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.

The Snowball Problem

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.

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Projectile Inside a Liquid

Let's discuss a problem useful for Physics Olympiad, based on Projectile Inside a Liquid.

The Problem: Projectile Inside a Liquid

A body of mass m is projected inside a liquid at an angle θ₀ with the horizontal at an initial velocity v₀. If the liquid develops a velocity-dependent force F= -kv where k is a positive constant, determine the x and y components of the velocity at any instant.

Solution:

A body of mass m is projected inside a liquid at an angle θ₀ with the horizontal at an initial velocity v₀. The liquid develops a velocity-dependent force F= -kv where k is a positive constant.

Hence,

m dv/dt= -kv

or, dv/dt= -k/m v

or, dv/v= -k/m dt

Integrating both sides,

∫dv/v = -k/m ∫dt

ln|v|= -kt/m+c (where c is a constant of integration)….. (i)

Now, for the x component of velocity,

ln v_x= -kt/m+ c

From the given problem, we have

v_x=v₀cosθ₀ at t=0

Applying the above condition in eqn.(i), we get

ln (v_x/ v₀cosθ₀)= -kt/m

or v_x=v₀cosθ₀e^-kt/m

For the y component, we have to consider the acceleration due to gravity g.

Hence,

m dv_y/dt= -kv_y-mg

or, dv_y/(kv_y+mg)= -k/m dt

Integrating both sides,

ln|kv_y+mg|=-kt/m+c

At t=0, v_y=v₀sinθ₀

Hence,

kv_y+mg=(kv₀sinθ₀+mg) e^-kt