
A bullet of mass \[50g\] is moving with a velocity of \[500{\text{ }}m{s^{ - 1}}\]. It penetrated \[10{\text{ }}cm\] into a still target and came to rest. Calculate the average retarding force offered by the target.
A.\[0N\]
B.\[625N\]
C.\[62500N\]
D.\[6.25N\]
Answer
460.2k+ views
Hint: Retarding forces are defined as the forces that resist relative motion. We can also call it by the name resisting forces. In the question, we are asked to find the average retarding force that is offered by the target. We need to simply use the equation of motion and find the deceleration and then we need to put this in the formula for the force to find the average retarding force.’
Complete answer:
Given in the question that the mass of the bullet is \[m = 50g\]
This bullet moves with a velocity of \[v = 500{\text{ }}m{s^{ - 1}}\]
Given that it is penetrated through a target of distance \[10{\text{ }}cm\]. Therefore the displacement s is equal to \[10{\text{ }}cm\].
Now from the equation of motion, we know that,
\[ \Rightarrow {v^2} = {u^2} + 2as\] …… (1)
Here,
\[v\] is the final velocity
\[u\] is the initial velocity
\[a\] is said to be the acceleration with which the object is moving
\[s\]is the displacement.
From equation (1) we are going to find the acceleration with which the bullet is moving. We know that after passing through the target the bullet comes to rest therefore the final velocity becomes \[v = 0m{s^{ - 1}}\]. Now substituting all the known values in the equation (1) we get,
\[ \Rightarrow 0 = {500^2} + 2a(\dfrac{{10}}{{100}})\]
\[ \Rightarrow a = - {\text{1250000m/s}}\]
The negative sign indicates that the bullet is decreasing in speed.
Now we know the formula for the force from Newton’s second law of motion.
\[ \Rightarrow {{\text{F}}_{av}}{\text{ = ma}}\] ….. (2).
Here \[{{\text{F}}_{av}}\] is the average retarding force.
Now substituting all the known values in the above equation we get,
\[ \Rightarrow {{\text{F}}_{av}}{\text{ = }}\dfrac{{{\text{50}}}}{{1000}}({\text{1250000)}}\]
\[ \Rightarrow {{\text{F}}_{av}}{\text{ = }}62500N\]
Therefore, the average retarding force is found to be \[62500N\]
Hence the correct option is C.
Note:
The opposite of acceleration is deceleration. We can only use deceleration in a special case of acceleration and it only applies to the objects slowing down. In our case the bullet is coming to rest after it passes through the target thereby the bullet is slowing down and thus we used the term deceleration. For finding deceleration we simply use the formula for acceleration with a negative sign.
Complete answer:
Given in the question that the mass of the bullet is \[m = 50g\]
This bullet moves with a velocity of \[v = 500{\text{ }}m{s^{ - 1}}\]
Given that it is penetrated through a target of distance \[10{\text{ }}cm\]. Therefore the displacement s is equal to \[10{\text{ }}cm\].
Now from the equation of motion, we know that,
\[ \Rightarrow {v^2} = {u^2} + 2as\] …… (1)
Here,
\[v\] is the final velocity
\[u\] is the initial velocity
\[a\] is said to be the acceleration with which the object is moving
\[s\]is the displacement.
From equation (1) we are going to find the acceleration with which the bullet is moving. We know that after passing through the target the bullet comes to rest therefore the final velocity becomes \[v = 0m{s^{ - 1}}\]. Now substituting all the known values in the equation (1) we get,
\[ \Rightarrow 0 = {500^2} + 2a(\dfrac{{10}}{{100}})\]
\[ \Rightarrow a = - {\text{1250000m/s}}\]
The negative sign indicates that the bullet is decreasing in speed.
Now we know the formula for the force from Newton’s second law of motion.
\[ \Rightarrow {{\text{F}}_{av}}{\text{ = ma}}\] ….. (2).
Here \[{{\text{F}}_{av}}\] is the average retarding force.
Now substituting all the known values in the above equation we get,
\[ \Rightarrow {{\text{F}}_{av}}{\text{ = }}\dfrac{{{\text{50}}}}{{1000}}({\text{1250000)}}\]
\[ \Rightarrow {{\text{F}}_{av}}{\text{ = }}62500N\]
Therefore, the average retarding force is found to be \[62500N\]
Hence the correct option is C.
Note:
The opposite of acceleration is deceleration. We can only use deceleration in a special case of acceleration and it only applies to the objects slowing down. In our case the bullet is coming to rest after it passes through the target thereby the bullet is slowing down and thus we used the term deceleration. For finding deceleration we simply use the formula for acceleration with a negative sign.
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