Question

First law of motion can be the mathematical formulation of the second law of motion. Justify this statement. A dumb bell mass $10kg$falls on the floor from a height of$80cm$. Calculate the change in momentum of the dumb-bell and the force of impact on the floor.

Answer 1

Hint As the first law of motion states that the body in rest or uniform motion will continue to do until an external four is applied on it.
The second law of motion states that the external force applied is equal to the rate of change of momentum.
Using this definition we will prove the result and solve the numerical.

Complete step by step solution: $\dfrac{\overrightarrow{db}}{\overrightarrow{de}}=\overrightarrow{F}=40N$
Derivation: As we know that Newton’s law states that a body stays at rest if it is at rest and moves with a constant velocity if already moving, until a net force is applied to it. In other words, the state of motion of a body changes only on application of a net non-zero force.
Newton’s second law states that the net force applied on a body is equal to the rate of change of its momentum
i.e. $\overrightarrow{F}=\dfrac{\overrightarrow{db}}{\overrightarrow{de}}$
Where, $\overrightarrow{F}$is net force
     \[\overrightarrow{P}\]is the momentum
\[\]\[\begin{align}
  & \therefore =\overrightarrow{F}\dfrac{\overrightarrow{d\theta }}{\overrightarrow{dt}}=\dfrac{d\left( \overrightarrow{mv} \right)}{dt}\therefore \left( \overrightarrow{P=}\overrightarrow{mv} \right) \\
 & \overrightarrow{F}=m\dfrac{dv}{dt}\left( \therefore 'n'\text{ is constant} \right) \\
\end{align}\]
$\text{if net force is zero}=\text{velocity is constant}$
I.e. a body at rest will be at rest and a body moving with constant verifications will continue to do so.
Hence, proved the first law mathematical formula from the second law.
Given: $\begin{align}
  & \text{mass of dumbell}=10kg \\
 & \text{Height from which it falls}=80cm=0\cdot 8m \\
\end{align}$
Acceleration in the downward direction$(a)=10m/{{s}^{2}}$
Initial velocity of the dumb bell$(u)=0$
Now, using the third equation of motion
     ${{v}^{2}}={{u}^{2}}+2as$
Putting the values we get
     $\begin{align}
  & {{v}^{2}}={{(0)}^{2}}+2(10)(0.8) \\
 & {{v}^{2}}=16 \\
 & v=4m/s \\
\end{align}$
Hence, the momentum with which it hits the flux is
$\begin{align}
  & p=mv \\
 & =10\times 4 \\
 & =p=40kg\text{ m}{{\text{s}}^{-1}} \\
\end{align}$
The Above momentum is actually the rate of change of momentum because velocity has changed from 0 to $4m/s$
     $\begin{align}
  & \therefore \dfrac{\overrightarrow{dp}}{\overrightarrow{dt}}=40kg\text{ }m{{s}^{-1}} \\
 & \text{Also, from second law of motion} \\
 & \text{ }\overrightarrow{F}=\dfrac{\overrightarrow{dp}}{\overrightarrow{dt}}=40kg\text{ }m/s \\
\end{align}$

Note:Concept of S.I. units should be cleaned. While solving numerical make sure that all units should be in their respective S.I. form. Various expressions like should be used carefully. Do not confuse yourself with.