
The engine of a train passes an electric pole with a velocity ′u′ and the last compartment of the train crosses the same pole with a velocity v. Then find the velocity with which the midpoint of the train passes the pole. Assume acceleration to be uniform.
Answer
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Hint: According to the question, we need to find the velocity with which the midpoint of the train passes the pole. Now to solve such a question it is important to have a knowledge of the three laws of motion and the three equations of motion. In this question we will use the third equation of motion , i.e., $ {v^2} - {u^2} = 2as $ .
Complete Step By Step Answer:
The three laws of motion are these:
1) Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
2) Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
3) Newton's 3rd law states that for every action there is an equal and opposite reaction.
Now according to question,
Let the length of the train be ‘l’
According to the Newton’s third equation of motion, $ {v^2} - {u^2} = 2as $
Here ‘s’ is equal to ‘l’ and thus we can rewrite the equation as,
$ {v^2} - {u^2} = 2al $
$ a = \dfrac{{{v^2} - {u^2}}}{{2l}} $
Now we need to find the solution about the midpoint and thus the length here will be $ \dfrac{l}{2} $ .
Putting $ \dfrac{l}{2} $ in the equation, we get,
$ v_{last}^2 = {u^2} + 2a \times \dfrac{l}{2} $
$ v_{last}^2 = {u^2} + a \times l $
Putting the value of ‘a’ in the above equation, we get,
$ v_{last}^2 = {u^2} + \dfrac{{{v^2} - {u^2}}}{{2l}} \times l $
$ v_{last}^2 = \dfrac{{2{u^2} + {v^2} - {u^2}}}{2} $
$ v_{last}^2 = \dfrac{{{u^2} + {v^2}}}{2} $
$ v_{last}^2 = \sqrt {\dfrac{{{v^2} + {u^2}}}{2}} $
So the final answer is $ v_{last}^2 = \sqrt {\dfrac{{{v^2} + {u^2}}}{2}} $ .
Note:
The motion of the train takes place due to Newton's third law of motion. The reason the train is able to move at all is due to the force the locomotive puts on the rails, which enables it to accelerate the train and move it in the forward direction.
Complete Step By Step Answer:
The three laws of motion are these:
1) Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
2) Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
3) Newton's 3rd law states that for every action there is an equal and opposite reaction.
Now according to question,
Let the length of the train be ‘l’
According to the Newton’s third equation of motion, $ {v^2} - {u^2} = 2as $
Here ‘s’ is equal to ‘l’ and thus we can rewrite the equation as,
$ {v^2} - {u^2} = 2al $
$ a = \dfrac{{{v^2} - {u^2}}}{{2l}} $
Now we need to find the solution about the midpoint and thus the length here will be $ \dfrac{l}{2} $ .
Putting $ \dfrac{l}{2} $ in the equation, we get,
$ v_{last}^2 = {u^2} + 2a \times \dfrac{l}{2} $
$ v_{last}^2 = {u^2} + a \times l $
Putting the value of ‘a’ in the above equation, we get,
$ v_{last}^2 = {u^2} + \dfrac{{{v^2} - {u^2}}}{{2l}} \times l $
$ v_{last}^2 = \dfrac{{2{u^2} + {v^2} - {u^2}}}{2} $
$ v_{last}^2 = \dfrac{{{u^2} + {v^2}}}{2} $
$ v_{last}^2 = \sqrt {\dfrac{{{v^2} + {u^2}}}{2}} $
So the final answer is $ v_{last}^2 = \sqrt {\dfrac{{{v^2} + {u^2}}}{2}} $ .
Note:
The motion of the train takes place due to Newton's third law of motion. The reason the train is able to move at all is due to the force the locomotive puts on the rails, which enables it to accelerate the train and move it in the forward direction.
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