Two masses 2m and m are connected by an inextensible light string. The string is passing over a light frictionless pulley. The mass 2m is resting on a surface and mass m is hanging in air as shown in figure. A particle of mass m strike the mass m from below in case (I) with a velocity v0 and in case (II) strikes the mass m with a velocity v0 from top and sticks to it.

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Two masses 2m and m are connected by an inextensible light string. The string is passing over a light frictionless pulley. The mass 2m is resting on a surface and mass m is hanging in air as shown in figure. A particle of mass m strike the mass m from below in case (I) with a velocity v₀ and in case (II) strikes the mass m with a velocity v₀ from top and sticks to it.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

The ratio of velocities of mass m just after collision in first and second cases is 1/2.

The conservation of linear momentum can be applied in both the cases just before and just after collision

The ratio of velocities of mass m just after collision in first and second case is 4.

The conservation of linear momentum can be applied in case I but cannot be applied in case II just before and just after collision.

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

In case I, Tension is zero. So, conservation of momentum is valid.

$(m + m) v_{1} = {mv}_{0} \Rightarrow v_{1} = \frac{v_{0}}{2}$

In case, II, impulse of tension needs to be considered.

$- \int T d t = (m + m) v_{2} - m v_{0} . . . . . . . . . (1) \int T d t = (2 m) v_{2} . . . . . . . . . . . . (2) \Rightarrow v_{2} = \frac{v_{0}}{4}$