Question

10,000 small balls, each weighing \(1 \mathrm{~g}\) strike one square \(\mathrm{cm}\) of area per second with a velocity \(100 \mathrm{~ms}^{-1}\) in a normal direction and rebound with the same velocity. The value of pressure on the surface will be (A) \(2 \times 10^{3} \mathrm{Nm}^{-2}\) (B) \(2 \times 10^{5} \mathrm{Nm}^{-2}\) (C) \(10^{7} \mathrm{Nm}^{-2}\) (D) \(2 \times 10^{7} \mathrm{Nm}^{2}\)

Short answer

The value of pressure on the surface is \(2 \times 10^5 \mathrm{Nm}^{-2}\) (Option B).

Step-by-step solution

Find the force exerted by a single ball

To find the force exerted by a single ball, we can use Newton's second law (F = ma). First, find the acceleration of the ball as it rebounds. The acceleration can be found using the formula: \(a = \frac{v^2 - u^2}{2s}\) where v is the final velocity, u is the initial velocity, and s is the distance. Since the balls are rebounding with the same velocity, the initial velocity is -100 m/s, and the final velocity is 100 m/s. \(a = \frac{(100 \mathrm{~ms}^{-1})^2 - (-100 \mathrm{~ms}^{-1})^2}{2 \times 0.01 \mathrm{~m}}\) \(a = \frac{(10000 - 10000) \mathrm{m^2s^{-2}}}{0.02 \mathrm{~m}}\) \(a = 10000 \times 10^3 \mathrm{ms}^{-2}\) Now, find the force exerted by a single ball using F = ma: \(F = m \times a\) \(F = (1 \times 10^{-3} \mathrm{~kg}) \times (10000 \times 10^3 \mathrm{ms}^{-2})\) \(F = 10 \mathrm{N}\)

Find the total force exerted on the surface

Since there are 10000 balls hitting the surface per second, we can find the total force by multiplying the force exerted by a single ball by the number of balls: Total Force = Force exerted by a single ball × number of balls Total Force = 10 N × 10000 Total Force = 100000 N

Find the pressure exerted on the surface

To find the pressure exerted on the surface, we can use the formula for pressure: Pressure = \(\frac{\text{Total Force}}{\text{Area}}\) Pressure = \(\frac{100000 \mathrm{N}}{0.0001 \mathrm{m^2}}\) Pressure = \(2 \times 10^5 \mathrm{Nm}^{-2}\)

Match the result with the given options

Comparing the calculated pressure with the provided options, we find that our answer matches option (B): Pressure = \(2 \times 10^5 \mathrm{Nm}^{-2}\)

Key concepts

Newton's Second Law

Newton's Second Law of Motion is one of the fundamental principles for understanding the behavior of objects. It states that the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. This can be expressed by the equation: \( F = ma \)- **\( F \)** is the force applied to the object (in Newtons).- **\( m \)** represents the mass of the object (in kilograms).- **\( a \)** is the acceleration (in meters per second squared).In the context of the exercise, the principle is applied to determine the force exerted by a single ball as it rebounds off a surface. Because both the mass of the ball and the change in velocity are provided, we can calculate the force using this principle.

Kinematics Equations

Kinematics equations are tools used to describe motion in terms of displacement, velocity, acceleration, and time without considering the causes of motion. They help solve problems involving uniformly accelerated motion.Considered in the exercise is:\[a = \frac{v^2 - u^2}{2s}\]- **\( a \)** is acceleration.- **\( v \)** is the final velocity.- **\( u \)** is the initial velocity.- **\( s \)** is the distance traveled.In the given problem, the ball rebounds with the same velocity but in the opposite direction, making the initial velocity -100 m/s, while the final is 100 m/s. This accelerates the ball as it changes direction, which is crucial for calculating the force using the principle discussed in Newton's Second Law.

Momentum Change

The concept of momentum change is also central to understanding the problem. Momentum, defined as the product of mass and velocity, describes the motion of an object. The change in momentum is particularly important when an object undergoes a change in velocity.In mathematical terms, momentum (p) can be expressed as:\( p = mv \)- **m** is mass.- **v** is velocity.When a ball hits a surface and rebounds, its momentum changes because its velocity changes direction. The net change in momentum aids in calculating the force exerted on the surface, as total force exerted involves the rate of momentum change. For this exercise, the force and subsequent pressure calculations rely on understanding momentum's response to the ball’s change in velocity.

Problem Solving in Physics

Problem solving in physics often involves breaking down a complex problem into smaller, more manageable parts. This requires a clear understanding of the principles involved and the ability to apply them. This exercise highlights a structured approach:

• Identify the applicable physical principles (like Newton's Second Law).
• Use relevant equations (like those from kinematics) to find necessary values.
• Step through calculations methodically, finding force, and finally pressure.
• Relate computed results to given options as part of verification.

Such a methodological approach is vital in physics problem solving. It not only facilitates calculating solutions accurately but builds a comprehensive understanding of the physics at play.