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A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest, assuming that it faces constant resistance to motion ?
[AIEEE 2005]
- a)3.0 cm
- b)Add Question2.0 cm
- c)1.5 cm
- d)1.0 cm
Correct answer is option 'D'. Can you explain this answer?
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To solve this problem, let's use the concept of linear motion with constant acceleration.
Let's assume the initial velocity of the bullet is v, and the final velocity (when it comes to rest) is 0. The distance traveled by the bullet before it comes to rest is the sum of the distances traveled while decelerating from the initial velocity to half of the initial velocity (v/2) and while decelerating from v/2 to 0.
We are given that the bullet loses half of its velocity (v/2) after penetrating 3 cm. Let's assume the time taken to decelerate from v to v/2 is t1, and the time taken to decelerate from v/2 to 0 is t2.
Using the equation of motion, v = u + at, where u is the initial velocity, a is the acceleration, and t is the time, we can write the following equations:
v/2 = v - at1 (Equation 1)
0 = v/2 - at2 (Equation 2)
From Equation 1, we can solve for t1:
t1 = v/2a (Equation 3)
From Equation 2, we can solve for t2:
t2 = v/2a (Equation 4)
Since the distance traveled is given by the equation s = ut + (1/2)at^2, we can calculate the distance traveled during the first deceleration:
s1 = v/2 * t1 + (1/2) * a * t1^2
Substituting the values from Equations 3 and 4, we get:
s1 = v/2 * (v/2a) + (1/2) * a * (v/2a)^2
s1 = v^2 / 4a + v^2 / 8a
s1 = 3v^2 / 8a
Similarly, we can calculate the distance traveled during the second deceleration:
s2 = v/2 * t2
Substituting the value from Equation 4, we get:
s2 = v/2 * (v/2a)
s2 = v^2 / 4a
The total distance traveled before coming to rest is the sum of s1 and s2:
s_total = s1 + s2
s_total = 3v^2 / 8a + v^2 / 4a
s_total = 3v^2 + 2v^2 / 8a
s_total = 5v^2 / 8a
Since we are given that the bullet faces constant resistance to motion, the acceleration (a) is constant. Therefore, the distance traveled before coming to rest (s_total) is directly proportional to the square of the initial velocity (v^2).
Since the initial velocity is halved after penetrating 3 cm, the distance traveled will be halved as well. Therefore, the bullet will penetrate an additional 1 cm before coming to rest.
Hence, the correct answer is option D) 1.0 cm.
Let's assume the initial velocity of the bullet is v, and the final velocity (when it comes to rest) is 0. The distance traveled by the bullet before it comes to rest is the sum of the distances traveled while decelerating from the initial velocity to half of the initial velocity (v/2) and while decelerating from v/2 to 0.
We are given that the bullet loses half of its velocity (v/2) after penetrating 3 cm. Let's assume the time taken to decelerate from v to v/2 is t1, and the time taken to decelerate from v/2 to 0 is t2.
Using the equation of motion, v = u + at, where u is the initial velocity, a is the acceleration, and t is the time, we can write the following equations:
v/2 = v - at1 (Equation 1)
0 = v/2 - at2 (Equation 2)
From Equation 1, we can solve for t1:
t1 = v/2a (Equation 3)
From Equation 2, we can solve for t2:
t2 = v/2a (Equation 4)
Since the distance traveled is given by the equation s = ut + (1/2)at^2, we can calculate the distance traveled during the first deceleration:
s1 = v/2 * t1 + (1/2) * a * t1^2
Substituting the values from Equations 3 and 4, we get:
s1 = v/2 * (v/2a) + (1/2) * a * (v/2a)^2
s1 = v^2 / 4a + v^2 / 8a
s1 = 3v^2 / 8a
Similarly, we can calculate the distance traveled during the second deceleration:
s2 = v/2 * t2
Substituting the value from Equation 4, we get:
s2 = v/2 * (v/2a)
s2 = v^2 / 4a
The total distance traveled before coming to rest is the sum of s1 and s2:
s_total = s1 + s2
s_total = 3v^2 / 8a + v^2 / 4a
s_total = 3v^2 + 2v^2 / 8a
s_total = 5v^2 / 8a
Since we are given that the bullet faces constant resistance to motion, the acceleration (a) is constant. Therefore, the distance traveled before coming to rest (s_total) is directly proportional to the square of the initial velocity (v^2).
Since the initial velocity is halved after penetrating 3 cm, the distance traveled will be halved as well. Therefore, the bullet will penetrate an additional 1 cm before coming to rest.
Hence, the correct answer is option D) 1.0 cm.
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