
When a force of $1N$ acts on \[1kg\] mass at rest for \[1s\], its momentum is \[P\]. When $1N$ force acts on\[1kg\] mass at rest through a distance \[1m\], its final momentum is \[P'\]. The ratio of \[P\] to \[P'\]is:
\[\begin{align}
& A)1:1 \\
& B)1:\sqrt{2} \\
& C)1:2 \\
& D)2:1 \\
\end{align}\]
Answer
576.6k+ views
Hint: The momentum of a body is given as the product of mass and the velocity of the body. We can also use the definition of force to know the change in momentum because force is termed as rate of change of momentum. Here, we will find the momentum of the body in both cases and take their ratio.
Formula used:
\[\begin{align}
& P=mv \\
& F=\dfrac{\Delta P}{\Delta t} \\
\end{align}\]
Complete answer: We will find the momentum of both cases separately and hence find the ratio.
For the first case, it is given that the force of $1N$ acts on \[1kg\] mass at rest for \[1s\]. We will use the equation relating force and momentum here.
Here, momentum \[P\] is given by \[F\Delta t=\Delta P\].
\[\Rightarrow {{P}_{f}}-{{P}_{i}}=F\left( {{t}_{f}}-{{t}_{1}} \right)\]
Here, initially the body is at rest. So the momentum will be 0.
\[\Rightarrow {{P}_{f}}=P=1\left( 1-0 \right)=1kg-m/s\]
So, the momentum of the body in first case is
\[P=1kg-m/s\]
Now, we will find the momentum for the second case. Here, only the distance is given. so we will use the formula \[{{v}^{2}}-{{u}^{2}}=2as\] to find the final velocity of the body. Initially the body is at rest and it is moved by \[1m\].
Also we need to find the acceleration here. It is given by the formula, \[F=ma\]. Here, force is $1N$and mass is \[1kg\].
\[\Rightarrow a=\dfrac{1}{1}=1m/{{s}^{2}}\]
Now, the final velocity will be
\[{{v}^{2}}-{{0}^{2}}=2as\]
\[{{v}^{2}}=2\times 1\times 1=2\]
\[v=\sqrt{2}m/s\].
Then the momentum of the body in the second case will be, \[P'=mv=1\times \sqrt{2}=\sqrt{2}kg-m/s\].
Now, the ratio of \[P\] to \[P'\] is given as\[\dfrac{P}{P'}=\dfrac{1}{\sqrt{2}}\]
\[P:P'=1:\sqrt{2}\].
So, the correct answer is “Option B”.
Note:
While doing this question, we must be very careful of which expression is to be used for finding the final velocity of the body if it is not given. We must look at the parameters that we have and choose the required expression accordingly. As momentum is defined in many ways, try to get a clear idea to use the appropriate equation at appropriate places.
Formula used:
\[\begin{align}
& P=mv \\
& F=\dfrac{\Delta P}{\Delta t} \\
\end{align}\]
Complete answer: We will find the momentum of both cases separately and hence find the ratio.
For the first case, it is given that the force of $1N$ acts on \[1kg\] mass at rest for \[1s\]. We will use the equation relating force and momentum here.
Here, momentum \[P\] is given by \[F\Delta t=\Delta P\].
\[\Rightarrow {{P}_{f}}-{{P}_{i}}=F\left( {{t}_{f}}-{{t}_{1}} \right)\]
Here, initially the body is at rest. So the momentum will be 0.
\[\Rightarrow {{P}_{f}}=P=1\left( 1-0 \right)=1kg-m/s\]
So, the momentum of the body in first case is
\[P=1kg-m/s\]
Now, we will find the momentum for the second case. Here, only the distance is given. so we will use the formula \[{{v}^{2}}-{{u}^{2}}=2as\] to find the final velocity of the body. Initially the body is at rest and it is moved by \[1m\].
Also we need to find the acceleration here. It is given by the formula, \[F=ma\]. Here, force is $1N$and mass is \[1kg\].
\[\Rightarrow a=\dfrac{1}{1}=1m/{{s}^{2}}\]
Now, the final velocity will be
\[{{v}^{2}}-{{0}^{2}}=2as\]
\[{{v}^{2}}=2\times 1\times 1=2\]
\[v=\sqrt{2}m/s\].
Then the momentum of the body in the second case will be, \[P'=mv=1\times \sqrt{2}=\sqrt{2}kg-m/s\].
Now, the ratio of \[P\] to \[P'\] is given as\[\dfrac{P}{P'}=\dfrac{1}{\sqrt{2}}\]
\[P:P'=1:\sqrt{2}\].
So, the correct answer is “Option B”.
Note:
While doing this question, we must be very careful of which expression is to be used for finding the final velocity of the body if it is not given. We must look at the parameters that we have and choose the required expression accordingly. As momentum is defined in many ways, try to get a clear idea to use the appropriate equation at appropriate places.
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