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Chapter 7: Problem 5

A particle falls freely from rest through a distance \(d\). Its speed is then: (a) \(\sqrt{g d}\) (b) \(-\sqrt{2 g u}\) (c) \(-\sqrt{\frac{g d}{2}}\) (d) \(\sqrt{2 g d}\).

Short Answer

Expert verified
The speed is \(\sqrt{2gd}\), so the correct answer is (d).

Step by step solution

01

- Understand the given information

The problem involves a particle falling freely, starting from rest. The particle falls through a distance of \(d\). We need to determine its speed at that point.
02

- Identify the relevant formula

For an object in free fall starting from rest, the relevant kinematic equation that relates distance \(d\), acceleration due to gravity \(g\), and velocity \(v\) is \(v^2 = u^2 + 2ad\). Since the initial velocity \(u\) is 0 (the particle starts from rest), the equation simplifies to \(v^2 = 2ad\).
03

- Simplify the formula

Here, \(u = 0\) because the particle starts from rest, and \(a = g\) as it is in free fall. Thus, the equation becomes: \[v^2 = 2gd\]
04

- Solve for the velocity \(v\)

To find the speed (velocity), take the square root of both sides of the equation: \[v = \sqrt{2gd}\]
05

- Match the correct answer

Comparing the result \(v = \sqrt{2gd}\) with the given options, we see that the correct answer matches option (d).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

kinematic equations
Kinematic equations are crucial in understanding motion in physics. These equations describe the relationships between displacement, velocity, acceleration, and time. One core equation used in free fall problems, like the one given in the exercise, is:
  • This specific equation is particularly useful as it links the final velocity ( v ) with the initial velocity ( u ), acceleration ( a ), and distance ( d ).
    In our exercise, the particle falls freely from rest, so u = 0. This simplifies the equation to v^2 = 2gd , where g represents acceleration due to gravity.
    • Understanding how and when to apply these equations can simplify solving many motion-related problems.
  • Accelerated down due to gravity with no initial force acting upon it, the equation becomes much simpler and easier to use since initial velocity ( u = 0.)
acceleration due to gravity
Acceleration due to gravity ( g ) is a fundamental concept in free fall motion. It represents the rate at which an object accelerates towards the Earth due to gravity. The standard value of g on the surface of the Earth is about 9.8 m/s^2.
This constant force acts on the particle causing it to accelerate downwards.

  • For instance, in our exercise, as the particle falls freely, it accelerates at 9.8 m/s^2 Going further, gravity affects all bodies equally regardless of their mass, meaning if you drop a pebble and a boulder, both would fall with the same acceleration due to gravity (Assuming no air resistance).
    This uniform acceleration greatly simplifies solving free fall problems, making it easier for students to apply kinematic equations effectively. Always remember, g stays constant as it measures how strong the pull of gravity is on any free-falling object.
    • initial velocity
    Initial velocity ( v0 ) is the velocity of an object before any external forces act upon it. In the case of free fall, if an object begins its motion from rest, the initial velocity ( u ) is zero.
    • In our exercise, since the particle is falling freely from rest,
    • it simplifies our calculations a lot as we don't need to worry about the starting speed. We start by analyzing how gravity pulls the particle from a stationary position.
      • Using the kinematic equation v^2 = u^2 + 2ad gets simplified to v^2 = 2gd because u = 0.
    • This results in the equation:
    • v = 鈳�2gd
    • as detailed in our solution. This is mostly why understanding whether an object starts from rest or not simplifies solving free fall problems.

      • Remember: Identifying an object's initial velocity is crucial for effectively using kinematic equations and arriving at the correct solutions in physics problems. Always check if the initial velocity is zero before proceeding with calculations.

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    Most popular questions from this chapter

    A particle of weight \(W\) is attached to a point \(C\) of an unstretched elastic string \(\mathrm{AB}\), where \(\mathrm{AC}=4 a / 3, \mathrm{CB}=4 a / 7\). The ends \(\mathrm{A}\) and \(\mathrm{B}\) are then attached to the extremities of a horizontal diameter of a fixed hemispherical bowl of radius \(a\). and the particle rests on the smooth inner surface, the angle BAC being \(30^{\circ}\). Show that the modulus of elasticity of the string is \(W\) and determine the reaction of the bowl on the particle. (U of L)

    A light elastic string, of unstretched length \(a\) and modulus of elasticity \(W\), is fixed at one end to a point on the ceiling of a room. To the other end of the string. is attached a particle of weight \(W\). A horizontal force \(P\) is applied to the particle and in equilibrium it is found that the string is stretched to three times its natural length. Calculate: (a) the angle the string makes with the horizontal, (b) the value of \(P\) in terms of \(W\). If, instead, \(P\) is not applied horizontally find the least value of \(P\) which in equilibrium will make the string have the same inclination to the horizontal as before. Deduce that the stretch length of the string is \(\frac{3}{2} a\) in this case and find the inclination of \(P\) to the vertical. (U of L)

    A ring is threaded on to a smooth wire in the form of a circle fixed in a vertical plane. The ring is projected from the lowest point on the wire with a velocity of \(4.2 \mathrm{~ms}^{-1}\), Jf the radius of the circular wire is \(0.6 \mathrm{~m}\), find the height above the centre at which the particle first comes to instantaneous rest. If, instead, the ring had been projected with a velocity of \(5.6 \mathrm{~ms}^{-1}\) describe its motion.

    One end \(\mathrm{O}\) of an elastic string OP is fixed to a point on a smooth plane inclined at \(30^{\circ}\) to the horizontal. A particle of mass \(m\) is attached to the end \(\mathrm{P}\) and is held at 0 . If the natural length of the string is \(a\) and its modulus is \(2 m g\), find: (a) the distance down the plane from \(\mathrm{O}\) at which the particle first comes to instantaneous rest after being released from rest at \(\mathrm{O}\). (b) the velocity of the particle as it passes through its equilibrium position.

    Prove that the potential energy of a light elastic string of natural length \(I\) and modulus \(\lambda\) when stretched to a length of \((l+x)\) is \(\frac{1}{2} \lambda \frac{x^{2}}{l}\). Two points \(\mathrm{A}\) and \(\mathrm{B}\) are in a horizontal line at a distance \(3 l\) apart. A particle Pof mass \(m\) is joined to \(\mathrm{A}\) by a light inextensible string of length \(4 l\) and is joined to \(\mathrm{B}\) by a light elastic string of natural length \(/\) and modulus \(\lambda\). Initially \(\mathrm{P}\) is held at a point \(C\) in \(A B\) produced such that \(B C=l\), both strings being just taut, and is then released from rest. If \(\lambda=\frac{m g}{4}\) show that when \(\mathrm{AP}\) is vertical the speed of the particle is \(2 \sqrt{g l}\) and find the instantaneous value of the tension in the elastic string in this position. (J.M.B)
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