302 - Mechanics part 1 Questions Answers
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
A bullet of mass m passes through a pendulum bob of mass M with velocity v and comes out of it with a velocity v/2. The minimum value of v so that tha bob mass complete one resolution in the vertical circle. (l= length of pendulum)
for completing revolution velocity of the bob at lowest point should be √5gl
so by momentum conservation mv = M√5gl + mv/2
now solve
A ball moving with speed v collides with a horizontal smooth surface at an angle ϴ with normal to surface. If coefficient of restitution is ‘e’ , then find velocity after rebounce making same angle theta with the normal.
let angle before collision is φ and after collision is θ with vertical then
for vertical direction velocity after rebound = evcosφ
for horizontal direction velocity after rebound = vsinφ
now use pythagorus theorem