SATHEE: Work Energy And Power Question 136

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 $μ$ 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

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

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

Further $μ$ and g are the same for the toy carts. So, $\frac{S_{1}}{S_{2}} = {(\frac{m_{2}}{m_{1}})}^{2}$

Work Energy And Power Question 136

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Answer:

Solution:

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Work Energy And Power Question 136

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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