Exams › JEE Advanced › Physics

In a head-on (collinear) collision, a particle moving with initial speed v0 hits a stationary particle of equal mass. If the total kinetic energy after the collision is 50% more than the initial kinetic energy, what is the magnitude of the relative velocity of the two particles after the collision?

1. sqrt(2) v0
2. v0/2
3. v0/sqrt(2)
4. v0/4

Correct answer: sqrt(2) v0

Solution

Momentum gives v1+v2 = v0; energy gives v1²+v2² = 1.5 v0². Then (v1-v2)² = 2(1.5 v0²) - v0² = 2 v0², so |v1-v2| = sqrt(2) v0.

Related JEE Advanced Physics questions

• The work done on a particle of mass m by a force, K [x / (x² + y²)^(3/2) î + y / (x² + y²)^(3/2) ĵ] (K being a constant of appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the x-y plane is
• A skier starts from rest on a smooth curved ramp that is at a height of R/4 above the base of a smooth fixed hemisphere of radius R. After descending the ramp the skier moves up over the hemisphere. At what angle theta (measured from the vertical through the top of the hemisphere) does the skier leave the surface of the hemisphere?
• A small ball strikes obliquely a smooth horizontal surface. The components of its velocity just before impact are: horizontal component (along surface) = 4 m/s, vertical component (downward, perpendicular to surface) = 2 m/s. The coefficient of restitution between ball and surface is 1/2. Find the horizontal component vₓ and vertical component v_y of velocity just after impact.
• A block A of mass m is connected to block B of mass m on a spring of constant k. The system is released from rest when the spring is at its natural length, with block A hanging above block B (B is on the ground). After block A undergoes a perfectly inelastic collision with the ground (e = 0), what is the minimum height h from which the system must be released so that block B is subsequently lifted off the ground?
• The power delivered by a motor to a body of mass 2 kg is given by P = v*(5 - v) watts, where v is the speed in m/s. If the body starts from rest, find the work done on the body during the first 2 seconds.
• A block of mass 1 kg rests on a plank of mass 2 kg. A spring is compressed between the block and one end of the plank, with the other spring end attached to the plank. The entire system is initially at rest on a frictionless surface. When the thread holding the spring compressed is cut, the spring releases. At the instant the spring returns to its natural length, the block moves at 6 m/s. What was the elastic potential energy stored in the spring?

⚔️ Practice JEE Advanced Physics free + battle 1v1 →