
A bullet of mass $20$ g travelling horizontally with a velocity of $100$ m/s strikes a wooden block and comes to rest in $0.05$ s. Calculate the magnitude of force in $N$ exerted by the wooden block on the bullet.
Answer
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Hint: Kinematics is a branch of physics that deals with the motion of points, bodies (objects), and structures of bodies (groups of objects) without taking into account the forces that drive them. Kinematics is often referred to as the "geometry of motion" and is often considered a branch of mathematics.
Formula used:
Formulas of Kinematics are:
$V = u + at \\
{V^2} = {u^2} + 2as \\
s = ut + \dfrac{1}{2}a{t^2}$
Complete step by step answer:
According to Newton’s second law: The second law states that the rate at which a body's momentum changes over time is proportional to the force applied and occurs in the same direction as the applied force.
$F = \dfrac{{dp}}{{dt}}$
Where $p$ is the mass momentum of the body.
The second law can be re-stated in terms of an object's acceleration for objects and structures with constant mass.
$F = \dfrac{{d(mv)}}{{dt}} \\
\Rightarrow F= m\dfrac{{dv}}{{dt}} \\
\Rightarrow F = ma$
Where \[F\] is the applied net force, $m$ is the body's mass, and $a$ is the body's acceleration. As a result, a proportional acceleration is generated by the net force applied to a body.
Given: Mass of the bullet $(m) = 20\,g= 0.02\,kg$
Initial velocity $(u) = 100$ m/s
Final velocity $(v) = 0$ (since the bullet has come to a rest)
Time taken to come to rest $(t) = 0.05s$
According to the first equation of kinematics:
$V = u + at$
Acceleration of the bullet (a)
$0 = 100 + a( - 0.05) \\
\Rightarrow a = - \dfrac{{100}}{{0.05}} \\
\Rightarrow a = - 2000\,m/{s^2} \\$
The negative sign here means that the bullet's velocity is decreasing.
From Newton’s law of motion we know that:
$\text{Force(F)} = \text{Mass} \times \text{Acceleration}$
$\Rightarrow F = 0.02 \times ( - 2000)$
$\therefore F = - 40\,N$
As a result, the force exerted on the bullet by the wooden block is 50 N.
Note: A condition that must be met in order to apply motion equations. It should be much slower than the speed of light in order to avoid relativistic effects. If $v$ is the object's velocity and $c$ is the speed of light, then Newton's laws of motion should apply if $v$ and $c$ are very close to $0$. It is not necessary to maintain constant acceleration or travel in a straight line.
Formula used:
Formulas of Kinematics are:
$V = u + at \\
{V^2} = {u^2} + 2as \\
s = ut + \dfrac{1}{2}a{t^2}$
Complete step by step answer:
According to Newton’s second law: The second law states that the rate at which a body's momentum changes over time is proportional to the force applied and occurs in the same direction as the applied force.
$F = \dfrac{{dp}}{{dt}}$
Where $p$ is the mass momentum of the body.
The second law can be re-stated in terms of an object's acceleration for objects and structures with constant mass.
$F = \dfrac{{d(mv)}}{{dt}} \\
\Rightarrow F= m\dfrac{{dv}}{{dt}} \\
\Rightarrow F = ma$
Where \[F\] is the applied net force, $m$ is the body's mass, and $a$ is the body's acceleration. As a result, a proportional acceleration is generated by the net force applied to a body.
Given: Mass of the bullet $(m) = 20\,g= 0.02\,kg$
Initial velocity $(u) = 100$ m/s
Final velocity $(v) = 0$ (since the bullet has come to a rest)
Time taken to come to rest $(t) = 0.05s$
According to the first equation of kinematics:
$V = u + at$
Acceleration of the bullet (a)
$0 = 100 + a( - 0.05) \\
\Rightarrow a = - \dfrac{{100}}{{0.05}} \\
\Rightarrow a = - 2000\,m/{s^2} \\$
The negative sign here means that the bullet's velocity is decreasing.
From Newton’s law of motion we know that:
$\text{Force(F)} = \text{Mass} \times \text{Acceleration}$
$\Rightarrow F = 0.02 \times ( - 2000)$
$\therefore F = - 40\,N$
As a result, the force exerted on the bullet by the wooden block is 50 N.
Note: A condition that must be met in order to apply motion equations. It should be much slower than the speed of light in order to avoid relativistic effects. If $v$ is the object's velocity and $c$ is the speed of light, then Newton's laws of motion should apply if $v$ and $c$ are very close to $0$. It is not necessary to maintain constant acceleration or travel in a straight line.
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