Two point masses connected by an ideal string are placed on a smooth horizontal surface as shown in the diagram. A sharp impulse of 10 kgms- 1 is given to the 5 kg mass at an angle of 60° to the line joining the masses. The velocity of the 10 kg mass just after the impulse will be

Question

Two point masses connected by an ideal string are placed on a smooth horizontal surface as shown in the diagram. A sharp impulse of 10 kgms- 1 is given to the 5 kg mass at an angle of 60° to the line joining the masses. The velocity of the 10 kg mass just after the impulse will be <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-pryazmqt1mzltnvg.jpg" /> (A) (2/3) ms- 1 (B) (1/3) m s- 1 (C) 2 ms- 1 (D) zero

Tardigrade Admin · Accepted Answer

Apply impulse equation, along string <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-pryazmqt1mzlaw0j.jpg" /> Impulse = Δ p 10 × (1/2)=(10 + 5)v v=(1/3)ms- 1

$\frac{2}{3} m s^{- 1}$

$\frac{1}{3} m s^{- 1}$

$2 m s^{- 1}$

zero

Apply impulse equation, along string

Impulse = $Δ p$
$10 \times \frac{1}{2} = (10 + 5) v$
$v = \frac{1}{3} m s^{- 1}$

Q. Two point masses connected by an ideal string are placed on a smooth horizontal surface as shown in the diagram. A sharp impulse of 10kgms−1 is given to the 5kg mass at an angle of 60∘ to the line joining the masses. The velocity of the 10kg mass just after the impulse will be

32​ms−1

31​ms−1

2ms−1

zero

Apply impulse equation, along string Impulse = Δp 10×21​=(10+5)v v=31​ms−1

$\frac{2}{3} m s^{- 1}$

$\frac{1}{3} m s^{- 1}$

$2 m s^{- 1}$

Apply impulse equation, along string

Impulse = $Δ p$
$10 \times \frac{1}{2} = (10 + 5) v$
$v = \frac{1}{3} m s^{- 1}$