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Chapter 3: Problem 14

A body of mass \(0.5 \mathrm{~kg}\) is projected under gravity with a speed of \(98 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) with the horizontal. The change in momentum (in magnitude) of the body is (a) \(24.5 \mathrm{~N}-\mathrm{s}\) (b) \(49.0 \mathrm{~N}-\mathrm{s}\) (c) \(98.0 \mathrm{~N}-\mathrm{s}\) (d) \(50.0 \mathrm{~N}-\mathrm{s}\)

Short Answer

Expert verified
The change in momentum (in magnitude) of the body is \(24.5 \mathrm{~N}-\mathrm{s}\) option (a).

Step by step solution

01

Find Initial Momentum

Since the body is initially projected with a speed of \(98 \mathrm{~m} / \mathrm{s}\), the initial momentum \(p_{i}\) can be calculated using the formula for momentum \(p=mv\). Hence, \(p_{i} = mv = 0.5 \mathrm{~kg} \times 98 \mathrm{~m} / \mathrm{s} = 49 \mathrm{~N}-\mathrm{s}\).
02

Resolve the Initial Momentum

Since the body is projected at an angle of \(30^{\circ}\) with the horizontal, the initial momentum needs to be resolved into horizontal (\(p_{ix}\)) and vertical (\(p_{iy}\)) components. Using the trigonometric relationships we find: \(p_{ix} = p_{i} \cos 30^{\circ} = 49 \mathrm{~N}-\mathrm{s} \times \sqrt{3}/2\) and \(p_{iy} = p_{i} \sin 30^{\circ} = 49 \mathrm{~N}-\mathrm{s} \times 1/2\).
03

Find Final Momentum

At the maximum height of the projectile, the vertical velocity (and hence momentum) will be zero, but the horizontal velocity (and momentum) remains unchanged due to the absence of any horizontal force. The final momentum \(p_{f}\) can therefore be calculated as follows: \[p_{fx} = p_{ix} = 49 \mathrm{~N}-\mathrm{s} \times \sqrt{3}/2\] and \[p_{fy} = 0\].
04

Calculate Change in Momentum

Finally, the change in momentum is the difference between the final and initial momentum. This difference is calculated vectorially, and it gives the magnitude as follows: \[\Delta p = \sqrt{(p_{fx} - p_{ix})^{2} + (p_{fy} - p_{iy})^{2}} = \sqrt{(49 \mathrm{~N}-\mathrm{s} \times \sqrt{3}/2 - 49 \mathrm{~N}-\mathrm{s} \times \sqrt{3}/2)^{2} + (0 - 49 \mathrm{~N}-\mathrm{s} \times 1/2)^{2}} = 24.5 \mathrm{~N}-\mathrm{s}\]

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Key Concepts

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

Projectile Motion
Projectile motion refers to the motion of an object that is thrown or projected into the air, subject to gravity's acceleration. It follows a curved path under the influence of gravity alone, assuming no air resistance. This path is typically in the shape of a parabola.

When dealing with projectile motion, it's important to separate the motion into horizontal and vertical components. The object's initial velocity, angle of projection, and the effects of gravity influence these components.
  • The horizontal component is usually influenced by the initial speed and angle, but remains constant throughout the motion since there are no horizontal forces.
  • The vertical component is affected by gravity, altering the velocity in the vertical direction.
By understanding these components, it becomes possible to predict the projectile's path, its maximum height, and the time it will be in the air.
Trigonometric Resolution
Trigonometric resolution involves breaking down forces or velocities into perpendicular components, typically along horizontal and vertical axes. This is fundamental to analyzing projectile motion.

Let's consider a body with velocity projected at an angle \( \theta \). To resolve this, utilize the basic trigonometric functions - sine and cosine.
  • Cosine \( \theta \) helps determine the horizontal component: \( v_x = v \cos \theta \).
  • Sine \( \theta \) helps find the vertical component: \( v_y = v \sin \theta \).
By applying these trigonometric functions, the vector is effectively split into two simple components. Knowing these components is crucial for further analysis, such as calculating momentum changes during projectile motion.
Kinematics
Kinematics is the study of motion without considering the forces that cause it. It focuses on describing the motion in terms of position, velocity, and acceleration. In the context of projectile motion:

  • The initial velocity of a projectile is split into horizontal and vertical components using trigonometry.
  • The horizontal motion is uniform, meaning constant velocity, since no forces influence it after the initial projection.
  • The vertical motion experiences a constant acceleration due to gravity, typically \(-9.8 \, \text{m/s}^2\). This affects the vertical position and velocity.
Using these principles, equations of motion can be applied to predict the trajectory and timing details of the projectile's motion. This enables determining key factors like the maximum height and range.
Conservation of Momentum
Conservation of momentum is a crucial principle in physics stating that when no external forces act on a system, the total momentum remains constant. For projectile motion, this principle plays out particularly for horizontal motion.

When a projectile is launched:
  • The horizontal momentum is conserved, as no external horizontal forces are acting on the body.
  • This means the horizontal velocity, and hence horizontal momentum, remains constant throughout the flight until it lands.
Conversely, the vertical momentum changes due to gravitational acceleration affecting the vertical velocity. Understanding the conservation of horizontal momentum aids in calculating resultant forces and subsequent motion accurately for systems in isolated conditions.

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Most popular questions from this chapter

A boy on a cycle pedals around a circle of 20 metres radius at a speed of 20 metres \(/ \mathrm{sec}\). The combined mass of the boy and the cycle is \(90 \mathrm{~kg}\). The angle that the cycle makes with the vertical so that it may not fall is \(\left(g=9.8 \mathrm{~m} / \mathrm{sec}^{2}\right)\) (a) \(60.25^{\circ}\) (b) \(63.90^{\circ}\) (c) \(26.12^{\circ}\) (d) \(30.00^{\circ}\)

The kinetic energy \(k\) of a particle moving along a circle of radius \(R\) depends on the distance covered. It is given as \(K . E .=a s^{2}\) where \(a\) is a constant. The force acting on the particle is [MNR 1992; JIPMER 2001, 2002] (a) \(2 a \frac{s^{2}}{R}\) (b) \(2 a\left(1+\frac{s^{2}}{R^{2}}\right)^{1 / 2}\) (c) \(2 a s\) (d) \(2 a \frac{R^{2}}{s}\)

The speed of a particle moving in a circle of radius \(0.1 m\) is \(v=1.0 t\) where \(t\) is time in second. The resultant acceleration of the particle at \(t=5 s\) will be (a) \(10 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(100 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(250 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(500 \mathrm{~m} / \mathrm{s}^{2}\)

Two masses \(m\) and \(M\) are connected by a light string that passes through a smooth hole \(O\) at the centre of a table. Mass \(m\) lies on the table and M hangs vertically. \(m\) is moved round in a horizontal circle with \(O\) as the centre. If \(l\) is the length of the string from \(O\) to \(m\) then the frequency with which \(m\) should revolve so that M remains stationary is (a) \(\frac{1}{2 \pi} \sqrt{\frac{M g}{m l}}\) (b) (c) (d) \(\frac{1}{\pi} \sqrt{\frac{m l}{M g}}\)

A flywheel rotates at constant speed of \(3000 \mathrm{rpm}\). The angle described by the shaft in radian in one second is (a) \(2 \pi\) (b) \(30 \pi\) (c) \(100 \pi\) (d) \(3000 \pi\)
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