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A particle falls from a height h upon a fixed horizontal plane and rebounds. If e is the coefficient of restitution, the total distance travelled before rebounding has stopped is
A body of mass 5 kg moving with a velocity 10m/s collides with another body of the mass 20 kg at, rest and comes to rest. The velocity of the second body due to collision is
Momentum conservation
⇒ v = 2.5 m/s
A ball of mass m moving with velocity V, makes a head on elastic collision with a ball of the same mass moving with velocity 2V towards it. Taking direction of V as positive, velocities of the two balls after collision are-
Due to elastic collision of bodies having equal mass, their velocities get interchanged.
A spacecraft of mass 'M' and moving with velocity 'v' suddenly breaks in two pieces. After the explosion one of the mass 'm' becomes stationary. What is the velocity of the other part of craft
By the principle of conservation of linear momentum,
⇒ ⇒
A tennis ball is released from height h above ground level. If the ball makes inelastic collision with the ground, to what height will it rise after third collision
after third collision [as n = 3]
A sphere collides with another sphere of identical mass. After collision, the two spheres move. The collision is inelastic. Then the angle between the directions of the two spheres is
Angle will be 90° if collision is perfectly elastic.
A particle of mass m moving eastward with a speed v collides with another particle of the same mass moving northward with the same speed v. The two particles coalesce on collision. The new particle of mass 2m will move in the north-easterly direction with a velocity
In an elastic collision between two equal masses moving at right angles with equal speeds, the final velocity of the combined mass is v/sqrt(2) at an angle of 45° from the original directions. This result follows from the conservation of momentum and kinetic energy in the collision.
A bullet of mass a and velocity b is fired into a large block of mass c. The final velocity of the system is
In an inelastic collision where a bullet of mass a and velocity b strikes a large block of mass c, the final velocity of the system is (a/(a+c))b. This result is derived from the conservation of momentum, considering the initial and final momenta of the system.
A bag (mass M) hangs by a long thread and a bullet (mass m) comes horizontally with velocity v and gets caught in the bag. Then for the combined (bag + bullet) system
A particle of mass m moving with velocity v strikes a stationary particle of mass 2m and sticks to it. The speed of the system will be
In an inelastic collision where the two bodies stick together, the total momentum before and after the collision must be conserved. Let the initial velocities be v and 0. By conservation of momentum, (mv + 2m0) = (m+2m)*v', where v' is the final velocity of the combined system. Solving this, we get v' = v/3.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.