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Physics Numerical Problem

A small mass of 10 KG is suspended from a spring balance. It is Pulled by Horizontal Spring so it makes 60º with the vertical. Find the new reading of the spring balance.

Motion of connected system – A numerical problem

What is the acceleration of the body m2 in the diagram below?

(The question was posted by Shuvam Shukla. Students and other visitors can post solution as comments to this post)

Physics Numerical Problems posted by Nabeela

The following questions were posted by Nabeela. We are just posting the question as such so that the visitors can post the answers.

  1. A mass-less string pulls a mass of 20kg upward against gravity. The string would break if subjected to a tension greater than 400N. What is the maximum acceleration with which the mass can be moved upward?
  2. A bullet of mass 20g is fired from a gun of mass 10kg with a velocity of 180m/s. Find the velocity of recoil of the gun. Find the force required to stop the gun before it moves 20cm
  3. A free body diagram of a mass on an inclined planeA body of mass 40kg is moving up an inclined plane with a uniform velocity when a force of 460N is applied. If the plane is inclined to the horizontal by an angle of 45o,calculate the coefficient of kinetic friction between the surfaces.
  4. If the coefficient of friction between the tyres of a truck and the road is k, show that the maximum stopping distance of the truck when moving with a velocity ‘v’ is v2/2 kg. Assume that the brakes are not applied
  5. A ball of mass 0.5kg moving with a speed of 20m/s collides with an identical ball at rest. After collision the direction of each ball makes an angle of 30o with the original direction. Find the speed of each ball after collision.
  6. What is the maximum horizontal distance that a ball thrown with a speed of 60 m/s can go without hitting the roof of a long hall 30m high?
  7. A ball is thrown horizontally strikes a wall 5m away. The height of the point struck by the ball is 1m lower than the height which it was thrown from. (1) With what velocity was it thrown (2) At what angle did the ball reach the wall?
  8. The time of flight of aprojectile is 10 seconds . its range on a horizontal plane is 100m . Calculate the angle of projection and the velocity of projection.

Distance of closest approach of two ships problem

Two ships are 10 km apart on a line running south to north. The one farther north is steaming west at 20 km/h. The other is steaming north at 20 km/h. What is their distance of closest approach? How long do they take to reach it?

Asked Gayathri

Solution and Answer:

The main hindrance to solving problem is that students fail to visualize the situation.

Let’s represent the condition using a diagram.

Let the first ship steaming north be named A and the other B. The relative velocity of A with respect to B is 20 √2 south west.  If we extend the line along the 45° south west from the initial position of A, the distance of closest approach is the perpendicular drawn from B to this line.

So, the distance of closest appraoch = 10 sin 45° = 10/√2 = 7.07 m

Time taken to reach the distance of closest approach = distance of closest approach / relative speed

= (10/√2)/(20 √2)

= 0.25 h

Four persons K, L , M, N are initially at the four corners of a square of side d Problem

Four persons K, L , M, N are initially at the four corners of a square of side ‘d’. Each person now starts moving with a uniform speed ‘v’ in such a way that K always moves directly towards L, L towards M, M towards N and N towards K. After what time will they meet?

Gayathri asked.

Answer: Many students do not understand the real situation initially. Every time the persons are approaching each other and hence they will be moving closer and closer as they continue their walking. Finally they’ll reach the centre of the square to meet each other. I’ve tried to visualize the situation below.

From the diagram, we can make out that the resultant displacement by each when they meet will be d/√2 and the component of velocity of each towards the final point (the centre of the square) is v/√2.

Therefore,the time taken = displacement by the component of velocity in its direction = (d/√2) / (v/√2) = d/v

Understanding Projectile Motion

IN A PROJECTILE MOTION WE TAKE MOTION IN A HORIZONTAL DIRECTION AS g=0 while in vertical motion as g=-9.8 meter / second square why?

Asks Atanu PAUL

English: motion of the projectile

Projectile Motion is an example for motion in two dimension. It can be studied easily by resolving into two components – The vertical motion is affected by gravity and the horizontal motion is not affected by gravity.

Therefore we take g=0 when we consider the horizontal component of motion as the projectile is not accelerated in the horizontal direction. The horizontal component of velocity remains constant through out its motion.

The following presentation will help you understand the concept better.

A PowerPoint Presentation on projectile_motion

Why light is not accelerated in vacuum?

If vacuum has no mass …. then why isn’t a light beam accelerated in space ?? – Asks Akil Raj

Answer:

The speed of light in a medium or in vacuum is constant. It is neither accelerated or decelerated while propagating. But, the speed of light in a (optically) denser medium is less than the speed of light in an optically rarer medium.

The speed of light is maximum in vacuum and it is constant and is equal to 299792458 m/s as theoretically established by electromagnetic wave theory too.

9If you look at it this way; “In vacuum there is no particle to oppose the propagation of light, therefore it should travel with a constant velocity, because to increase or decrease the velocity some external influence is essential)

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