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From an automatic gun a man fires 240 bullet per minute with a speed of 360 km/ h. If each weighs 20 g, the power of the gun is
Power of gun = Total K.E.of fired bullet /time = $ n \times {1 \over 2 } mv^2 /t $
A body of mass M is moving with a uniform speed of 10 m/s on frictionless surface under the influence of two forces $ F_1 and F_2 $ . . The net power of the system is $ F_1 ---> M <---F-2 $
Work done by forces = 0 ;W = p/t
A body is moved along a straight line by a machine delivering a constant power. The velocity gained by the body in time t is proportional to.....
$ { 1 \over 2} mv^2 = kt $ ( K,constant t ) , $ v = \sqrt { 2k \over m } t^{1/2} $, $ dx = \sqrt { 2k \over m } t^{1/2} dt $ Take integration
The coefficient of restitution e for a perfectly elastic collision is
The coefficient of restitution (e) measures the elasticity of a collision between two bodies. For a perfectly elastic collision, e is equal to 1. This means that there is no loss of kinetic energy in the collision, and the bodies rebound with the same relative velocity with which they approached each other.
Two balls at same temperature collide. What is conserved
when two balls at the same temperature collide some fraction of their kinetic energy appear in other forms of energy like heat energy,sound energy.hence neither temperature,not velocity or kinetic energy will remain conserved.the only quantity which will remain conserved is their momentum.
A rubber ball is dropped from a height of 5 m on a planet where the acceleration due to gravity is not known. On bouncing, it rises to 1.8 m. The ball losses its velocity on bouncing by a factor of
$ { v_2 \over v_1 } = \sqrt {h_2 \over h_1} $; i.e. fractional loss in velocity $ = 1 - { v_2 \over v_1} $
Two solid rubber balls P and Q having masses 200 g and 400 g respectively are moving in opposite directions with velocity of P equal to 0.3 m/s. After collision the two balls come to rest, then the velocity of Q is
$ m_p v_p + m_Q v_Q = 0 $
A ball is allowed to fall from a height 20m . If there is 30% loss of energy due to impact, then after one impact ball will go up to
mgh ' = 70% of mgh
If a skater of weight 4 kg has intial speed 40 m/s and 2nd one of weight 6 kg has 6 m/s. After collision, they have speed (couple) 6 m/s. Then the loss in K.E. is.....
Loss in K.E. = (initial K.E. - Final K.E.) of system = $ { 1 \over 2} m_1 u_1^2 + { 1 \over 2} m_2 u_2^2 - {1 \over 2( m_1 +m_2 )v^2 $
A metal ball of mass 2 kg moving with a velocity of 36 km/h has a 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 low of conservation of momentum $ 2 \times 10 = (2+3) v $ $ \therefore v = 4 m/s $ Loss in K.E = $ { 1 \over 2} (2) (10)^2 - { 1 \over 2} (5) (4) ^2 $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.