Categories

## Tossing a Coin in a Moving Train

Try this beautiful problem, useful for Physics Olympiad, based on Tossing a Coin in a Moving Train.

The Problem: Tossing a Coin in a Moving Train

A person is sitting in a moving train and is facing the engine. He tosses up a coin that falls behind him. He concludes that the train is moving. A person is sitting in a moving train and is facing the engine. He tosses up a coin that falls behind him. He concludes that the train is moving-

(a) forward with increasing speed

(b) forward with decreasing speed

(c) backward with increasing speed

(d) backward with decreasing speed

Discussion:

When the person throws up a coin in a train moving with uniform speed, the coin goes up with the initial velocity of the train. Hence, due to the inertia of motion, the coin will travel the same distance as the train and will fall in the hands of the person.

Now, if the train is accelerating i.e. the velocity of the train is increasing in the forward direction, the coin will fall behind the person as the distance covered by the train will be greater than that of the coin.

Also, if the train is going backward with decreasing speed, the distance covered by the coin will be greater than the distance covered by the train as the velocity of the train is gradually decreasing. Hence for this case also, the coin will fall behind the person.

Categories

## Vertical Motion

Try this problem, useful for Physics Olympiad based on Vertical Motion.

The Problem: Vertical Motion

A boy is standing on top of a tower of height 85m and throws a ball in the vertically upward direction at 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=10m/s2, speed of sound in air=340m/s)
(A)    6m/s
(B)    8m/s
(C)    10m/s
(D)    12m/s

Solution:

Time taken by sound= (\frac{85}{340})=0.25secs
Time taken by the ball= 5.25-0.25=5s
Now,
Let us assume that the ball is thrown upwards with a velocity u. We know, (s=ut+\frac{1}{2}gt^2) where s is the distance covered,u is the initial velocity, g is the acceleration due to gravity and t is the time taken. Here, time taken t=5. Since the ball is thrown upwards, we have $$85=-5u+\frac{1}{2}\times10\times25$$
Hence, u is 8m/s.

Categories

## Relative Velocity of Canoe in River

A canoe has a velocity of 0.40m/s southeast relative to the earth. The canoe is on a river that is flowing 0.50m/s east relative to the earth. Find the velocity (magnitude and direction ) of the canoe relative to the river.

Discussion:

We apply the relative velocity relation. The relative velocities are (\vec{v_{CE}}), the canoe relative to the earth, (\vec{v_{RE}}), the velocity of the river with respect to Earth and (\vec{v_{CR}}), the velocity of the canoe relative to the earth.
$$\vec{v_{CE}}=\vec{v_{CR}}+\vec{v_{RE}}$$
Hence $$\vec{v_{CR}}=\vec{v_{CE}}-\vec{v_{RE}}$$
The velocity components of (
\vec{v_{CR}}) are $$-0.5+\frac{0.4}{\sqrt{2}}=-0.217m/s$$( in the east direction)
Now, for the velocity component in the south direction
$$\frac{0.4}{\sqrt{2}}=0.28$$ (in the south direction)
Now, the magnitude of the velocity of canoe relative to river $$\sqrt{(-0.217)^2+(0.28)^2}=0.356m/s$$
If we consider (\theta) as the angle between the canoe and the river,the direction of the canoe with respect to the river can be given by
$$\theta=52.5^\circ$$ ( in south west direction)

Categories

## Position of a Particle

Try this problem useful for the Physics Olympiad based on the Position of a Particle.

The problem:

A particle of mass m is subject to a force $$F(t)=me^{-bt}$$. The initial position and speed are zero. Find (x(t)).

Solution: In the given problem $$\ddot{x}=e^{-bt}$$
Integrating this with respect to time gives $$v(t)=-\frac{e^{-bt}}{b}+A$$ ( A is the constant of integration)
We integrate again with respect to x.
$$x(t)=\frac{e^{-bt}}{b^2}+At+B$$ ( B is the constant of integration)
The initial condition ( v(0)=0),gives (\frac{-1}{b}+A=0) $$\Rightarrow A= \frac{1}{b}$$
The intial condition $$x(0)=0$$, gives $$\frac{1}{b^2}+B=0$$ $$\Rightarrow B=-\frac{1}{b^2}$$

Hence,

$$x(t)=\frac{e^{-bt}}{b^2}+\frac{1}{b}t-\frac{1}{b^2}$$

Categories

## Instantaneous Velocity and Acceleration

Let’s discuss a problem useful from Physics Olympiad based on Instantaneous Velocity and Acceleration.

The Problem:

Let (\vec{v}) and (\vec{a}) be instantaneous velocity and the acceleration respectively of a particle moving in a plane. The rate of change of speed (dv/dt) of the particle is:
(a) (|a|)
(b) ((v.a)/|v|)
(c) the component of (\vec{a}) in the direction of (\vec{v})
(d) the component of (\vec{a}) perpendicular to (\vec{v})

Solution:

Let us consider (v^2=v_x^2+v_y^2).
We differentiate the above equation.
(\frac{dv}{dt})=((v_xa_x+v_ya_y)v)=(\frac{v.a}{v}).
Hence, the correct option will be B along with C since the component of a is in the direction of v.

Categories

## 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.

Categories

## Instantaneous Acceleration

Let’s discuss a problem useful for Physics Olympiad based on Instantaneous Acceleration.

The Problem:

Velocity-displacement curve of a particle moving in a straight line is as shown

• 2m/s2
• 5 m/s2
• 1 m/s2
• Zero

Discussion:

Acceleration a=v dv/ds

From, the graph, we can see a=vtanθ

tanθ=1/4

Putting the values, we get acceleration a= 1m/s2

Categories

## 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

v22x

=2(α2/2)x

Acceleration= α2/2

The initial velocity u is therefore zero and the acceleration is constant.

• v=u+at= (α2/2)t
• The acceleration= α2/2
• V=α√s

Average velocity=(0+v)/2=α√s/2