Answer
Verified
450.3k+ views
Hint: We will use the law of conservation of momentum which is given as initial momentum of the system will be equal to final momentum of the system to determine the final velocity of the combined block. Then we will use the expression for kinetic energy to find the kinetic energy of the given system using the velocity we found. Kinetic energy is directly proportional to mass and square of the velocity of the body.
Formula used:
\[\begin{align}
& {{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}} \\
& K.E=\dfrac{1}{2}m{{v}^{2}} \\
\end{align}\]
Complete step by step answer:
Firstly we will draw a diagram for understanding the situation better.
Here, the mass of the bullet is taken as \[m\].
Mass of the block is taken as \[M\].
Velocity of the bullet is taken as \[v\].
And velocity of the block is zero as it is initially at rest.
Now, we will use the law of conservation of momentum to find the velocity of the combined block whose mass will be \[M+m\]. That is,
\[mv+M\times 0=\left( m+M \right){{v}_{f}}\]
\[\begin{align}
& \Rightarrow mv=\left( m+M \right){{v}_{f}} \\
& \Rightarrow {{v}_{f}}=\dfrac{m}{\left( m+M \right)}v \\
\end{align}\]
So, the final velocity of the combined block is found to be, \[{{v}_{f}}=\dfrac{m}{\left( m+M \right)}v\].
Now, we will use this velocity to find the kinetic energy of the combined block. Kinetic energy is given as,
\[K.E=\dfrac{1}{2}m{{v}^{2}}\]
We have the mass of the combined block as \[M+m\].
\[\begin{align}
& \Rightarrow K.E=\dfrac{1}{2}\left( m+M \right){{\left( \dfrac{m}{\left( m+M \right)}v \right)}^{2}} \\
&\Rightarrow K.E=\dfrac{1}{2}\left( m+M \right)\times \dfrac{{{m}^{2}}}{{{\left( m+M \right)}^{2}}}{{v}^{2}} \\
& \Rightarrow K.E=\dfrac{1}{2}m{{v}^{2}}\times \dfrac{m}{\left( m+M \right)} \\
\end{align}\]
Hence, we have found the kinetic energy of composite block as\[\dfrac{1}{2}m{{v}^{2}}\times \dfrac{m}{\left( m+M \right)}\].
Hence, the correct answer is option A.
Note:
We must be very careful while doing problems like this. The sign of velocity has a major role in determining the momentum. In this case the velocity of all the objects and systems is in the same direction which we have taken as positive. But we must remember that we will need to use sign convention if any of the colliding particles or the composite system has a change in direction in velocity.
Formula used:
\[\begin{align}
& {{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}} \\
& K.E=\dfrac{1}{2}m{{v}^{2}} \\
\end{align}\]
Complete step by step answer:
Firstly we will draw a diagram for understanding the situation better.
Here, the mass of the bullet is taken as \[m\].
Mass of the block is taken as \[M\].
Velocity of the bullet is taken as \[v\].
And velocity of the block is zero as it is initially at rest.
Now, we will use the law of conservation of momentum to find the velocity of the combined block whose mass will be \[M+m\]. That is,
\[mv+M\times 0=\left( m+M \right){{v}_{f}}\]
\[\begin{align}
& \Rightarrow mv=\left( m+M \right){{v}_{f}} \\
& \Rightarrow {{v}_{f}}=\dfrac{m}{\left( m+M \right)}v \\
\end{align}\]
So, the final velocity of the combined block is found to be, \[{{v}_{f}}=\dfrac{m}{\left( m+M \right)}v\].
Now, we will use this velocity to find the kinetic energy of the combined block. Kinetic energy is given as,
\[K.E=\dfrac{1}{2}m{{v}^{2}}\]
We have the mass of the combined block as \[M+m\].
\[\begin{align}
& \Rightarrow K.E=\dfrac{1}{2}\left( m+M \right){{\left( \dfrac{m}{\left( m+M \right)}v \right)}^{2}} \\
&\Rightarrow K.E=\dfrac{1}{2}\left( m+M \right)\times \dfrac{{{m}^{2}}}{{{\left( m+M \right)}^{2}}}{{v}^{2}} \\
& \Rightarrow K.E=\dfrac{1}{2}m{{v}^{2}}\times \dfrac{m}{\left( m+M \right)} \\
\end{align}\]
Hence, we have found the kinetic energy of composite block as\[\dfrac{1}{2}m{{v}^{2}}\times \dfrac{m}{\left( m+M \right)}\].
Hence, the correct answer is option A.
Note:
We must be very careful while doing problems like this. The sign of velocity has a major role in determining the momentum. In this case the velocity of all the objects and systems is in the same direction which we have taken as positive. But we must remember that we will need to use sign convention if any of the colliding particles or the composite system has a change in direction in velocity.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Write a letter to the principal requesting him to grant class 10 english CBSE