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Chapter 2: Problem 41

A particle of mass \(m\) comes down on a smooth inclined plane from point \(B\) at a height of \(h\) from rest. The magnitude of change in momentum of the particle between position \(A\) (just before arriving on horizontal surface) and \(C\) (assuming the angle of inclination of the plane as \(\theta\) with respect to the horizontal) is (1) 0 (2) \(2 m \sqrt{(2 g h)} \sin \theta\) (3) \(2 m \sqrt{(2 g h)} \sin \left(\frac{\theta}{2}\right)\) (4) \(2 m \sqrt{(2 g h)}\)

Short Answer

Expert verified
The magnitude of change in momentum is \(2m\sqrt{2gh}\) (option 4).

Step by step solution

01

Calculate the velocity at position A

The particle starts from rest at height \( h \), so the initial potential energy is \( mgh \). By conservation of energy, all this energy converts into kinetic energy at position \( A \), just before the particle reaches the horizontal (still on the incline): \( \frac{1}{2} mv^2 = mgh \). Solving for velocity \( v \), we get \( v = \sqrt{2gh} \).
02

Determine the initial momentum at A

The momentum of the particle at position \( A \) is given by the product of its mass and velocity. Therefore, the momentum at \( A \) is \( p_A = m \times \sqrt{2gh} \).
03

Calculate the velocity at position C

When the particle reaches the horizontal surface at position \( C \), its direction would be horizontal. Hence, the only change during the descent on the incline is in direction, not speed, because we assume a smooth (frictionless) motion. The velocity remains \( \sqrt{2gh} \).
04

Determine the final momentum at C in the horizontal direction

In the horizontal direction, the velocity through the change in direction will be the same due to the frictionless condition, \( v \). However, compared to the inclined plane motion, the horizontal component of the velocity before reaching the horizontal surface was \( v \sin \theta = \sqrt{2gh} \sin \theta \). So the momentum at \( C \) after reaching the horizontal is \( p_C = mv = m \times \sqrt{2gh} \). Since \( \sin \theta = 1 \), it is absolute.
05

Find the magnitude of change in momentum

The change in momentum is calculated by \( |p_A - p_C| \). Since they are equal in magnitude in opposite directions in the plane but simplified findings align, \( |p_A| = m \times \sqrt{2gh} \) and \( |p_C| = m \times \sqrt{2gh} \). Thus, the magnitude of change in momentum is \( 2m \sqrt{2gh} \).
06

Select the correct option

From the options given in the question, option (4), \( 2m \sqrt{2gh} \), matches the calculated magnitude of change in momentum.

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is fundamental to understanding momentum change in mechanics.
It states that energy cannot be created or destroyed, only transformed from one form to another.
This means that the total energy within a closed system remains constant over time.

When a particle slides down an inclined plane, as in our exercise, it initially possesses potential energy because of its height.
  • Potential energy at the top of the incline is given by the expression: \( PE = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.
  • As the particle slides down, this potential energy is converted into kinetic energy.
  • The equation \( KE = \frac{1}{2}mv^2 \) is used to calculate kinetic energy at the bottom of the incline, where the velocity is maximum because all potential energy has been transformed.
Understanding this transformation helps explain how the particle's speed changes but not its total energy.
Inclined Plane Mechanics
Inclined planes are physical surfaces where the angle of elevation affects how forces act upon a body.
In our scenario, this angle is denoted by \( \theta \).
When a particle moves down a smooth inclined plane, gravity, the only force acting, does so at an angle.

Here's what happens:
  • The gravitational force along the plane is \( mg \sin \theta \), where \( \theta \) is the angle of inclination.
  • This force component causes the particle to accelerate downwards without any horizontal friction.
  • The change in momentum is significant in inclined plane mechanics, although the smoothness of the plane ensures no energy is lost to frictional forces.
This illustrates why analyzing the angle and its components is crucial in such problems.
Kinetic Energy Conversion
In this context, kinetic energy conversion occurs as the particle descends the inclined plane.
Initially at rest, all of the particle鈥檚 energy is potential.
As it moves down the plane, the energy converts fully from potential to kinetic.

Here's a deeper look at this conversion process:
  • The initial velocity at the top allows us to compute the potential energy, \( mgh \).
  • Upon reaching the base, this converts to kinetic energy: \( KE = \frac{1}{2} mv^2 \), and equating it gives \( v = \sqrt{2gh} \).
  • This motion illustrates pure energy conversion facilitated by gravity's influence and absence of friction.
Therefore, kinetic energy conversion is the key to understanding how velocity changes while maintaining energy balance in such mechanics problems.

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Most popular questions from this chapter

Column I contains physical quantity/process while column contains formula/principle. Match columns I and II such that the formula/principle is correct corresponding to the quantity in column I. $$ \begin{array}{|l|l|l|} \hline {\text { Column I }} & {\text { Column II }} \\ \hline \text { i. } \quad \text { Momentum } & \text { a. } \quad m\left(v_{2}-v_{1}\right) \\ \hline \text { ii. } \text { Impulse } & \text { b. } \begin{array}{l} \text { only momentum is } \\ \text { conserved } \end{array} \\ \hline \text { iii. Elastic collision } & \text { c. } \begin{array}{l} \text { momentum and kinefie } \\ \text { energy both are conserved } \end{array} \\ \hline \text { iv. Inelastic collision } & \text { d. } m v \\ \hline \end{array} $$

A body of mass \(3 \mathrm{~kg}\) collides elastically with another body at rest and then continues to move in the original direction with one half of its original speed. What is the mass of the target body? (1) \(1 \mathrm{~kg}\) (2) \(1.5 \mathrm{~kg}\) (3) \(2 \mathrm{~kg}\) (4) \(5 \mathrm{~kg}\)

A \(5 \mathrm{~kg}\) sphere is connected to a fixed point \(O\) by an inextensible string of length \(5 \mathrm{~m}\). The sphere is resting on a horizontal surface at a distance \(4 \mathrm{~m}\) from \(O\). Sphere is given a vertical velocity \(v_{0}\), and it moves freely till it reaches point \(P\), when string becomes taut. Find the maximum allowable velocity \(v_{0}\) (in \(\mathrm{m} / \mathrm{s}\) ), if impulse of tension in string is not to exceed \(6 \mathrm{~N} \mathrm{~s}\)

A small ball thrown at an initial velocity \(u=25 \mathrm{~m} / \mathrm{s}\) directed at an angle \(\theta=37^{\circ}\) above the horizontal collides elastically with a vertical massive smooth wall moving with a uniform horizontal velocity \(u / 5\) towards the ball. After collision with the wall the ball returns to the point from where it was thrown. Determine the time \(t\) (in \(\mathrm{s}\) ) from the beginning of motion of the ball to the moment of its impact with the wall. (Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) )

Block \(A\) moving on frictionless horizontal plane collides head-on with block \(B\) initially at rest. The collision is NOT \((0
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