### Welcome to Talha's Physics Academy

To Help Teachers and Students.

Talha's Physics Academy is an exploration environment for concepts in physics which employs free Physics Books and other linking strategies to facilitate smooth navigationThe entire environment is interconnected with thousands of links, reminiscent of a neural network.

New content for Talha's Physics Academy will be posted as it is developed,It is my intent to keep this material continuously available except for brief maintenance times.

All the Branches of Physics are covered.

Share with others.

### Why a cricketer lowers his hands while catching a ball?

• We know how to calculate the kinetic energy of moving objects -- isn't that enough? No. It turns out that many situations involving collisions do not obey the simple conservation of Mechanical Energy. Why not? Because it takes energy to bend, break, mutilate and deform objects, energy which disappears from the kinetic and gravitational potential energy.

• But a different quantity is conserved, even during collisions. The linear momentum of an object is defined as
```                   p  =  (mass) * (velocity)
```
It is a vector quantity, and the total linear momentum of a bunch of objects will remain the same, before and after a collision.

• Momentum is connected to force by impulse, which is simply
```            impulse   =  (force) * (time)
```
if the force has a constant magnitude during its action. If the force changes with time, then one must integrate to find the impulse:
```                         /
impulse   =  |  (force) dt```
```                         /
```

• The Momentum-Impulse Theorem states that the change in momentum of an object is equal to the impulse exerted on it:
```           (change in momentum)    =  (impulse)

p      -   p          =  (force) * (time)
final      initial

m*v      -  m*v         =  (force) * (time)
final       initial```

Impulse  It is effect of force acting for a short interval of time.
According to Newton’s 2nd law
Rate of change of momentum = Force applied.
If time period dt is  increased than Impact force decreases.

Why a cricketer lowers his hands while catching a ball

If suppose a 0.16 kg cricket ball hits a bat at 100 km/hr and then comes off the bat at 100 km/hr in 0.001 sec. The average force on ball is 8800 N which is enough to lift 880 kg of mass off the ground.  That is why it hurts to get struck by ball. Therefore while catching cricketer lowers his hands to increase time interval of change in momentum thereby impact force reduces and the chances of any injury and avoiding reaction of hand that can drop the catch .

### HEAVY OR LIGHT BAT?

Momentum:

•Momentum is total quantity of motion possessed by body
•  It is product of mass and velocity
Collisions and the Conservation of Momentum
The impact between bat and ball is a collision between two objects, and in its simplest analysis the collision may be taken to occur in one-dimension. In reality most collisions between bat and ball (especially the ones I am able to make) are glancing collisions which require a two-dimensional analysis. It turns out, in fact, that a glancing blow is necessary to impart spin to the ball which allows it to travel farther.[5] Maybe I'll write about this more interesting, but more difficult problem later, but for right now I'll keep things simple and look at the collision in one-dimension only. The ball, m1, and bat, m2, both have initial velocities before the collision (subscript "b"), with the ball's velocity being negative. After the collision (subscript "a") both bat and ball have positive velocities. The before and after velocities and the masses of bat and ball may be related to each other through the physical relationship known as the conservation of linear momentum. Linear momentum is the product of the mass and velocity of an object, p=mv. If the net force acting on a system of objects is zero then the total momentum of the system is constant. While the bat and ball are in contact the player is exerting a force on the bat; the force needed to swing the bat. So, in a completely correct analysis, momentum is not constant because of this force exerted by the player swinging the bat. However, the force on the bat by the player is very much smaller than the forces between bat and ball during the collision, and the contact time between ball and bat is very short (less than 1 millisecond). This allows us to ignore the force on the bat by the player during the collision between ball and bat without significantly affecting our results. If we ignore the force by the player on the bat, we can express the conservation of linear momentum by setting the total momentum before the collision equal to the total momentum after the collision.
m1v1b + m2v2b = m1v1a + m2v2a

Usually when a student encounters the conservation of momentum in a physics course the masses of both objects are given, along with the initial velocities before the collision. A typical homework or quiz question would be to determine the final velocities of the two objects after the collision. When one is searching for two unknown quantities one must have two equations. So, we need more than just the conservation of momentum. For our student in a physics course this second equation is usually the conservation of energy. The conservation of energy relates the change in kinetic energy (associated with motion), the change in potential energy (associated with springs and position), and any work done by nonconservative forces (like friction) which act on the system. The change in kinetic energy includes information about the velocities of the ball and bat before and after the collision. During the collision the ball undergoes a significant amount of compression, and damping forces convert much of the ball's initial kinetic energy into heat. The change in potential energy and work done by friction describe how much of the initial energy is lost during compression of the bat and ball. The manner in which these energies are related during the bat-ball collision is rather complicated. However, the effective relationship between the elastic properties of the ball and the relative velocities of bat and ball may be summarized in terms of the coefficient of restitution, (e)

The coefficient of restitution of a baseball or softball decreases with increasing incoming ball speed (v1b). Modern baseballs are manufactured to have a coefficient of restitution of 0.55 for a 90mph pitch speed, while softballs are manufactured to have e=0.44 for pitch speeds of 60 mph. Assuming a constant pitch speed, we can combine two equations above and do a little algebra to solve for the velocity of the baseball after the collision:

This equation tells us how the batted ball velocity (v1a)depends on the mass of the ball (m1) and bat (m2), the elasticity of the ball (e), the pitched ball speed (v1b) and the bat swing speed (v2b). The properties of the ball may be treated as constants since they don't change during a turn at bat. The hitter has no control over the pitched ball speed, and while it may vary considerably from pitch to pitch we'll assume that it is a constant. The only two remaining variables which determine the final velocity of the ball are the mass of the bat, m2 and the initial speed of the bat, v2b. If we know these two parameters, we can predict the batted ball speed. As we will see, however, the problem is complicated somewhat by the fact that the speed with which a player can swing a bat depends on the weight of the bat.

The crowd loves a batter who can hit sixes. If you want to hit the ball as fast and far as possible, should you use a light or heavy bat? That's an age old question with plenty of answers, but which is the correct answer? Light bats can be swung faster than heavy bats, but only about 10% faster (for the usual range of bat weights). Imagine hypothetically that the bat weighs 10 grams - light as a feather. If you swing it as fast as possible, you might get the tip to travel at say 160 km/hr. Now double the weight to 20 gm. This time the tip travels at about 159 km/hr. The problem here is that your arms weigh about 8 kg all up, so the extra 0.01 kg is hardly noticeable. Most of the effort needed to swing a bat goes into swinging the arms. That's why light bats can be swung only about 10% faster than heavy bats.

If a light bat was swung at the same speed as a heavy bat and both hit the same ball, the heavy bat would pack more power since it has more energy and more momentum. But light bats can be swung 10% faster. If a bat is swung 10% faster, the ball comes off the bat about 7.5% faster. That almost makes up for the fact that light bats are basically less powerful when swung at the same speed as heavy bats. The end result is that heavy bats are about 1% more powerful than light bats. Having a heavy bat is a definite advantage if you swing all bats at the same medium speed, but if you need to move the bat quickly into position to strike the ball, a light bat will get there faster. Heavy for a 10 year old might be light for a 100 kg cricketer, so the real answer for raw bat power is to use a bat that is as heavy as feels comfortable to swing

### PROJECTILE MOTION

PROJECTILE
•                       Projectile is the name given to a body thrown with some initial velocity, and then allowed to move in two dimensions under the action of gravity alone, without being propelled by any engine or fuel. The path followed by a projectile is called its trajectory.
Examples of projectiles
•       A cricket ball hit by a bat.
•       A javelin or hammer thrown by an athlete.

•       A bullet fired from a rifle.
•       A piece of stone thrown in any direction.

ASSUMPTIONS IN PROJECTILE MOTION

There is no frictional resistance of air.
•       The effect due to rotation of earth and curvature of the earth is negligible.
•       The acceleration due to gravity is constant in magnitude and direction at all points of the motion of projectile.

ANGULAR PROJECTION
•       OX,is a horizontal line on ground and
•        OY is a vertical line perpendicular to ground.
•        Suppose a cricket ball be projected from the point O with velocity u, making an angle Î¸  with the horizontal direction OX
Resolving velocity  into two rectangular components, we get
(i) u cosÎ¸ , along OX
(ii) u sin Î¸  along OY.
•        As these two component velocities act at right angles to each other, therefore they are independent of each other.
The horizontal component velocity u cos Î¸  is constant throughout the motion as there is no accelerating force in the horizontal direction.
The vertical component velocity u sinÎ¸   decreases continuously with height, from O to H, due to downward force of gravity and becomes zero at H.

EQUATION OF TRAJECTORY:

Path of projectile
•       Suppose at any time t, the object reaches at P (x, y) clearly,
•       x= horizontal distance traveled by object in time t
•       y = vertical distance traveled by object in time t.
This is an equation of a parabola

Time of flight

•       . It is the total time for which the object is in flight (i.e. remains in air).  It is denoted by T.
•       The total time of flight consists of two parts
•       Total time of flight = time of ascent + time of descent

Maximum height
It is the maximum vertical height attained by the object above the point of projection during its flight. It is denoted by h.

 Horizontal Range : It is the horizontal distance covered by the object between its point of projection and the point of hitting the ground.  It is denoted by R.
Clearly, the horizontal range is the horizontal distance covered by the object with uniform velocity u cosÎ¸  in the time equal to total time of flight T.

Maximum Horizontal Range

we note that for a given speed u of the object, the value of horizontal range depends upon angle of projection  as g is constant at a place. Therefore horizontal range R will be maximum if
Sin 2Î¸ = maximum = 1
= sin 90°
Or 2Î¸ =90° or Î¸ =45°
This concept has been used by athletes in long jump, javelin throw, cricket ball throw etc

## Book Description

Schaum's Easy Outline of College Physics is a pared-down, simplified, and tightly focused review of the topic. With an emphasis on clarity and brevity, it features a streamlined and updated format and the absolute essence of the subject, presented in a concise and readily understandable form. Graphic elements such as sidebars, reader-alert icons, and boxed highlights stress selected points from the text, illuminate keys to learning, and give you quick pointers to the essentials.
• Expert tips for mastering college physics
• Last-minute essentials to pass the course
• Appropriate for the following courses: College Physics, Introduction to Physics, Physics I and II, Noncalculus Physics, Advanced Placement H.S. Physics
• Easy-to-follow review of college physics
• Supports all the major textbooks for college physics courses

### How To Solve Physics Problems - R. Oman, D. Oman

275,000 students in noncalculus physics; Required pre-med course; Super-accessible, straightforward help; Student-grabbing graphics and style; Icons for important concepts; 1-2-3 help with problems.