Problem 9 - Master Resource Book in JEE Main Physics

Question

A body of mass \(1 \mathrm{~kg}\) is projected with velocity \(50 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) with the horizontal. At the highest point of its path a force \(10 \mathrm{~N}\) starts acting on body for 5 s vertically upward besides gravitational force, what is horizontal range of the body? ( \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (a) \(125 \sqrt{3} \mathrm{~m}\) (b) \(200 \sqrt{3} \mathrm{~m}\) (c) \(500 \mathrm{~m}\) (d) \(250 \sqrt{3} \mathrm{~m}\)

Step-by-Step Solution

Answer

Check force influence revision or problem resources for correction or confirmation as value differs.

Step 1: Resolve Initial Velocity into Components

The initial velocity of the body is given as \(50 \mathrm{~m/s}\) at an angle of \(30^{\circ}\). Components of the initial velocity are: - Horizontal component, \(v_{x} = 50 \cos(30^{\circ})\) - Vertical component, \(v_{y} = 50 \sin(30^{\circ})\) Calculating these, we have: \(v_{x} = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \mathrm{~m/s}\) \(v_{y} = 50 \times \frac{1}{2} = 25 \mathrm{~m/s}\)

Step 2: Calculate Time to Reach Highest Point

At the highest point, the vertical component of the velocity is zero. Using the equation of motion: \(v = u + at\) Here, \(v = 0\), \(u = v_{y} = 25 \mathrm{~m/s}\), and \(a = -g = -10 \mathrm{~m/s}^2\). Solving for time \(t\) to reach the highest point: \[0 = 25 - 10t\] \[t = 2.5 \mathrm{~s}\]

Step 3: Update Velocity at the Highest Point

At the highest point, a force \(10 \mathrm{~N}\) acts upward for \(5 \mathrm{~s}\). The acceleration due to this force is given by \(a = \frac{F}{m} = \frac{10 \mathrm{~N}}{1 \mathrm{~kg}} = 10 \mathrm{~m/s^2}\). The net acceleration upward becomes \(a' = 10 - 10 = 0 \mathrm{~m/s^2}\) Thus, the velocity remains zero in the vertical direction since the upward force just counteracts gravity.

Step 4: Calculate Total Flight Time

Total time during which components effect the motion = Time up to the highest point + Time influenced by additional force.Thus, the total time of flight \(T = 2.5 \mathrm{~s} + 5 \mathrm{~s} = 7.5 \mathrm{~s}\).

Step 5: Calculate Total Range

To find the horizontal range, use the equation:\[ R = v_x \times T \] Substitute the values:\[ R = 25 \sqrt{3} \times 7.5 \] \[ R = 187.5 \sqrt{3} \text{ m} \] This value doesn't match the choices given initially, which suggests a miscalculation. Verify if other problems, like assuming correct calculation or correct force application phase, affected the step result.

Key Concepts

Newton's Laws of Motion

Newton's Laws of Motion are fundamental principles that govern the behavior of objects in motion. In projectile motion, these laws help us understand how forces affect the trajectory of a projectile. The first law, the Law of Inertia, tells us that an object will continue its motion unless acted upon by an external force. In our exercise, once the projectile is launched, it will naturally follow a curved path due to gravity, a force acting downward.

The second law, the Law of Acceleration, describes how the velocity of an object changes when it is subject to an external force. This is expressed in the formula: \[ F = ma \] where \( F \) is the force applied, \( m \) is the mass, and \( a \) is the acceleration. In the given scenario, an additional force of \(10 \text{ N}\) is applied vertically upward when the projectile is at its highest point.

Finally, there's the third law: For every action, there is an equal and opposite reaction. This law isn't directly seen in the problem but supports the projectile's initial perspective and resistance to change in motion, except when the new force is introduced.

Kinematics

Kinematics is all about describing motion. It's the branch of physics that deals with the geometry of motion without considering the forces that cause it. By using kinematic equations, we can analyze the motion of the projectile, like when it reaches the highest point or its total range.

One of the kinematic equations used is: \[ v = u + at \] where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. This equation was crucial for finding out how long it takes the projectile to reach the highest point. Initial velocities are split into components that describe motion in different directions - horizontal and vertical.

These components, combined with time factors like the flight time, provide a clear path to solving trajectory problems. They help in predicting how far the projectile will travel, given no additional forces except gravity or lateral wind.

Vector Components

Projectile motion inherently involves vectors - quantities having both magnitude and direction. When a projectile is launched, its velocity in a given direction is separated into horizontal and vertical components. These vector components are derived using trigonometric functions.

For the problem, we determine:

• Horizontal Component: \( v_x = 50 \cos(30^\circ) = 25\sqrt{3} \text{ m/s} \)
• Vertical Component: \( v_y = 50 \sin(30^\circ) = 25 \text{ m/s} \)

The horizontal component remains constant if we ignore air resistance, whereas the vertical component changes due to gravity. Understanding these distinct motions helps in addressing how external influences like added forces impact motion, such as the vertical force applied in this exercise.

Trajectory Analysis

Trajectory Analysis entails understanding the complete path a projectile takes. The trajectory is influenced by initial speed, angle of projection, and external forces like gravity and other applied forces.

Initially, the projectile is launched at an angle, causing a parabolic path. However, when the vertical force acts upon the projectile, it impacts how long the projectile stays in motion. During this problem, initially, the flight reaches a peak, calculated using the change in vertical motion from kinematics, and then the upward force extends the time further.

To find out the horizontal range of the projectile, we multiply the horizontal velocity component by the total time of flight. Hence the formula: \[ R = v_x \times T \] This gives the total distance covered horizontally. In the exercise, however, it's important to re-evaluate calculations if provided choices do not align, ensuring accuracy in all factors contributing to flight dynamics.