(C) The spring exerts an equal and opposite impulse on both carts,so the magnitude of the momentum $p$ imparted to each cart is the same $(p_1 = p_2 =

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$\frac{S_1}{S_2} = -\left(\frac{m_2}{m_1}\right)^2$

(C) The spring exerts an equal and opposite impulse on both carts,so the magnitude of the momentum $p$ imparted to each cart is the same $(p_1 = p_2 = p)$.Using the work-energy theorem,the work done by friction equals the change in kinetic energy: $\mu mg S = \frac{p^2}{2m}$.Rearranging for displacement $S$,we get $S = \frac{p^2}{2\mu g m^2}$.Since $p$,$\mu$,and $g$ are the same for both carts,the displacement $S$ is inversely proportional to the square of the mass: $S \propto \frac{1}{m^2}$.Therefore,the ratio of displacements is $\frac{S_1}{S_2} = \frac{m_2^2}{m_1^2} = \left(\frac{m_2}{m_1}\right)^2$.Since the displacements are in opposite directions,the ratio is $\frac{S_1}{S_2} = -\left(\frac{m_2}{m_1}\right)^2$.

Class 11 Physics Newton's Laws of Motion and Friction Mix Examples-Newton's Laws of Motion and Friction

Medium

$A$ spring is compressed between two toy carts of masses $m_1$ and $m_2$. When the toy carts are released,the spring exerts an equal and opposite force on each cart for the same time $t$. If the coefficient of friction $\mu$ between the ground and the carts is the same,find the ratio of the displacements of the two toy carts.

A
$\frac{S_1}{S_2} = \frac{m_2}{m_1}$
B
$\frac{S_1}{S_2} = \frac{m_1}{m_2}$
C
$\frac{S_1}{S_2} = -\left(\frac{m_2}{m_1}\right)^2$
D
$\frac{S_1}{S_2} = -\left(\frac{m_1}{m_2}\right)^2$

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Newton's Laws of Motion and Friction

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Mix Examples-Newton's Laws of Motion and Friction

$A$ block of mass $25 \ kg$ is pulled along a horizontal surface by a force at an angle $45^{\circ}$ with the horizontal. The friction coefficient between the block and the surface is $0.25$. The block travels at a uniform velocity. The work done by the applied force during a displacement of $5 \ m$ of the block is (in $J$)

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$A$ racing car travels on a track (without banking) $ABCDEFA$. $ABC$ is a circular arc of radius $2R$. $CD$ and $FA$ are straight paths of length $R$ and $DEF$ is a circular arc of radius $R = 100 \, m$. The coefficient of friction on the road is $\mu = 0.1$. The maximum speed of the car on straight paths is $50 \, m/s$. Find the minimum time for completing one round.

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$A$ block of mass $m$ is lying on a rough inclined plane having an inclination $\alpha = \tan^{-1}(\frac{1}{5})$. The inclined plane is moving horizontally with a constant acceleration of $a = 2 \text{ ms}^{-2}$ as shown in the figure. The minimum value of the coefficient of friction,so that the block remains stationary with respect to the inclined plane,is (Take $g = 10 \text{ ms}^{-2}$):

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Masses of $10\, kg$ and $20\, kg$ are connected by a massless spring as shown in the figure. $A$ force of $200\, N$ acts on the $20\, kg$ mass. At the instant shown,the $10\, kg$ mass has an acceleration of $12\, m/s^2$. What is the acceleration of the $20\, kg$ mass in $m/s^2$?

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Two blocks $A$ and $B$ of masses $5 \ kg$ and $10 \ kg$ respectively are placed as shown in the figure. The coefficient of friction between all the surfaces is $0.2$. $A$ force $F$ is applied on block $A$ at an angle of $30^{\circ}$ with the horizontal. If the system moves with constant acceleration,find the acceleration of block $B$. (Assume $g = 10 \ m/s^2$ and $F = 100 \ N$) (in $m/s^2$)

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Masses of $10\, kg$ and $20\, kg$ are connected by a massless spring as shown in the figure. $A$ force of $200\, N$ acts on the $20\, kg$ mass. At the instant shown,the $10\, kg$ mass has an acceleration of $12\, m/s^2$. What is the acceleration of the $20\, kg$ mass in $m/s^2$?

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