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A cricket ball of mass \(150 \mathrm{~g}\). is moving with a velocity of \(12 \mathrm{~m} / \mathrm{s}\) and is hit by a bat so that the ball is turned back with a velocity of \(20 \mathrm{~m} / \mathrm{s}\). If duration of contact between the ball and the bat is \(0.01 \mathrm{sec}\). The impulse of the force is (A) \(7.4 \mathrm{NS}\) (B) \(4.8 \mathrm{NS}\) (C) \(1.2 \mathrm{NS}\) (D) \(4.7 \mathrm{NS}\)

Short Answer

Expert verified
The impulse of the force is (B) \(4.8 \mathrm{NS}\).

Step by step solution

01

Convert mass to kg

The mass of the cricket ball is given in grams (150g). We need to convert this to kilograms (kg) to work in the SI unit system. To convert grams to kilograms, divide by 1000. \(Mass (kg) = \frac{150}{1000} = 0.15 \ kg\)
02

Calculate initial and final momentum

Now, we can calculate the initial and final momentum of the cricket ball: \(Initial \ momentum = mass \times initial \ velocity\) \(Final \ momentum = mass \times final \ velocity\) Plug in the values for mass and velocities: \(Initial \ momentum = 0.15 \ kg \times 12 \ m/s = 1.8 \ kg \cdot m/s\) \(Final \ momentum = 0.15 \ kg \times -20 \ m/s = -3 \ kg \cdot m/s\) Notice that we used a negative sign for the final velocity since the ball is turned back.
03

Calculate the impulse

Now, we can calculate the impulse using the impulse-momentum theorem: \(Impulse = Final \ momentum - Initial \ momentum\) Plug in the values for initial and final momentum: \(Impulse = -3 \ kg \cdot m/s - 1.8 \ kg \cdot m/s = -4.8 \ kg \cdot m/s\)
04

Convert impulse to Ns (Newton-seconds)

Since impulse is the product of force and time, the units of impulse are Newton-seconds (Ns). In this case, our impulse is already in Ns, since we have kg·m/s (which is equivalent to Ns). Impulse = -4.8 Ns The negative sign means that the direction of the impulse is opposite to the initial direction of the ball.
05

Choose the correct answer

The correct answer is: (B) \(4.8 \mathrm{NS}\) Note that the answer is given as a positive value, which is referring to the magnitude of the impulse.

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Key Concepts

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

Understanding Momentum
Momentum is a fundamental concept in physics that describes the quantity of motion an object possesses. It is determined by two key factors: an object's mass and its velocity. The formula for momentum is given by the product of mass and velocity:
  • \( \text{Momentum} = \text{mass} \times \text{velocity} \)
In our cricket ball example, the mass is 0.15 kg, and we calculate momentum by multiplying the mass with its velocity. The momentum has both magnitude and direction, making it a vector quantity. This means changing either the mass, velocity, or both can affect momentum, significantly impacting motion during events like a cricket ball being hit by a bat.
Momentum is crucial when analyzing systems where forces are exchanged, as it helps to predict the motion after events such as collisions or other interactions.
Newton's Third Law
Newton's Third Law of Motion provides insight into the interaction of forces during events. It states: "For every action, there is an equal and opposite reaction." In simpler terms, when one body exerts a force on another, the second body exerts an equivalent force in the opposite direction.
When a cricket ball is hit by a bat, the bat applies force to the ball, changing its momentum. At the same time, the ball applies an opposite force back onto the bat. This interaction can be understood through Newton's Third Law, explaining why both parties (bat and ball) experience a force even if we are primarily concerned with the ball's motion in this case.
  • Action: Force applied by the bat on the ball
  • Reaction: Equal force exerted by the ball back on the bat
SI Units in Physics
Physics primarily uses the International System of Units (SI Units) for consistency and standardization worldwide. The mass of objects is measured in kilograms (kg), velocity in meters per second (m/s), and force in newtons (N).
Converting to SI units is essential, as it allows for straightforward calculations and comparisons across various physics problems. In the exercise given, the mass of the ball was converted from grams to kilograms to fit the SI measurement system, ensuring clarity and accuracy during calculations and their application.
  • Mass: grams (g) to kilograms (kg)
  • Velocity: meters per second (m/s)
  • Impulse and force: Newton-seconds (Ns)
Velocity Change
Velocity change is significant when analyzing motion and impact, as a change in velocity indicates an alteration in momentum. In the case of the cricket ball, it initially moves in one direction and, after being hit by the bat, changes direction with a different speed.
The change in velocity reflects the impulse transferred to the ball due to the force from the bat. This is calculated by finding the difference in initial and final velocities:
  • Initial velocity: 12 m/s
  • Final velocity: -20 m/s (indicating a reversal in direction)
  • Velocity change magnitude: \( \Delta v = v_{final} - v_{initial} = -20 \, \text{m/s} - 12 \, \text{m/s} \)
Evaluating the change in velocity is key to understanding the broader dynamics at play during interactions like the hit of the ball.
Force-Impulse Relation
The relationship between force, impulse, and momentum change is fundamental in mechanics. Impulse can be described as the effect of a force acting over a period of time. It is the change in momentum of an object and is related to force by the equation:
  • \( \text{Impulse} = \text{Force} \times \text{Time} \)
  • \( \text{Impulse} = \Delta \text{Momentum} \)
In the exercise, the impulse experienced by the cricket ball is calculated by subtracting initial momentum from final momentum:
  • Impulse: -4.8 Ns (indicating a change in direction)
Since force applied over time leads to a momentum change, understanding this relationship helps in predicting how objects move and interact in various scenarios, such as in sports or real-life applications.

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Most popular questions from this chapter

Match the column \begin{tabular}{l|l} Column - I & Column - II \end{tabular} (a) Body line on a (p) is a self adjusting horizontal surface \(\quad\) force (b) Static friction (q) is a maximum value of static friction (c) Limiting friction (r) is then limiting friction (d) Dynamic friction (s) force of friction \(=0\) (A) \(\mathrm{a}-\mathrm{s}, \mathrm{b}-\mathrm{p}, \mathrm{c}-\mathrm{q}, \mathrm{d}-\mathrm{r}\) (B) \(\mathrm{a}-\mathrm{p}, \mathrm{b}-\mathrm{q}, \mathrm{c}-\mathrm{r}, \mathrm{d}-\mathrm{s}\) (C) \(\mathrm{a}-\mathrm{s}, \mathrm{b}-\mathrm{r}, \mathrm{c}-\mathrm{q}, \mathrm{d}-\mathrm{p}\) (D) \(\mathrm{a}-\mathrm{r}, \mathrm{b}-\mathrm{q}, \mathrm{c}-\mathrm{p}, \mathrm{d}-\mathrm{s}\)

which of the following statement is correct? (A) A body has a constant velocity but a varying speed. (B) A body has a constant speed but a varying value of acceleration. (C) A body has a constant speed and zero acceleration. (D) A body has a constant speed but velocity is zero.

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