Find ratio of the displacement of block A along x and y-directions S y S x after the time block A leaves the surface of B till the time it reaches ground.

On B: 2 5 m g − f = 2 m a 2 with f = μ m g = m a 1 , so a 1 = 2 g and a 2 = g . L = 2 1 ( a 2 − a 1 ) t 2 ⇒ t = 2 g L ; hence v A = 2 g × 2 g L = gL = 10 m/s. Fall time: t ′ = g 2 h ; s x = v A t ′ = 10 g 2 h = 1 m. Wind gives a y = 2 g : s y = 2 1 ⋅ 2 g ⋅ g 2 h = 2 h = 4 1 m. s y s x = 4 - (C)

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PhysicsNLMChallenging application of NLMHard2 minQB

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Passage / Comprehension

Two blocks A and B of masses m and 2 m are initially at rest. Length of block B is L and block A is placed at the right end corner of block B and the friction coefficient between them is $μ = 1/2$ . At $t = 0$ a constant force $F = \frac{5 mg}{2}$ begins to act on block B towards right. Just when the block A leaves B, wind begins to blow along y-direction which exerts a constant force $\frac{m g}{2}$ on A. Assume the size of block A is small compared to B and neglect any rotational effects and toppling of block B. (Given $h = 1/2 m, L = 1 m$ and $g = 10 m / s^{2}$ )\\

Find ratio of the displacement of block A along x and y-directions $\frac{S _{x}}{S _{y}}$ after the time block A leaves the surface of B till the time it reaches ground.

Options:

Answer:

Solution:

On B: $\frac{5 m g}{2} - f = 2 m a_{2}$ with $f = μ m g = m a_{1}$ , so $a_{1} = \frac{g}{2}$ and $a_{2} = g$ .

$L = \frac{1}{2} (a_{2} - a_{1}) t^{2} \Rightarrow t = 2 \frac{L}{g}$ ; hence $v_{A} = \frac{g}{2} \times 2 \frac{L}{g} = gL = 10$ m/s.

Fall time: $t^{'} = \frac{2 h}{g}$ ; $s_{x} = v_{A} t^{'} = 10 \frac{2 h}{g} = 1$ m.

Wind gives $a_{y} = \frac{g}{2}$ : $s_{y} = \frac{1}{2} \cdot \frac{g}{2} \cdot \frac{2 h}{g} = \frac{h}{2} = \frac{1}{4}$ m.

$\frac{s _{x}}{s _{y}} = 4$ - (C)

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Challenging application of NLM

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