
An impulse \[I\] given to a body changes its velocity from \[{v_1}\] to \[{v_2}\]. The increase in the kinetic energy of the body is given by:
A. \[I\left( {{v_1} + {v_2}} \right)\]
B. \[I\left( {{v_1} + {v_2}} \right)/2\]
C. \[I\left( {{v_1} - {v_2}} \right)\]
D. \[I\left( {{v_1} - {v_2}} \right)/2\]
Answer
560.4k+ views
Hint: The impulse of the body is the change in linear momentum of the body. Use the formula for change in kinetic energy and then substitute the expression for impulse in change in kinetic energy.
Formula used:
\[I = \Delta p\], where, I is the impulse and \[\Delta p\] is a change in linear momentum.
The momentum of the body of mass m moving with velocity v is,
\[p = mv\]
The kinetic energy of the body of mass m is,
\[K.E. = \dfrac{1}{2}m{v^2}\]
Complete step by step answer:
We know that the impulse of the body is the change in its momentum. Therefore, we can express the impulse of body of mass m as follows,
\[I = m{v_2} - m{v_1}\]
\[ \Rightarrow I = m\left( {{v_2} - {v_1}} \right)\] …… (1)
Now, we know the change in kinetic energy of the body of mass m moving from velocity \[{v_1}\] to \[{v_2}\] is can be written as,
\[\Delta K.E. = \dfrac{1}{2}mv_2^2 - \dfrac{1}{2}mv_1^2\]
\[ \Rightarrow \Delta K.E. = \dfrac{1}{2}m\left( {v_2^2 - v_1^2} \right)\]
We know that, \[\left( {{a^2} - {b^2}} \right)\] can be expressed as \[\left( {a - b} \right)\left( {a + b} \right)\].
Therefore, we write the above equation as,
\[\Delta K.E. = \dfrac{1}{2}m\left( {{v_2} - {v_1}} \right)\left( {{v_2} + {v_1}} \right)\]
Using equation (1), we can write the above equation as follows,
\[\Delta K.E. = \dfrac{{I\left( {{v_2} + {v_1}} \right)}}{2}\]
So, the correct answer is “Option B”.
Additional Information:
We know that the impulse is the amount of force which is applied on the body for a specific interval of time. When we strike the nail with a hammer and keep the hammer for a certain interval of time on the nail, it increases the impulse on the nail and a greater amount of force can be applied on the nail.
We know that impulse is the product of applied force and time interval. Therefore,
\[I = F\Delta t\]
Note:
To solve such types of questions, the key point is to remember all three Newton’s laws of motion, work-energy principle, and relation between work done and applied force. Remembering all these concepts will ease in converting the formulae for such questions.
Formula used:
\[I = \Delta p\], where, I is the impulse and \[\Delta p\] is a change in linear momentum.
The momentum of the body of mass m moving with velocity v is,
\[p = mv\]
The kinetic energy of the body of mass m is,
\[K.E. = \dfrac{1}{2}m{v^2}\]
Complete step by step answer:
We know that the impulse of the body is the change in its momentum. Therefore, we can express the impulse of body of mass m as follows,
\[I = m{v_2} - m{v_1}\]
\[ \Rightarrow I = m\left( {{v_2} - {v_1}} \right)\] …… (1)
Now, we know the change in kinetic energy of the body of mass m moving from velocity \[{v_1}\] to \[{v_2}\] is can be written as,
\[\Delta K.E. = \dfrac{1}{2}mv_2^2 - \dfrac{1}{2}mv_1^2\]
\[ \Rightarrow \Delta K.E. = \dfrac{1}{2}m\left( {v_2^2 - v_1^2} \right)\]
We know that, \[\left( {{a^2} - {b^2}} \right)\] can be expressed as \[\left( {a - b} \right)\left( {a + b} \right)\].
Therefore, we write the above equation as,
\[\Delta K.E. = \dfrac{1}{2}m\left( {{v_2} - {v_1}} \right)\left( {{v_2} + {v_1}} \right)\]
Using equation (1), we can write the above equation as follows,
\[\Delta K.E. = \dfrac{{I\left( {{v_2} + {v_1}} \right)}}{2}\]
So, the correct answer is “Option B”.
Additional Information:
We know that the impulse is the amount of force which is applied on the body for a specific interval of time. When we strike the nail with a hammer and keep the hammer for a certain interval of time on the nail, it increases the impulse on the nail and a greater amount of force can be applied on the nail.
We know that impulse is the product of applied force and time interval. Therefore,
\[I = F\Delta t\]
Note:
To solve such types of questions, the key point is to remember all three Newton’s laws of motion, work-energy principle, and relation between work done and applied force. Remembering all these concepts will ease in converting the formulae for such questions.
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