Question

In figure shown, both blocks are released from rest. The time to cross each other is
(A) 2 second
(B) 3 second
(C) 1 second
(D) 4 second ​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

u = 0 m/s.

Length of 4 kg block = 2m.

Length of 1 kg block = 4m.

Weight of 4kg block = 4 × g = 4g N.

Weight of 1kg block = 1 × g = 1g N.

Let the system has an acceleration "a".

(The formula used is Weight = mass × acceleration due to gravity)

I.e (W = mg).

$\huge{\bold{\underline{Explanation:-}}}$

Relative displacement (S)

As these are moving in opposite direction their relative displacement will get added up.

$\large{\bold{S = S_{4 \: kg \: block} + S_{1 \: Kg \: block}}}$

As sum of length of blocks will be required as they need to just cross each other.

$\large{S = 2m + 4m}$

$\large{\boxed{S = 6 \: meters}}$

F.B.D of block 4 kg.

Newton's second law

$\large{\bold{F = ma}}$

Substituting the values

$\large{4g - T = 4a}$

(Here the Force is F = (4g - T))

As acceleration is downwards.

Simplifying,

$\large{T = 4g - 4a ----(1)}$

F.B.D of block 1kg.

From Newton's second law

$\large{\bold{F = ma}}$

Substituting the values

$\large{T - 1g = 1a}$

(Here the Force is (F = T - 1g))

As acceleration is upwards.

Substituting the equation (1) in the above equation.

$\large{4g - 4a - 1g = 1a}$

Simplifying

$\large{3g = 5a}$

$\large{a = \frac{3g}{5}}$

$\large{\boxed{ a = 0.6 \: g}}$

Both the bodies move with the same acceleration but in opposite direction,

Therefore relative acceleration will get added up.

$\large{\bold{a_r = a + a}}$

$\large{\bold{a_r = 2a}}$

Substituting the values

$\large{a_r = 2 \times 0.6g}$

$\large{\boxed{a_r = 1.2 \: g}}$

From the second kinematic equation

$\large{\bold{S = ut + \frac{1}{2}at^2}}$

Substituting the values.

$\large{S = 0 + \frac{1}{2}a_rt^2}$

As initial velocity is zero.

$\large{6 = \frac{1}{2} \times 1.2 \times g \times t^2}$

As g = 10 m/s².

$\large{6 = \frac{1}{2} \times 1.2 \times 10 \times t^2}$

Simplifying,

$\large{6 = \frac{1}{2} \times 12 \times t^2}$

$\large{6 = \frac{1}{ \cancel{2}} \times \cancel{12} \times t^2}$

$\large{6 = t^2 \times 6}$

$\large{t^2 = \frac{6}{6}}$

$\large{t = \sqrt{1}}$

$\huge{\boxed{\boxed{t = 1 \: second}}}$

So, they will cross each other at t = 1 second(Option - c).

#refer the attachment for figure .