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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?
Solution :
Given Variables & Initial Setup
To determine the force exerted by the nail on the hammer, we utilize Newton's Second Law of Motion. First, we identify the physical quantities provided in the SI system:
- Mass of the hammer ($m$) = $500 \text{ g} = 0.5 \text{ kg}$ [Conversion factor: $1 \text{ kg} = 1000 \text{ g}$]
- Initial velocity of the hammer ($u$) = $50 \text{ m s}^{-1}$
- Final velocity of the hammer ($v$) = $0 \text{ m s}^{-1}$ [Since the nail stops the hammer]
- Time taken to stop ($t$) = $0.01 \text{ s}$
Step 1: Calculating the Change in Momentum
Momentum ($p$) is defined as the product of mass and velocity. We calculate the initial momentum ($p_i$) and final momentum ($p_f$):
$p_i = m \times u = 0.5 \text{ kg} \times 50 \text{ m s}^{-1} = 25 \text{ kg m s}^{-1}$
$p_f = m \times v = 0.5 \text{ kg} \times 0 \text{ m s}^{-1} = 0 \text{ kg m s}^{-1}$
Change in momentum ($\Delta p$) = $p_f - p_i = 0 - 25 = -25 \text{ kg m s}^{-1}$
Step 2: Applying Newton's Second Law
Per Newton’s Second Law, the force ($F$) exerted is equal to the rate of change of momentum:
$F = \frac{\Delta p}{t}$ [By definition of Force, $F = ma = m \frac{(v - u)}{t}$]
$F = \frac{-25 \text{ kg m s}^{-1}}{0.01 \text{ s}}$
$F = -2500 \text{ N}$
Step 3: Interpretation of Results
The negative sign indicates that the direction of the force is opposite to the initial direction of the hammer's motion. The magnitude of the force exerted by the nail on the hammer is $2500 \text{ N}$.
Final Solution: The force exerted by the nail on the hammer is 2500 N in the direction opposite to the motion of the hammer.
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.)
- 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.
CBSE Solutions for Class 9 Science Force and Laws of Motion
Chapters in CBSE - Class 9 Science
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