Two particles of masses m and 2m are connected by a string of length L and placed at rest over a smooth horizontal surface. The particles are then given velocities as indicated in the figure shown. The tension developed in the string will be

Question

Two particles of masses m and 2m are connected by a string of length L and placed at rest over a smooth horizontal surface. The particles are then given velocities as indicated in the figure shown. The tension developed in the string will be <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-39rfr8j9bsuz.png" /> (A) (m v2/2 L) (B) (3 m v2/4 L) (C) (4 m v2/3 L) (D) (8 m v2/3 L)

Tardigrade Admin · Accepted Answer

Vc=(2 m v - m v/(. 2 m + m .))=(v/3) Vr e l=v-(v/3)=((2 v/3)) r1=(m L/3 m)=(L/3) ∴ T=((2 m) × ((2 v/3))2/((L)3))=(8 m v2/3 L)

Q. Two particles of masses $m$ and $2 m$ are connected by a string of length $L$ and placed at rest over a smooth horizontal surface. The particles are then given velocities as indicated in the figure shown. The tension developed in the string will be

$\frac{m v ^{2}}{2 L}$

$\frac{3 m v ^{2}}{4 L}$

$\frac{4 m v ^{2}}{3 L}$

$\frac{8 m v ^{2}}{3 L}$

$V_{c} = \frac{2 m v - m v}{( 2 m + m )} = \frac{v}{3}$
$V_{re l} = v - \frac{v}{3} = (\frac{2 v}{3})$
$r_{1} = \frac{m L}{3 m} = \frac{L}{3}$
$∴ T = \frac{( 2 m ) \times ( \frac{2 v}{3} ) ^{2}}{( \frac{L}{3} )} = \frac{8 m v ^{2}}{3 L}$

Q. Two particles of masses m and 2m are connected by a string of length L and placed at rest over a smooth horizontal surface. The particles are then given velocities as indicated in the figure shown. The tension developed in the string will be

2Lmv2​

4L3mv2​

3L4mv2​

3L8mv2​