Answer
77.7k+ views
Hint: This problem can be solved by first finding out the negative acceleration (or deceleration) of the bullet by using the given information and applying the equations of uniform motion. Then, we will use Newton’s second law of motion which says that the force on a body is the product of the mass and the acceleration produced by the force, to find out the average resistive force.
Formula used:
$\dfrac{{{v}^{2}}-{{u}^{2}}}{2s}=a$
Where v, u, s, and a are the final and initial velocity, displacement, and acceleration of the body respectively.
$F=ma$
Where F, m and a are the force on the body, its mass and the acceleration of the body.
Complete step by step answer:
First, we will identify the information given to us in the question and understand it.
Given, the initial velocity of the bullet (u) = $90 ms^{-1}$.
Final velocity of the bullet (v) = 0 ms-1 since it comes to a stop
Displacement of the bullet inside the block (s) = 60 cm or 0.6m $\left( \because 100cm=1m \right)$
Mass of the body (m) = 0.04 kg
Now, to find out the acceleration of the bullet due to the block, we will use the equation of motion
$\dfrac{{{v}^{2}}-{{u}^{2}}}{2s}=a$
$\therefore a=\dfrac{{{0}^{2}}-{{90}^{2}}}{2\times 0.06}=-\dfrac{8100}{0.12}=-67500m{{s}^{-2}}$ --(1)
The acceleration is negative since the bullet is decelerating due to the retarding or resistive force of the block.
Now, using Newton’s second law of motion which says that the force on a body is the product of the mass and the acceleration produced by the force, we will find out the force applied by the block on the bullet. Thus,
$F=ma$
$=0.04\times \left( -67500 \right)=-2700N$
The force is negative since it is a resistive force (it tries to decelerate the bullet opposing its motion).
Thus, the magnitude of the resistive force exerted by the block on the bullet is 2700N.
Note: Students must not get confused by seeing the negative signs of acceleration and force. It only means that the body is decelerating due to the resistive or retarding force applied to it. It is a good practice to explain the reason behind a force or acceleration being negative, in examinations. In fact, forces should always be written with correct signs in front of them to let the reader understand that the force is a resistive one.
Formula used:
$\dfrac{{{v}^{2}}-{{u}^{2}}}{2s}=a$
Where v, u, s, and a are the final and initial velocity, displacement, and acceleration of the body respectively.
$F=ma$
Where F, m and a are the force on the body, its mass and the acceleration of the body.
Complete step by step answer:
First, we will identify the information given to us in the question and understand it.
Given, the initial velocity of the bullet (u) = $90 ms^{-1}$.
Final velocity of the bullet (v) = 0 ms-1 since it comes to a stop
Displacement of the bullet inside the block (s) = 60 cm or 0.6m $\left( \because 100cm=1m \right)$
Mass of the body (m) = 0.04 kg
Now, to find out the acceleration of the bullet due to the block, we will use the equation of motion
$\dfrac{{{v}^{2}}-{{u}^{2}}}{2s}=a$
$\therefore a=\dfrac{{{0}^{2}}-{{90}^{2}}}{2\times 0.06}=-\dfrac{8100}{0.12}=-67500m{{s}^{-2}}$ --(1)
The acceleration is negative since the bullet is decelerating due to the retarding or resistive force of the block.
Now, using Newton’s second law of motion which says that the force on a body is the product of the mass and the acceleration produced by the force, we will find out the force applied by the block on the bullet. Thus,
$F=ma$
$=0.04\times \left( -67500 \right)=-2700N$
The force is negative since it is a resistive force (it tries to decelerate the bullet opposing its motion).
Thus, the magnitude of the resistive force exerted by the block on the bullet is 2700N.
Note: Students must not get confused by seeing the negative signs of acceleration and force. It only means that the body is decelerating due to the resistive or retarding force applied to it. It is a good practice to explain the reason behind a force or acceleration being negative, in examinations. In fact, forces should always be written with correct signs in front of them to let the reader understand that the force is a resistive one.
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