Work Energy And Power Question 136

Question: A spring is compressed between two toy carts of masses $ m_1 $ and $ m_2 $ . When the toy carts are released, the spring exerts on each toy cart equal and opposite forces for the same small time $ t $ . If the coefficients of friction $ \mu $ between the ground and the toy carts are equal, then the magnitude of displacements of the toy carts are in the ratio

Options:

A) $ \frac{S_1}{S_2}=\frac{m_2}{m_2} $

B) $ \frac{S_1}{S_2}=\frac{m_1}{m_2} $

C) $ \frac{S_1}{S_2}={{( \frac{m_2}{m_1} )}^{2}} $

D) $ \frac{S_1}{S_2}={{( \frac{m_1}{m_2} )}^{2}} $

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Minimum stopping distance =s Work done against the friction

$ =W=\mu mgs $ Initial momentum gained by both toy carts will be same because same force acts for same time initial kinetic energy of the toy cart = $ ( \frac{p^{2}}{2m} ) $

Therefore, $ \mu mgs=\frac{p^{2}}{2m} $ or $ s=( \frac{p^{2}}{2\mu gm^{2}} ) $ For the two toy carts, momentum is numerically the same.

Further $ \mu $ and g are the same for the toy carts. So, $ \frac{S_1}{S_2}={{( \frac{m_2}{m_1} )}^{2}} $



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