Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Two carts of masses 200 kg and 300 kg on horizontal rails are pushed apart. Suppose the coefficient of friction between the carts and the rails are the same. If the 200 kg cart travels a distance of 36 m and stops, then the distance travelled by the cart weighing 300 kg is:

32 m

24 m

16 m

12 m

Step-by-Step Solution

Conservation of Momentum: When the carts are pushed apart, the total momentum of the system remains zero. Thus, the magnitudes of their momenta are equal: $p_1 = p_2$ $m_1 v_1 = m_2 v_2$ [Source 61].
Work-Energy Theorem with Friction: The kinetic energy acquired by each cart is dissipated by the work done against friction to bring it to a stop. $K = \text{Work} = f \cdot s$ Since friction $f = \mu N = \mu mg$ , we have: $\frac{1}{2}mv^2 = \mu m g s$ Alternatively, expressed using momentum ( $K = p^2/2m$ ): $\frac{p^2}{2m} = \mu m g s$ $s = \frac{p^2}{2 \mu g m^2}$ [Source 77, 78].
Proportionality: Since $p$ , $\mu$ , and $g$ are constant for both carts: $s \propto \frac{1}{m^2}$ $\frac{s_2}{s_1} = \left( \frac{m_1}{m_2} \right)^2$
Calculation: Given $m_1 = 200 \text{ kg}$ , $s_1 = 36 \text{ m}$ , and $m_2 = 300 \text{ kg}$ : $s_2 = 36 \times \left( \frac{200}{300} \right)^2$ $s_2 = 36 \times \left( \frac{2}{3} \right)^2 = 36 \times \frac{4}{9} = 16 \text{ m}$

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