
A bullet of mass 10g is fired from a rifle with a velocity of 800 m/s. After passing through a mud wall 1 m thick, the velocity drops to 100m/s. The average resistance of the wall is
A. 750 N
B. 1250 N
C. -3150 N
D. 2250 N
Answer
572.7k+ views
Hint: Try using one of the three equations of motion. You can choose the correct equation for this question by looking at the given parameters. Also, the resistance we are talking about in this question is the force applied by the wall which can easily be calculated by using F = ma
Formula used: For solving the given question, we will be using the following formulae:
\[{{v}^{2}}-{{u}^{2}}=2as\]
F = ma
Complete step-by-step solution:
Now, before we start solving the question, let us take a look at the given parameters
m = 10 gm, where m is the mass of the bullet
u = 800 m/s, where u is the initial velocity of the bullet
v = 100 m/s, where v is the final velocity of the bullet
s = 1 m, where s is the distance covered by the bullet while penetrating the mud wall, i.e., the thickness of the mud wall
Now, as we discussed in the ‘formula used’ part,
By the third equation of motion.
\[{{v}^{2}}-{{u}^{2}}=2as\]
Now, using the given values in this formula
We have,
\[{{100}^{2}}-{{800}^{2}}=2a(1)\]
Where a is the acceleration of the bullet after hitting the wall
$10000-640000=2a$
$\Rightarrow-630000=2a$
$\Rightarrow a=-315000m/{{s}^{2}}$
Now, using the formula F = ma
We have
$F=-315000\times 10\times {{10}^{-3}}$ (Since 10 gm = $10\times {{10}^{-3}}kg$)
F = -3150 N
So, The average resistance of the wall is -3150 N, i.e., Option C
Note: In the real-life cases, we can not calculate the correct answer with this method as there will be some air resistance and energy loss by the friction due to the wall on bullet in real scenarios.
Formula used: For solving the given question, we will be using the following formulae:
\[{{v}^{2}}-{{u}^{2}}=2as\]
F = ma
Complete step-by-step solution:
Now, before we start solving the question, let us take a look at the given parameters
m = 10 gm, where m is the mass of the bullet
u = 800 m/s, where u is the initial velocity of the bullet
v = 100 m/s, where v is the final velocity of the bullet
s = 1 m, where s is the distance covered by the bullet while penetrating the mud wall, i.e., the thickness of the mud wall
Now, as we discussed in the ‘formula used’ part,
By the third equation of motion.
\[{{v}^{2}}-{{u}^{2}}=2as\]
Now, using the given values in this formula
We have,
\[{{100}^{2}}-{{800}^{2}}=2a(1)\]
Where a is the acceleration of the bullet after hitting the wall
$10000-640000=2a$
$\Rightarrow-630000=2a$
$\Rightarrow a=-315000m/{{s}^{2}}$
Now, using the formula F = ma
We have
$F=-315000\times 10\times {{10}^{-3}}$ (Since 10 gm = $10\times {{10}^{-3}}kg$)
F = -3150 N
So, The average resistance of the wall is -3150 N, i.e., Option C
Note: In the real-life cases, we can not calculate the correct answer with this method as there will be some air resistance and energy loss by the friction due to the wall on bullet in real scenarios.
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