575 - Physics Questions Answers
A shell at rest on a smooth horizontal surface explodes into two fragments of masses m1 and m2. If just after explosion m1 moves with speed u, then work done by internal forces during explosion is
By momentum conservation velocity of m2 = m1u/m2
now sum of kinetic energies = Net work done by internal forces
A body of mass m falls from height h on ground. If 'e' be the coefficient of restitution of collision between the body and ground, then the distance travelled by body before it comes to rest is
if a body falls from height h then height attained after rebound will be e2h
by using this fact
total distance covered before stop = h+2e2h+2e2e2h+2e2e2e2h+..................................................
so d = h+2e2h[1+e2+e4+e6+.........................]
now solve
A particle is moving in a circular path of radius r under the action of a force F. If at an instant velocity of particle is v, and speed of particle is increasing, then
(i) F.v=0 (ii) F.v > 0 (iii) F.v < 0
centripetal acc. is the need for circular motion but speed is increasing so tangential acc will also be present so angle between velocity and acc will be acute so F.v > 0
A ball of mass m moving with speed v collides with a smooth horizontal surface at angle θ with it as shown in figure. The magnitude of impulse imparted to surface by ball is
From where θ is measured, horizontal or vertical ?
A particle of mass m moving from origin along x-axis and its velocity varies with position x as v= k√x. The work done by force acting on it during first 't' seconds is
v = k√x
so a = dv/dt = [k/2√x]*dx/dt = [k/2√x]*v = [k/2√x]*[k√x] = k2/2 = constant
and dx/dt = k√x
so dx/√x = kdt
integrate within o to t
finally work = Fx because F is constant
A particle of mass m is projected with speed u at angle θ with horizontal from ground. The work done by gravity on it during its upward motion is
H = u2sin2θ/2g
so work done by gravity = mgH
A body of mass m is projected from ground with speed u at an angle θ with the horizontal. The power delivered by gravity to it at half of maximum height from the ground is
at half height, vertical velocity v2 = u2sin2θ - 2gH/2, from this equation calculate v
then P = F.v = -mg* vertical component of velocity
this will be instataneous power
A ball of mass m moving with speed v collides perfectly inelastically with another ball of mass m at rest. The magnitude of impulse imparted to the first ball is
inelastic collision means velocity of both ball will be same after collision
so by momentum conservation mv = 2mv' so v' = v/2
now initial momentum + impulse imparted = final momentum
mv + I = mv/2
so I = -mv/2
Two balls of equal mass undergo head on collision while each was moving with speed 6 m/s. If the coefficient of restitution is 1/3 , the speed of each ball after impact will be
according to the given condition u1 = 6, u2 = -6
let after collision velocities are v1, v2
then by momentum conservation m1u1 + m2u2 = m1v1 + m2v2
and v2 - v1 = e(u2 - u1)
now solve
The position x of a particle moving along x-axis at time t is given by the equation t = √x + 2 , where x is in metres and t in seconds. Find the work done by the force in first four seconds.
t = √x + 2
implies x = (t-2)2
so dx/dt = v = 2(t-2)
and dv/dt = a = 2
so F = ma = 2m
and x(initial) = 4 m
x (final) = 4 m
so displacement = 0 m
so work = 0 J