
Assume that every projectile fired by the toy cannon experiences a constant net force F along the entire length of barrel. If a projectile of mass m leaves the barrel of the cannon with a speed v, at what speed will a projectile of mass $2m$ leave the barrel?
A. $\dfrac{v}{2}$
B. $\dfrac{v}{{\sqrt 2 }}$
C. $v$
D. $2v$
E. $4v$
Answer
515.1k+ views
Hint: n physics, projectile is the motion of a body when given a certain force and it follows a particular trajectory. Here, we will use the general formulas of force on a body which is defined as the product of mass of the body and its acceleration, mathematically written as $F = ma$ and we will use the newton’s equation of motion as ${v^2} - {u^2} = 2aS$ where, $v$ is the final velocity along the distance $S$ and $u$ is the acceleration of the body.
Complete step by step answer:
Let us suppose $F$ be the net force acting on both bodies of mass $m(and)2m$ , let their accelerations be ${a_m}(and){a_{2m}}$ respectively, so using
$F = ma$ We have,
$\Rightarrow F = m{a_m}$
And
$F = 2m{a_{2m}}$
Since both forces are equal so we write
$m{a_m} = 2m{a_{2m}}$
$\Rightarrow \dfrac{{{a_m}}}{{{a_{2m}}}} = 2 \to (i)$
Now for the length of $L$ mass $m$ leaves with velocity $v$ with an acceleration of ${a_m}$ and having initial velocity of $u = 0$ so by using, ${v^2} - {u^2} = 2aS$ we can write,
${v^2} - 0 = 2{a_m}L$
$\Rightarrow {v^2} = 2{a_m}L$
Similarly, for the length of $L$ mass $2m$ leaves with velocity say $v'$ with an acceleration of ${a_{2m}}$ and having initial velocity of $u = 0$ so by using, ${v^2} - {u^2} = 2aS$ we can write,
$v{'^2} - 0 = 2{a_{2m}}L$
$\Rightarrow v{'^2} = 2{a_{2m}}L$
Now, divide the equation $v{'^2} = 2{a_{2m}}L$ by the equation ${v^2} = 2{a_m}L$ we get,
$\dfrac{{v{'^2}}}{{{v^2}}} = \dfrac{{{a_{2m}}}}{{{a_m}}}$
From the equation $(i)$ put $\dfrac{{{a_{2m}}}}{{{a_m}}} = \dfrac{1}{2}$
$v{'^2} = {v^2} \times \dfrac{1}{2}$
$\therefore v' = \dfrac{v}{{\sqrt 2 }}$
So, the mass $2m$ body leaves the barrel with a velocity of $v' = \dfrac{v}{{\sqrt 2 }}$.
Hence, the correct option is B.
Note: It should be remembered that, while using the newton’s equation of motion ${v^2} - {u^2} = 2aS$ , the initial velocity of the projectile body when enters in a barrel is taken as zero and other equation of motion are written as $v = u + at$ and $S = ut + \dfrac{1}{2}a{t^2}$ ,these equations of motion is widely used in classical kinematics to deal with physical problems of physics.
Complete step by step answer:
Let us suppose $F$ be the net force acting on both bodies of mass $m(and)2m$ , let their accelerations be ${a_m}(and){a_{2m}}$ respectively, so using
$F = ma$ We have,
$\Rightarrow F = m{a_m}$
And
$F = 2m{a_{2m}}$
Since both forces are equal so we write
$m{a_m} = 2m{a_{2m}}$
$\Rightarrow \dfrac{{{a_m}}}{{{a_{2m}}}} = 2 \to (i)$
Now for the length of $L$ mass $m$ leaves with velocity $v$ with an acceleration of ${a_m}$ and having initial velocity of $u = 0$ so by using, ${v^2} - {u^2} = 2aS$ we can write,
${v^2} - 0 = 2{a_m}L$
$\Rightarrow {v^2} = 2{a_m}L$
Similarly, for the length of $L$ mass $2m$ leaves with velocity say $v'$ with an acceleration of ${a_{2m}}$ and having initial velocity of $u = 0$ so by using, ${v^2} - {u^2} = 2aS$ we can write,
$v{'^2} - 0 = 2{a_{2m}}L$
$\Rightarrow v{'^2} = 2{a_{2m}}L$
Now, divide the equation $v{'^2} = 2{a_{2m}}L$ by the equation ${v^2} = 2{a_m}L$ we get,
$\dfrac{{v{'^2}}}{{{v^2}}} = \dfrac{{{a_{2m}}}}{{{a_m}}}$
From the equation $(i)$ put $\dfrac{{{a_{2m}}}}{{{a_m}}} = \dfrac{1}{2}$
$v{'^2} = {v^2} \times \dfrac{1}{2}$
$\therefore v' = \dfrac{v}{{\sqrt 2 }}$
So, the mass $2m$ body leaves the barrel with a velocity of $v' = \dfrac{v}{{\sqrt 2 }}$.
Hence, the correct option is B.
Note: It should be remembered that, while using the newton’s equation of motion ${v^2} - {u^2} = 2aS$ , the initial velocity of the projectile body when enters in a barrel is taken as zero and other equation of motion are written as $v = u + at$ and $S = ut + \dfrac{1}{2}a{t^2}$ ,these equations of motion is widely used in classical kinematics to deal with physical problems of physics.
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