
A hypothetical train moving with a speed of \[0.6c\] passes by the platform of a small stationary without being slowed down. The observers on the platform note that the length of the train is just equal to the length of the platform which is \[200\,{\text{m}}\]. (a) Find the rest length of the train. (b) Find the length of the platform as measured by the observers in the train.
Answer
563.1k+ views
Hint:Use the formula for the observed length in the reference length in which the object is moving. This formula gives the relation between the observed length, actual length in the reference frame in which the object is at rest, velocity of the object whose length is to be measured and speed of light. Use this equation to calculate the rest length of the train and length of the platform measured by the observers in the train.
Formula used:
The observed length \[L'\] in the reference length in which the object is moving is given by
\[L' = L\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} \] …… (1)
Here, \[L\] is the actual length in the reference frame in which the object is at rest, \[v\] is the velocity of the object whose length is to be measured and \[c\] is the speed of light.
Complete step by step answer:
We have given that the speed of the hypothetical train is \[0.6c\].
\[v = 0.6c\]
The length of the train for the observer on the platform is \[200\,{\text{m}}\].
\[L' = 200\,{\text{m}}\]
Let \[L'\] be the rest length of the train. The rest length of the train is given by
\[L = \dfrac{{L'}}{{\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} }}\]
Substitute \[200\,{\text{m}}\] for \[L'\] and \[0.6c\] for \[v\] in the above equation.
\[L = \dfrac{{200\,{\text{m}}}}{{\sqrt {1 - \dfrac{{{{\left( {0.6c} \right)}^2}}}{{{c^2}}}} }}\]
\[ \Rightarrow L = \dfrac{{200\,{\text{m}}}}{{\sqrt {1 - {{\left( {0.6} \right)}^2}} }}\]
\[ \Rightarrow L = \dfrac{{200\,{\text{m}}}}{{\sqrt {1 - 0.36} }}\]
\[ \Rightarrow L = \dfrac{{200\,{\text{m}}}}{{\sqrt {0.64} }}\]
\[ \Rightarrow L = \dfrac{{200\,{\text{m}}}}{{0.8}}\]
\[\therefore L = 250\,{\text{m}}\]
Hence, the rest length of the train is \[250\,{\text{m}}\].
(b) Let now calculate the length of the platform measured by the observers in the train.The length of the platform measured by the observers in the train is
\[L' = L\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} \]
Substitute \[200\,{\text{m}}\] for \[L\] and \[0.6c\] for \[v\] in the above equation.
\[L' = \left( {200\,{\text{m}}} \right)\sqrt {1 - \dfrac{{{{\left( {0.6c} \right)}^2}}}{{{c^2}}}} \]
\[ \Rightarrow L' = \left( {200\,{\text{m}}} \right)\sqrt {1 - {{\left( {0.6c} \right)}^2}} \]
\[ \Rightarrow L' = \left( {200\,{\text{m}}} \right)\sqrt {1 - 0.36} \]
\[ \Rightarrow L' = \left( {200\,{\text{m}}} \right)\sqrt {0.64} \]
\[ \Rightarrow L' = \left( {200\,{\text{m}}} \right)\left( {0.8} \right)\]
\[ \therefore L' = 160\,{\text{m}}\]
Hence, the length of the platform measured by the observers in the train is \[160\,{\text{m}}\].
Note: The students should not get confused between the different signs for the length of the train measured by the observer on the platform. For the first case, we have used the sign \[L'\] for length of the train measured by the observer on the platform and for the second case we have used sign \[L\] for the length of the train measured by the observer on the platform.
Formula used:
The observed length \[L'\] in the reference length in which the object is moving is given by
\[L' = L\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} \] …… (1)
Here, \[L\] is the actual length in the reference frame in which the object is at rest, \[v\] is the velocity of the object whose length is to be measured and \[c\] is the speed of light.
Complete step by step answer:
We have given that the speed of the hypothetical train is \[0.6c\].
\[v = 0.6c\]
The length of the train for the observer on the platform is \[200\,{\text{m}}\].
\[L' = 200\,{\text{m}}\]
Let \[L'\] be the rest length of the train. The rest length of the train is given by
\[L = \dfrac{{L'}}{{\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} }}\]
Substitute \[200\,{\text{m}}\] for \[L'\] and \[0.6c\] for \[v\] in the above equation.
\[L = \dfrac{{200\,{\text{m}}}}{{\sqrt {1 - \dfrac{{{{\left( {0.6c} \right)}^2}}}{{{c^2}}}} }}\]
\[ \Rightarrow L = \dfrac{{200\,{\text{m}}}}{{\sqrt {1 - {{\left( {0.6} \right)}^2}} }}\]
\[ \Rightarrow L = \dfrac{{200\,{\text{m}}}}{{\sqrt {1 - 0.36} }}\]
\[ \Rightarrow L = \dfrac{{200\,{\text{m}}}}{{\sqrt {0.64} }}\]
\[ \Rightarrow L = \dfrac{{200\,{\text{m}}}}{{0.8}}\]
\[\therefore L = 250\,{\text{m}}\]
Hence, the rest length of the train is \[250\,{\text{m}}\].
(b) Let now calculate the length of the platform measured by the observers in the train.The length of the platform measured by the observers in the train is
\[L' = L\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} \]
Substitute \[200\,{\text{m}}\] for \[L\] and \[0.6c\] for \[v\] in the above equation.
\[L' = \left( {200\,{\text{m}}} \right)\sqrt {1 - \dfrac{{{{\left( {0.6c} \right)}^2}}}{{{c^2}}}} \]
\[ \Rightarrow L' = \left( {200\,{\text{m}}} \right)\sqrt {1 - {{\left( {0.6c} \right)}^2}} \]
\[ \Rightarrow L' = \left( {200\,{\text{m}}} \right)\sqrt {1 - 0.36} \]
\[ \Rightarrow L' = \left( {200\,{\text{m}}} \right)\sqrt {0.64} \]
\[ \Rightarrow L' = \left( {200\,{\text{m}}} \right)\left( {0.8} \right)\]
\[ \therefore L' = 160\,{\text{m}}\]
Hence, the length of the platform measured by the observers in the train is \[160\,{\text{m}}\].
Note: The students should not get confused between the different signs for the length of the train measured by the observer on the platform. For the first case, we have used the sign \[L'\] for length of the train measured by the observer on the platform and for the second case we have used sign \[L\] for the length of the train measured by the observer on the platform.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Why cannot DNA pass through cell membranes class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

