JEE Advanced (Fill in the Blanks): Momentum & Impulse | Chapter-wise Tests for JEE Main & Advanced PDF Download

Fill in the Blanks

Q.1. A particle of mass 4 m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with a speed v each in mutually perpendicular directions. The total energy released in the process of explosion is ............              (1987 - 2 Marks)

Ans. 

Solution. 2mv' cos θ = mv   ... (i) 

2mv' sin θ = mv

Putting this value in equation (i), we get

Q.2. The magnitude of the force (in newtons) acting on a body varies with time t (in micro seconds) as shown in the fig  AB, BC and CD are straight line segments. The magnitude of the total impulse of the force on the body from t = 4 ms to t = 16ms is ...............Ns.                     (1994 - 2 Marks)

Ans.0.005 Ns

Solution. KEY CONCEPT : Area under the F – t graph gives the impulse imparted to the body.

The magnitude of total impulse of force on the body from
t = 4 µs to t = 16 µs
= area (BCDFEB)
= area of  BCFEB + area CDFC

Integer Value Correct Type

Q.1. A bob of mass m, suspended by a string of length l₁, is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspended by a string of length l₂, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio

Ans. 5

Solution. Velocity at the highest point of bob tied to string ℓ₁ is acquired by the bob tied to string ℓ₂ due to elastic head-on collision of equal masses