(A) Given: Mass of the bullet $= m$, initial velocity $= v_1$, final velocity $= v_2$, and distance traveled $= L$.Since the bullet experiences a resi

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$F = \frac{m}{2L}(v_1^2 - v_2^2)$

(A) Given: Mass of the bullet $= m$, initial velocity $= v_1$, final velocity $= v_2$, and distance traveled $= L$.Since the bullet experiences a resistive force $F$, it undergoes uniform deceleration $a = F/m$.Using the third equation of motion, $v^2 - u^2 = 2as$, where $v = v_2$, $u = v_1$, $a = -F/m$, and $s = L$:$v_2^2 - v_1^2 = 2(-F/m)L$$v_2^2 - v_1^2 = -\frac{2FL}{m}$$F = \frac{m(v_1^2 - v_2^2)}{2L}$Since $v_1 > v_2$, the force $F$ is positive.

Class 11 Physics Work, Energy, Power and Collision Work Energy Theorem and Conservation of Mechanical Energy

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$A$ bullet of mass $m$ enters a wooden block of length $L$ at a speed $v_1$ and emerges out of the block with a speed $v_2$. If $F$ is the average force which impeded its motion through the wooden block, then (Assume uniform deceleration inside the block):

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A
$F = \frac{m}{2L}(v_1^2 - v_2^2)$
B
$F = \frac{m}{2L}(v_2^2 - v_1^2)$
C
$F = \frac{m}{2L}(v_1^2 + v_2^2)$
D
$F = \frac{m}{L}(v_1^2 - v_2^2)$

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Work, Energy, Power and Collision

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Work Energy Theorem and Conservation of Mechanical Energy

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$A$ bullet of mass $m$ enters a wooden block of length $L$ at a speed $v_1$ and emerges out of the block with a speed $v_2$. If $F$ is the average force which impeded its motion through the wooden block, then (Assume uniform deceleration inside the block):

Similar Questions

$A$ body of mass $2 \, kg$ is projected vertically upwards with a kinetic energy of $490 \, J$. At what height will its kinetic energy become half of its initial value? $(g = 9.8 \, m/s^2)$

$A$ bullet fired from a gun into a large wooden block loses half of its velocity after traveling $6 \ cm$ inside the block. How much further distance (in $cm$) will it travel before coming to rest?

$A$ stone with weight $w$ is thrown vertically upward into the air from ground level with initial speed $v_0$. If a constant force $f$ due to air drag acts on the stone throughout its flight,the maximum height attained by the stone is

$A$ ball of mass $1 \ kg$ moves in a straight line with velocity $v = c x^\alpha$, where $c = 1$ ($SI$ unit) and $\alpha$ is a constant. If the work done by the net force during its displacement from $x = 0$ to $x = 4 \ m$ is $128 \ J$, then the value of $\alpha$ is

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