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QA4:
A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Solution :
Step 1: Given Variables & Unit Conversions
First, we identify the physical quantities provided and convert them into SI units (kilograms for mass, meters per second for velocity, and seconds for time).
- Mass ($m$) = $1200\text{ kg}$
- Initial velocity ($u$) = $90\text{ km/h} = 90 \times \frac{5}{18}\text{ m/s} = 25\text{ m/s}$
- Final velocity ($v$) = $18\text{ km/h} = 18 \times \frac{5}{18}\text{ m/s} = 5\text{ m/s}$
- Time interval ($t$) = $4\text{ s}$
Step 2: Calculating Acceleration
Acceleration ($a$) is defined as the rate of change of velocity [Per Newton’s First Law of Motion and Kinematic Definitions].
$a = \frac{v - u}{t}$
$a = \frac{5\text{ m/s} - 25\text{ m/s}}{4\text{ s}}$
$a = \frac{-20\text{ m/s}}{4\text{ s}} = -5\text{ m/s}^2$
[The negative sign indicates retardation or deceleration].
Step 3: Calculating Change in Momentum
Momentum ($p$) is the product of mass and velocity. The change in momentum ($\Delta p$) is given by $m(v - u)$ [Per Newton’s Second Law of Motion].
$\Delta p = m \times v - m \times u$
$\Delta p = 1200\text{ kg} \times (5\text{ m/s} - 25\text{ m/s})$
$\Delta p = 1200\text{ kg} \times (-20\text{ m/s})$
$\Delta p = -24000\text{ kg}\cdot\text{m/s}$
Step 4: Calculating Magnitude of the Force
The magnitude of the force ($F$) required is calculated using the formula $F = ma$ [Per Newton’s Second Law of Motion: $F = \frac{\Delta p}{\Delta t}$].
$F = 1200\text{ kg} \times (-5\text{ m/s}^2)$
$F = -6000\text{ N}$
The magnitude of the force is $6000\text{ N}$ (acting in the direction opposite to the motion).
Visual Representation of Velocity Change
Final Solution: The acceleration is $-5\text{ m/s}^2$, the change in momentum is $-24000\text{ kg}\cdot\text{m/s}$, and the magnitude of the force required is $6000\text{ N}$.
More Questions from Class 9 Science Force and Laws of Motion Additional Exercises
- QA1(a): The following is the distance-time table of an object in motion: (a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
- QA1(b): The following is the distance-time table of an object in motion: (b) What do you infer about the forces acting on the object?
- QA2: Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 $m s^{-2}$. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
- QA3: A hammer of mass 500 g, moving at 50 $m s^{-1}$, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
CBSE Solutions for Class 9 Science Force and Laws of Motion
Chapters in CBSE - Class 9 Science
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