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The potential energy of a projectile at its highest point is (1/2)th the value of its initial kinetic energy. Therefore its angle of projection is ......
$ Hmax = { V_o^2 Sin^2 \theta_0 \over 2g } , U = mg Hmax = { mv_0^2 sin^2 \theta_0 \over 2} = {1 \over2} Ko $
The potential energy of 2kg particle, free to move along x axis is given by $ U (X) = \left( { x^4 \over 4 } - { x^2 \over 2} \right) J $ .. If its mechanical energy is 2 J, its maximum speed is……….. m/s
K.E. is maximum than P.E. minimum. $ so { du \over dx } = 0 \Rightarrow x = 0 OR $ $ For x = \pm 4 U(x) = - { 1 \over 4} = Umin$
If the K.E. of a body is increased by 44%, its momentum will increase by.......
$E_1 = E and E_2 = 1.44E $ $ Than P \alpha \sqrt E $
A bullet of mass 0.10 kg moving with a speed of 100 m/s enters a wooden block and is stopped after a distance of 0.20m. what is the average resistive force exerted by the block on the bullet ?
$Fd = {1 \over 2} m ( v_2^2 - v_2^2 ) $
A sphere of mass m moving the velocity v enters a hanging bag of sand and stops. If the mass of the bag is M and it is raised by height h, then the velocity of the sphere was
By conservation of linear momentum$ \Rightarrow mv = (m+M) V_{sys} $ By convervation of low of energy. $ {1 \over 2} (m+M) v^2 sys = (m+M)gh $ Find Vsys and put it in sin egn(1)
If the water falls from a dam into a turbine wheel 19.6m below, then the velocity of water at the turbine is ...... $( g = 9.8 m/s^2) $
P.E. of water = K.E. at turbine. mgh =$ {1 \over 2} mv^2 $
An ice-cream has a marked value of 700kcal. How many kilo-watt-hour of energy will it deliver to the body as it is digested (J = 4.2 J/cal)
$ 1 k cal = 10^3 calorie = 4200J = { 4200 \over 3.6 \times 10^6 } kwh $ $\therefore 700K cal = { 700 \times 4200 \over 3.6 \times 10^6 } kwh $
A bomb of mass 10 kg explodes into 2 pieces of mass 4 kg and 6 kg. The velocity of mass 4 kg is 1.5 m/s, the K.E. of mass 6 kg is .......
$ m_1 v_1 + m_2 v_2 =0 and ,than K.E = {1 \over 2} m_2 v_2^2 $
A bomb of mass 3.0 kg explodes in air into two pieces of masses 2.0 kg and 1.0 kg. The smaller mass goes at a speed of 80m/s. The total energy imparted to the two fragments is
$ m_1 v_1 = m_2 v_2 =0 $ $ Total energy of system = {1\over 2}m_1 v_1^2 + {1 \over 2} m_2 v_2^2 $
The bob of simple pendulum (mass m and length l) dropped from a horizontal position strike a block of the same mass elastically placed on a horizontal frictionless table. The K.E. of the block will be
The collision between bob and block is elastic P.E.= K.E.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.