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}} $
BETA
