
A body exerts an impulse I on a body which is changing its speed from u to v. The force and the impulse of the body are along the same line. The work by the force is
A. \[\left[ {I\left( {{v^2} - {u^2}} \right)} \right]/2\]
B. \[\left[ {I\left( {v + u} \right)} \right]/2\]
C. \[\left[ {I\left( {v - u} \right)} \right]/2\]
D. \[\left[ {I\left( {{v^2} + {u^2}} \right)} \right]/2\]
Answer
573.9k+ views
Hint:Determine the value of the impulse on the body using formula of impulse which equals change in momentum. Also, use the work-energy theorem. Determine the change in kinetic energy of the body and substitute it in the work energy theorem. Now substitute the derived value of impulse in this equation.
Formula used:
The impulse \[I\] on an object is
\[\Rightarrow I = \Delta P\] …… (1)
Here, \[\Delta P\] is the change in momentum of the object.
The momentum \[P\] of an object is
\[\Rightarrow P = mv\] …… (2)
Here, \[m\] is the mass of the object and \[v\] is the velocity of the object.
The expression for work-energy theorem is
\[\Rightarrow W = \Delta K\] ….. (3)
Here, is the work done due to a force and is the change in kinetic energy of the object.
The kinetic energy \[K\] of an object is
\[\Rightarrow K = \dfrac{1}{2}m{v^2}\] …… (4)
Here, \[m\] is the mass of the object and \[v\] is the velocity of the object.
Complete step by step solution:
We have given that the impulse on the body is \[I\] and the velocity of the body changes from \[u\] to \[v\]. Hence, the initial speed of the body is \[u\] and the final speed is \[v\].
Let us determine the impulse \[I\] on the body. Let \[m\] be the mass of the body.
According to equation (1), the initial momentum \[{P_i}\] of the body becomes
\[\Rightarrow{P_i} = mu\]
According to equation (1), the final momentum \[{P_f}\] of the body becomes
\[\Rightarrow{P_f} = mv\]
Substitute \[{P_f} - {P_i}\] for \[\Delta P\] in equation (1).
\[\Rightarrow I = {P_f} - {P_i}\]
Substitute \[mv\] for \[{P_f}\] and \[mu\] for \[{P_i}\] in the above equation.
\[\Rightarrow I = mv - mu\]
\[ \Rightarrow I = m\left( {v - u} \right)\]
Hence, the impulse on the body is \[m\left( {v - u} \right)\].
According to equation (4), the initial kinetic energy \[{K_i}\] of the body becomes
\[\Rightarrow{K_i} = \dfrac{1}{2}m{u^2}\]
According to equation (4), the final kinetic energy \[{K_f}\] of the body becomes
\[\Rightarrow{K_f} = \dfrac{1}{2}m{v^2}\]
The change in kinetic energy of the body is
\[\Rightarrow\Delta K = {K_f} - {K_i}\]
Substitute \[\dfrac{1}{2}m{v^2}\] for \[{K_f}\] and \[\dfrac{1}{2}m{u^2}\] for \[{K_i}\] in the above equation.
\[\Rightarrow\Delta K = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2}\]
Hence, the work done by the force can be determined by using equation (3).
Substitute \[\dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2}\] for \[\Delta K\] in equation (3).
\[\Rightarrow W = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2}\]
\[ \Rightarrow W = \dfrac{1}{2}m\left( {{v^2} - {u^2}} \right)\]
\[ \Rightarrow W = \dfrac{1}{2}m\left[ {\left( {v + u} \right)\left( {v - u} \right)} \right]\]
Substitute \[I\] for \[m\left( {v - u} \right)\] in the above equation.
\[ \therefore W = \left[ {I\left( {v + u} \right)} \right]/2\]
Therefore, the work done by the force is \[\left[ {I\left( {v + u} \right)} \right]/2\].
Hence, the correct option is B.
Note: The students should always read the question carefully. In the present question, the direction of the velocity and the force are along the same line. Hence, the values of both the velocities are taken positively. If the force and velocity were opposite in direction, then the initial velocity should be negative.
Formula used:
The impulse \[I\] on an object is
\[\Rightarrow I = \Delta P\] …… (1)
Here, \[\Delta P\] is the change in momentum of the object.
The momentum \[P\] of an object is
\[\Rightarrow P = mv\] …… (2)
Here, \[m\] is the mass of the object and \[v\] is the velocity of the object.
The expression for work-energy theorem is
\[\Rightarrow W = \Delta K\] ….. (3)
Here, is the work done due to a force and is the change in kinetic energy of the object.
The kinetic energy \[K\] of an object is
\[\Rightarrow K = \dfrac{1}{2}m{v^2}\] …… (4)
Here, \[m\] is the mass of the object and \[v\] is the velocity of the object.
Complete step by step solution:
We have given that the impulse on the body is \[I\] and the velocity of the body changes from \[u\] to \[v\]. Hence, the initial speed of the body is \[u\] and the final speed is \[v\].
Let us determine the impulse \[I\] on the body. Let \[m\] be the mass of the body.
According to equation (1), the initial momentum \[{P_i}\] of the body becomes
\[\Rightarrow{P_i} = mu\]
According to equation (1), the final momentum \[{P_f}\] of the body becomes
\[\Rightarrow{P_f} = mv\]
Substitute \[{P_f} - {P_i}\] for \[\Delta P\] in equation (1).
\[\Rightarrow I = {P_f} - {P_i}\]
Substitute \[mv\] for \[{P_f}\] and \[mu\] for \[{P_i}\] in the above equation.
\[\Rightarrow I = mv - mu\]
\[ \Rightarrow I = m\left( {v - u} \right)\]
Hence, the impulse on the body is \[m\left( {v - u} \right)\].
According to equation (4), the initial kinetic energy \[{K_i}\] of the body becomes
\[\Rightarrow{K_i} = \dfrac{1}{2}m{u^2}\]
According to equation (4), the final kinetic energy \[{K_f}\] of the body becomes
\[\Rightarrow{K_f} = \dfrac{1}{2}m{v^2}\]
The change in kinetic energy of the body is
\[\Rightarrow\Delta K = {K_f} - {K_i}\]
Substitute \[\dfrac{1}{2}m{v^2}\] for \[{K_f}\] and \[\dfrac{1}{2}m{u^2}\] for \[{K_i}\] in the above equation.
\[\Rightarrow\Delta K = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2}\]
Hence, the work done by the force can be determined by using equation (3).
Substitute \[\dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2}\] for \[\Delta K\] in equation (3).
\[\Rightarrow W = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2}\]
\[ \Rightarrow W = \dfrac{1}{2}m\left( {{v^2} - {u^2}} \right)\]
\[ \Rightarrow W = \dfrac{1}{2}m\left[ {\left( {v + u} \right)\left( {v - u} \right)} \right]\]
Substitute \[I\] for \[m\left( {v - u} \right)\] in the above equation.
\[ \therefore W = \left[ {I\left( {v + u} \right)} \right]/2\]
Therefore, the work done by the force is \[\left[ {I\left( {v + u} \right)} \right]/2\].
Hence, the correct option is B.
Note: The students should always read the question carefully. In the present question, the direction of the velocity and the force are along the same line. Hence, the values of both the velocities are taken positively. If the force and velocity were opposite in direction, then the initial velocity should be negative.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

