Question

# A system of two blocks A and B are connected by an inextensiible massless string as shown in figure. The pulley is massless and frictionless. Initially, the system is at rest. A bullet of mass m moving with a velocity u as shown in the figure hits block B and gets embedded into it. Then find out the impulse imparted by tension force to the block of mass 3m, immediately after the collision takes place.

A
54mu
B
45mu
C
25mu
D
35mu

Solution

## The correct option is D 35muTension in both strings will be equal, hence impulse due to tension on both blocks will be the same. J1=J2=−J (assuming −ve y direction upwards) After collision, mass of block B becomes 2m due to embedded bullet. ⇒The system of (blocks+bullet) will move with common speed v, to satisfy string constraint. Block A moves upwards with speed v and B downwards with speed v Applying impulse-momentum theorem on both blocks, considering vertically downwards as +ve y axis: For block A: −J=Pf−Pi=3m(vf−vi) =3m(−v−0)=−3mv ...(i) For block B: −J=Pf−Pi=2m(v)−[(m×0)+(m×u)] =2mv−mu ...(ii) Equating Eq. (i) and (ii): −3mv=2mv−mu ⇒5mv=mu ∴v=u5 Hence, impulse on block A: =−J=2mv−mu=25mu−mu =−35mu ∴|J|=35muPhysics

Suggest Corrections

2

Similar questions
View More

People also searched for
View More