barn. Will he be hit by the snowball?
Solution:
The snowball follows the projectile motion.
In part(a), the vertical motion determines the time in the air. The acceleration $$a_x=0$$, $$a_y=+9.80m/s^2$$ and $$v_x=v_0cos\theta_0=7cos40^\circ=5.36m/s$$ Vertical component of velocity $$ v_y=v_0sin\theta_0=4.50m/s$$ Now, distance from ground \(s=14m\). $$ 14=(4.50)t+(\frac{1}{2}\times9.8\times t^2)$$ We use Sreedhar-Acharya's method to solve for t. $$ t=\frac{(-4.5 \pm \sqrt{4.5^2-4(4.9)(-14)}}{2(4.9))}$$ The positive root for \( t=1.29s\) The horizontal distance$$ v_xt+(1/2)a_x t^2= 6.91m$$
(b) $$ x-x_0=v_xt+\frac{1}{2}a_xt^2$$ gives $$ t=\frac{x-x_0}{v_x}=\frac{4.0}{5.36}=0.746s$$ In this time the snowball travels downward a distance $$ y-y_0=v_y+\frac{1}{2}a_y t^2=6.08m$$ and is therefore $$ 14.0-6.08=7.9m$$ above the ground. The snowball passes well above the man and does not hit him.
Let's discuss a beautiful problem useful for Physics Olympiad based on the Conical Pendulum.
The Conical Pendulum Problem:
An inventor designs a pendulum clock using a bob with mass m at the end of a thin wire of length L. Instead of swinging back and forth, the bob is to move in a horizontal circle with constant speed v, with the wire making a fixed angle \(\beta\) with the vertical direction. This is called a conical pendulum because the suspending wire traces out a cone. Find the tension F in the wire and the period T( the time for one revolution of the bob). Solution:
Conical Pendulum
SET UP: To find the tension F and period R, we need two equations. These will be the horizontal and vertical components of Newton's second law applied to the bob. We'll find the radial acceleration of the bob using one of the circular motion equations. The figure shows the free-body diagram and coordinate system for the bob at a particular instant. There are just two forces on the bob: the weight mg and the tension F in the wire. Note that the entre of the circular path is in the same horizontal plane as the bob, not at the top end of the wire. The horizontal component of tension is the force that produces the radial acceleration \(a_rad\). Execution: The bob has zero vertical acceleration.
Newton's second law says that $$ \Sigma F_x=Fsin\beta=ma_r$$ $$ \Sigma F_y=Fcos\beta+(-mg)=0$$ These are two equations for the two unknowns F and \(\beta\). The equation for \(\Sigma F_y\) gives $$ F=mg/cos\beta$$. That's our target expression for F in terms of \(\beta\). Substituting this result into the equation for \(\Sigma F_x\) and using \(sin\beta/ cos\beta =tan\beta\), we find $$ a_r=g tan\beta$$ To relate \(\beta\) to the period T, we use the equation $$ T=2\pi R/v$$ in terms of the period , $$ a_R=4\pi^2R/T^2$$ so, T=\(2 \pi \sqrt{\frac{R}{gtan\beta}}\) Now, $$ R=Lsin\beta$$. We substitute this and use \(sin\beta / tan \beta=cos\beta\):
So, $$ T= 2\pi\sqrt{\frac{Lcos\beta}{g}}$$ Evaluation: For a given length L, as the angle \(\beta\) increases, \(cos\beta\) decreases. A conical pendulum will never make a very good clock because the period depends on the angle \(\beta\) in such a direct way.
Let's discuss a beautiful problem useful for Physics Olympiad based on Suitcase Falling from an Airplane.
The Problem: Suitcase Falling from an Airplane
An airplane is flying with a velocity of \(90m/s\) at an angle of \(23^\circ\) above the horizontal. When the plane is \(114m\) directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment. How far from the dog will the suitcase land? Ignore air resistance.
Solution:
The suitcase moves in projectile motion. The initial velocity of the suitcase is equal to the velocity of the airplane. To find the time, it takes to reach the ground the y-component of the velocity $$ v_{0y}=v_0sin23^\circ$$ The acceleration $$ a_y=-9.8m/s^2$$ since +y is taken to be upward. Now, the vertical distance from the plane to the dog s=114m. Putting these values in the equation of motion to find the time t, we have $$ 114=90sin23^\circ+\frac{1}{2}\times(-9.8)\times t^2$$ This gives, $$ t=9.60s$$ TThe distance the suitcase travels horizontally is $$v_{0x}t=(v_0 cos23^\circ)t=795m$$
Let's discuss a beautiful problem useful for Physics Olympiad based on Mass supported by a Hollow Cylinder.
The Problem:
A mass m is supported by a massless string wound on a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?
Solution:
For the mass m, the force equation stands as:
$$mg-T=ma....(i)$$
For the cylinder, the force equation is:
$$ T.R=mR^2(a/R)$$
Hence,
from above equation $$ T=ma $$
Now, putting the value of T in equation (i), we get
$$ma=2mg$$
$$\Rightarrow a=g/2$$
Speed of ball thrown down from a height
A boy is standing on top of a tower of height 85m and throws a ball in the vertically upward direction with a certain speed. If 5.25 secs later he hears the ball hitting the ground, then the speed with which the boy threw the ball is ( take g=910m/s2, speed of sound in air=340m/s)
6m/s
8m/s
10m/s
12m/s
Solution:
Time taken by sound= 85/340=0.25secs
Time taken by the ball= 5.25-0.25=5 sec
Now,
we write the equation $$ s=ut+1/2ft^2 $$
Here s=-85 and time t =5 secs
Hence,
$$ -85=5u-1/21025$$
$$\Rightarrow u=8m/s$$
Masses over a frictionless pulley
Let's discuss a beautiful problem useful for Physics Olympiad based on masses over a frictionless pulley.
The Problem: Masses over a frictionless pulley
Two bodies A and B hanging in the air are tied to the ends of a string which passes over a frictionless pulley. The masses of the string and the pulley are negligible and the masses of two bodies are 2kg and 3kg respectively. (Assume g=(10m/s^2)). Body A moves upwards under a force equal to
(a)30N
(b)24N
(c)10N
(d)4N Solution:
The masses of two bodies are 2kg and 3kg respectively. The acceleration of A is
$$ a=\frac{3-2}{3+2}g=2m/s^2$$
The net force on A will be F=ma=(2\times2)=4N.
Downward Acceleration
A soldier carrying a gun with parachute A soldier with a machine gun, falling from an airplane gets detached from his parachute. He is able to resist the downward acceleration if he shoots 40 bullets a second at the speed of 500m/s. If the weight of a bullet is 49gm, what is the weight of the man with the gun? Ignore resistance due to air and assume the acceleration due to gravity g=9.8m/s2
Solution:
Let M be the mass of the soldier and Mg be the mass of the gun.
To nullify the downward acceleration, 40 bullets are shot at the speed of 500m/s.
The force equation can be written as:
$$(M+M_g)9.8=40\times500\times49\times10^-3$$
$$\Rightarrow (M+M_g)=100Kg$$
Therefore,
the weight of the man with the gun=100Kg.
Frictional force between pen and paper
A pen of mass 'm' is lying on a piece of paper of mass M placed on a rough table. If the cofficient of friction between the pen and paper and, the paper and the table are (\mu_1) and (\mu_2) respectively, what is the minimum horizontal force with which the paper has to be pulled for the pen to start slipping?
Solution:
For pen to start slipping, maximum horizontal acceleration for pen and paper to start slipping is (f=\mu_1)g
Therefore, (a=\mu_1 g) is the common acceleration for pen and paper.
If (f_1) and (f_2) be the frictional forces for pen and paper respectively, the net force for the system
$$ F=f_1+f_2+Ma
F=\mu_1mg+\mu_2mg+Ma$$
Now, (a=\mu_1 g)
Hence, $$ F=(m+M)(\mu_1+\mu_2)g$$
Motion under Constant Gravity
Let's discuss a beautiful problem useful for Physics Olympiad based on Motion under Constant Gravity.
The Problem: Motion under Constant Gravity
A person throws vertically up n balls per second with the same velocity. He throws a ball whenever the previous one is at its highest point. The height to which the balls rise is
(a) g/n2
(b) 2gn
(c) g/2n2
(d) 2gn2
Solution:
We know v=u-gt. v is zero at the highest point. Time t taken by one ball to reach maximum height is 1/n.
Hence, we have
u=gt
or, u=g/n……. (i)
now, from the relation v2=u2-2gh. Again, v=0
u2=2gh……. (ii)
Putting the value of u from equation (i) in above relation (ii), we get
h=g/2n2.
Velocity and Acceleration
Let's discuss a beautiful problem from Physics Olympiad based on Velocity and Acceleration.
The Problem: Velocity and Acceleration
A particle is moving in positive x-direction with its velocity varying as v= α√x. Assume that at t=0, the particle was located at x=0. Determine the
the time dependence of velocity
acceleration
the mean velocity of the particle averaged over the time that the particle takes to cover the first s metres of the path.
Discussion:
v=α√x.
Squaring both sides
v2=α2x
=2(α2/2)x
Acceleration= α2/2
The initial velocity u is therefore zero and the acceleration is constant.
