/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 589 A player caught a cricket ball o... [FREE SOLUTION] | 91Ó°ÊÓ

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A player caught a cricket ball of mass \(150 \mathrm{gm}\) moving at a rate of \(20 \mathrm{~m} / \mathrm{s}\) If the catching process is Completed in \(0.1\) sec the force of the flow exerted by the ball on the hand of the player ..... N \(\\{\mathrm{A}\\} 3\) \(\\{B\\} 30\) \(\\{\mathrm{C}\\} 150\) \(\\{\mathrm{D}\\} 300\)

Short Answer

Expert verified
The force exerted by the ball on the hand of the player during the catching process is \(30 \mathrm{N}\). Therefore, the correct answer is B.

Step by step solution

01

Convert the mass to kilograms

The mass of the ball is given in grams. We need to first convert it to kilograms: \(150 \mathrm{gm} = 150 \times 10^{-3} \mathrm{kg}\).
02

Calculate the initial momentum of the ball

The initial momentum of the ball can be calculated using the formula: Momentum = mass × initial velocity Initial momentum = \( 150 \times 10^{-3} \mathrm{kg} \times 20 \mathrm{~m} / \mathrm{s} = 3 \mathrm{kg ~m} / \mathrm{s}\).
03

Calculate the final momentum of the ball

Since the player catches the ball and stops it, the final velocity of the ball is 0. Therefore, the final momentum of the ball will be: Final momentum = mass × final velocity Final momentum = \( 150 \times 10^{-3} \mathrm{kg} \times 0 \mathrm{~m} / \mathrm{s} = 0 \mathrm{kg ~m} / \mathrm{s}\).
04

Calculate the change in momentum

To calculate the change in momentum, we'll subtract the final momentum from the initial momentum: Change in momentum = Initial momentum - Final momentum Change in momentum = \(3 \mathrm{kg ~m} / \mathrm{s} - 0 \mathrm{kg ~m} / \mathrm{s} = 3 \mathrm{kg ~m} / \mathrm{s}\).
05

Calculate the force exerted by the ball on the hand of the player

The force required to change the momentum of the ball can be calculated using the formula: Force = change in momentum / Time taken for the change in momentum Force = \( \frac{(3 \mathrm{kg ~m} / \mathrm{s})}{0.1 \mathrm{s}} = 30 \mathrm{N} \). The force exerted by the ball on the hand of the player during the catching process is \(30 \mathrm{N}\). Therefore, the correct answer is B.

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics that describes the motion of an object. It is defined as the product of an object's mass and its velocity. Key properties of momentum include:
  • Vector Quantity: Momentum has both magnitude and direction, making it a vector quantity. It points in the same direction as the object's velocity.

  • Conservation: In an isolated system, without external forces, the total momentum remains constant.

In this exercise, the initial momentum of the cricket ball was calculated using the formula:\[ \text{Momentum} = \text{mass} \times \text{velocity} \]Given that the mass was converted to kilograms, the initial velocity was used to find a momentum of 3 kg m/s. This highlights how mass and speed contribute to how momentum describes the ball's current movement. Once the ball is caught and brought to a halt, no momentum remains, which is a crucial factor when analyzing the force exerted in the catching process.
Force Calculation
Calculating force can reveal the interaction between objects, as described in Newton's Second Law of Motion. Force is defined as the rate of change of momentum. To find the force exerted by the ball when caught, the change in momentum and the time over which this change occurs must be known. The formula used to calculate force is:\[ \text{Force} = \frac{\text{Change in Momentum}}{\text{Time}} \]Here, the change in momentum was 3 kg m/s (from initial momentum of 3 kg m/s to final momentum of 0 kg m/s), and the process took 0.1 seconds. Plugging these values into the formula gives the force as 30 N. This calculation shows that the force exerted is directly proportional to the rate at which momentum changes, or how quickly the action occurs.
Impulse
Impulse is closely related to momentum and describes the effect of a force over time. Essentially, impulse quantifies the change in momentum resulting from a force application over a specific duration.Impulse can be calculated by:\[ \text{Impulse} = \text{Force} \times \text{Time} = \text{Change in Momentum} \]In the exercise, when the player catches the ball, the impulse provided equals the change in momentum (3 kg m/s) experienced during the catching period of 0.1 seconds. This effectively means that the same impulse could have been calculated directly using the change in momentum, demonstrating the interchangeability and close connection of these concepts in momentum-related problems.
Units Conversion
Units conversion is an essential step in solving physics problems to ensure consistency across calculations. Here, the mass of the cricket ball was initially given in grams and needed to be converted into kilograms to match the SI unit system commonly used in physics equations.To convert grams to kilograms, divide the mass value by 1000, since 1000 grams equals 1 kilogram:\[ 150 \text{ gm} = 150 \times 10^{-3} \text{ kg} \]This conversion ensures that calculations like momentum and force, which use the standard units of kg, m/s, and N (newton), are accurate. Proper conversion is crucial in physics to avoid incorrect results and maintain the accuracy of derived quantities across all formulas.

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Most popular questions from this chapter

Match list I with list II and select the correct answer $$ \begin{aligned} &\begin{array}{|l|l|} \hline \text { List-I } & \begin{array}{l} \text { List - II } \\ \text { System } \end{array} & \text { Moment of inertia } \\ \hline \text { (x) A ring about it axis } & \text { (1) }\left(\mathrm{MR}^{2} / 2\right) \\ \hline \text { (y) A uniform circular disc about it axis } & \text { (2) }(2 / 5) \mathrm{MR}^{2} \\ \hline \text { (z) A solid sphere about any diameter } & \text { (3) }(7 / 5) \mathrm{MR}^{2} \\ \hline \text { (w) A solid sphere about any tangent } & \text { (4) } \mathrm{MR}^{2} \\ \cline { 2 } & \text { (5) }(9 / 5) \mathrm{MR}^{2} \\ \hline \end{array}\\\ &\text { Select correct option }\\\ &\begin{array}{|l|l|l|l|l|} \hline \text { Option? } & \mathrm{X} & \mathrm{Y} & \mathrm{Z} & \mathrm{W} \\\ \hline\\{\mathrm{A}\\} & 2 & 1 & 3 & 4 \\ \hline\\{\mathrm{B}\\} & 4 & 3 & 2 & 5 \\ \hline\\{\mathrm{C}\\} & 1 & 5 & 4 & 3 \\ \hline\\{\mathrm{D}\\} & 4 & 1 & 2 & 3 \\ \hline \end{array} \end{aligned} $$

A body of mass \(\mathrm{m}\) is tied to one end of spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one centimetre. If the angular velocity is doubted, the elongation in the spring is \(5 \mathrm{~cm}\). The original length of spring is... \(\\{\mathrm{A}\\} 16 \mathrm{~cm}\) \(\\{B\\} 15 \mathrm{~cm}\) \(\\{\mathrm{C}\\} 14 \mathrm{~cm}\) \(\\{\mathrm{D}\\} 13 \mathrm{~cm}\)

The moment of inertia of a meter scale of mass \(0.6 \mathrm{~kg}\) about an axis perpendicular to the scale and passing through \(30 \mathrm{~cm}\) position on the scale is given by (Breath of scale is negligible). \(\\{\mathrm{A}\\} 0.104 \mathrm{kgm}^{2}\) \\{B \(\\} 0.208 \mathrm{kgm}^{2}\) \(\\{\mathrm{C}\\} 0.074 \mathrm{kgm}^{2}\) \(\\{\mathrm{D}\\} 0.148 \mathrm{kgm}^{2}\)

A car is moving at a speed of \(72 \mathrm{~km} / \mathrm{hr}\) the radius of its wheel is \(0.25 \mathrm{~m}\). If the wheels are stopped in 20 rotations after applying breaks then angular retardation produced by the breaks is \(\ldots .\) \(\\{\mathrm{A}\\}-25.5 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{C}\\}-33.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{D}\\}-45.52 \mathrm{rad} / \mathrm{s}^{2}\)

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