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

(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.

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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)

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

(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.

v=u+at= (α²/2)t

The acceleration= α²/2
V=α√s

Average velocity=(0+v)/2=α√s/2

Motion of an Elevator

Try this problem useful for the Physics Olympiad based on Motion of an Elevator.

The Problem: Motion of an Elevator

An elevator of mass M is accelerated upwards by applying a force F. A mass m initially situated at a height

of 1m above the floor of the elevator is falling freely. It will hit the floor of the elevator after a time equal to

√(2M/F+mg)
√(2M/F-mg)
√2M/F
√2M/(F+Mg)

Discussion:

Acceleration of elevator a_e= F/M ( in upward direction)

Acceleration due to gravity is in downward direction so acceleration of mass a_s= -g

Acceleration of mass with respect to elevator

= a_s-a_e

=(F/M)+g

=(F+Mg)/M

We know,

s=ut+(1/2)at²

From the given problem, we have s=1m

so,

1=(F+Mg)t²/M

t=√2M/(F+Mg)