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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.
A moving body of mass m and velocity 3 km/h collides with a rest body of mass 2m and sticks to it. Now the combined mass starts to move. What will be the combined velocity
Converting the given velocities to SI units, the initial velocity of the moving body is (3 km/h) = (3/3.6) m/s = 5/6 m/s. Using conservation of momentum, (m*(5/6) + 2m0) = 3mv', where v' is the final velocity of the combined mass. Solving this, we get v' = (5/18) m/s = 1 km/h.
If a skater of weight 3 kg has initial speed 32 m/s and second one of weight 4 kg has 5 m/s. After collision, they have speed (couple) 5 m/s. Then the loss in K.E. is
Loss in K.E. = (initial K.E. – Final K.E.) of system
= 986.5 J
A metal ball of mass 2 kg moving with a velocity of 36 km/h has an head on collision with a stationary ball of mass 3 kg. If after the collision, the two balls move together, the loss in kinetic energy due to collision is
By law of conservation of momentum
⇒ V = 4 m/s
Loss in K.E.
A body of mass 2kg is moving with velocity 10 m/s towards east. Another body of same mass and same velocity moving towards north collides with former and coalsces and moves towards north-east. Its velocity is
Initial momentum =
Final momentum = 2m × V
By the law of conservation of momentum
⇒
In the problem v = 10 m/s (given)
∴
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.