A bullet of mass m moving with a horizontal velocity n strikes a stationary block of mass M suspended by a spring of length L. The bullet gets embedded in the block. Prove that the maximum angle made the string after impact is
The situationn is shown in figure. Let n' be the velocity of the block with the embedded bullet after impact.
Applying the law of conservation of momentum, i.e.,
Momentum before impact = Momentum after impact
mn = (M + m)n'......(1)
Applying the law of conservation of mechanical energy, i.e., K.E. of the combined mass = P.E. at the highest point
h = L -L cosq = L(1-cosq).
From equations (2) and (3), we get
Substituting the value of n' from equation (1), we get
A sphere has a perfectly elastic oblique collision with another identical sphere which is initially at rest. The angle between their velocities after the collision is
Law of conservation of momentum gives
mu1 = mv1cosq1 + mv2cosq2 .......(1)
and 0 mv1sinq1 - mv2sinq1 ....(2)
Energy conservation gives
Squaring and adding (1) and (2) and comparing with (3) we get θ1 + θ2 = 90o
A girl throws a ball with initial velocity u at an inclination of 45o. The ball strikes the smooth vertical wall at a horizontal distance d from girl and after rebounding returns to her hand. What is the coefficient of restitution between wall and ball?
As the wall is smooth, the vertical component of the impulse it receives is zero. The total time of flight is given by
The time of flight t1 from G to A =
The time of flight l2 from A to G =
where e = coefficient of restitution.
Solving we get e =
A uniform chain of mass m and length I hangs on a thread and touches the surface of a table by its lower end. Find the force exerted by the table on the chain when half of its length has fallen on the table. The fallen part does not form heap.
Weight of the portion BC of the chain lying on the table, W = mg/2(downwards)
F1 = lv2 (where, λ = m/l, is mass per unit length of chain)
Net force eerted by the chain on the table is
So, from Newton's thrid law the force exerted by the table on the chain will be 3/2 mg (vertically upwards).
It l1 , l2 and l3 are the moments of inertia about, a diameter of a thin hollow sphere, solid sphere and disc having same mass and same radius respectively, then which one of the following is correct?
m = mass
r = radius
A uniform rod of length l is free to rotate in a vertical plane about a fixed horizontal axis through B. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle q, its angular velocity w is given as
When the rod rotates through an angle q, the centre of gravity falls through a distance h from DBG'C,
The decrease in P.E. is equal to the kinetic energy of rotation
From equation (1) and (2), we get
Two bodies with moment of inertia l1 and l2 and (l1>l2) have equal angular momenta. If K.E. of rotation are E1 and E2 then
Frin eqs, (1) and (2), we get
The object in Figure might be a billiard ball a steel ball bearing, or the world's largest ball of string. Find the moment of the inertia about an axis through the center of the sphere.
We divide the sphere into thin disks of thickness dx. The radius r or the disk shown is
Its volume is
dV = pr2 dx = p(R2-x2)dx
and its mass is
dm = r dV = pr (R2-x2) dx
We will find the moment of inertia of a sphere about an axis through its center
The mass element is a disk with thickness dx.
The moment of inertia of a disk of radius r and mass dm is
Integrating this expression from x = 0 to x = R gives the moment of inertia of the light hemisphere. From symmetry, the total I for the entire sphere is just twice of this:
Carrying out the integration, we obtain
The mass M of the sphere of the volume V = 4pR3/3 is
By comparing the expression for I and M, we find
Figure shows a hollow, uniform cylinder with length L, inner radius R1, and outer radius R2, It might be a steel cylinder in a printing press or a sheet- steel rolling mill. The moment of inertia about the axis of symmetry of the cylinder is,
We choose a thin cylindrical shell of radius r, thickness dr, and length L as a volume element. All parts of this element are at very nearly the same distance from the axis. The volume of the element is very nearly equal to that of a flat sheet with thickness dr, length L, and width 2pr(the circumference of the shell).
dm = p dV = p (width x length x thickness)
= r(2prL dr) .....(1)
The moment of inertia is given by
If we express the moment of inertia in terms of the total mass M of the body.
Q m = rV ......(3)
and the volume of the cylindrical shell is
[Volume of the cylinder V = pR2h] ...(4)
From equation (3) and (4) we get
using equation (5) in equation (2), we get
Moment of inertia of a solid cylinder of mass M height 2h and radius 2r about an axis (shown in the figure by dahsed line) passing through its centre of mass and perpendicular to its symmetry axis is
Let radius & height of solid cylinder is 2r and 2h. Assume that it is made up by small disc having radius 2r. Now we take a disc of width of dx and at x distance from zz' axis volume of thin disc is = π(2r)2dx
so, mass dm =
where is mass per unit volume
so moment of inertia about the diameter AB is
By theorem of "parallel axes" moment of inertia w.r.t. axis xz is
So, Total moment of inertia,
Consider the following statements with reference to X-rays:
1. They produce heat when absorbed by matter.
2. They are generated when fast-moving electrons strike a metal target.
3. They can penetrate through a thin sheet of aluminium.
Which of the statements given above are correct?
The following statements are reference to X-rays:
1. They produce heat when absorbed by matter : X-ray beam attenuation. As the x-ray beam passes through tissue, photons get absorbed so there is less energy; this is known as attenuation. It turns out that higher energy photons travel through tissue more easily than low-energy photons (i.e. the higher energy photons are less likely to interact with matter).
2. They are generated when fast-moving electrons strike a metal target : An electromagnetic wave called x-ray is produced when fast moving electrons impact on high atomic weight substances. However, when these fast moving electrons strike on a metal target, a photoelectric effect occurs.
3. They can penetrate through a thin sheet of aluminium : X-Rays will Pass Straight through Aluminium of Almost ANY Thickness as it's Density is so Low !! But the Crystalline Structure of Diamond makes it Totally Opaque to X-Rays !! X-Rays do not "see"! But yes, they can pass easily through a single sheet of aluminum foil.
An elastic collision conserves.
We know that in elastic collision both kinetic energy and momentum remains conserved.
Two particles A and B initially at rest, moves towards each other under a mutual force of attraction. If at an instant the speed of A is V and that of B is 2V then the speed of the centre of mass of the system is
According to principle of Conservation of momentum under mutual attractive force. The momentum of centre of mass remains constant, since the system is initially at rest, hence is momentum is zero.
Radioactive decays into by emitting a positron and a neutrino. The positron and the neutrino are observed t move at right angles to each other and carry momenta 2 x 10-22 and 5 x 10-23 kg msec-1 respectively. Then the momentum of the recoiling nucleus is
Let the positron more along the +ve x-direction and the neutrino along +y direction with momenta
A uniform chain of mass m and length I hangs on a thread and touches the surface of a table by its lower end. find the force exerted by the table on the chain when half of its length has fallen on the gable. The fallen part does not from heap.
Weight of the portion BC of the chain lying on the table W = mg/2 (downwards)
is mass per unit length of chain
Net force exerted by the chain on the table is
So, from Newton's third law the force exerted by the table on the chain will be
3/2 mg (vertically upwards)
Two objects of masses m1 = 200 gm and m2 = 500gm possess velocities just prior to a collision during which they become permanently attached to each other.
(b) Final momentum = (m1+m2) V = 0.7V =
Now applying the law of conservation of momentum
where v is the velocity of the combined particle in the laboratory system. Note that v is identical to V, i.e.,
A particle is released at x = 1 in a force field which one of the following statement is true?
Since, is exact differential work done by this force is independent of path. So the force is conservative
Torque about origin is zero, so angular momentum of particle is constant about origin.
Equilibrium position of particle is given by
F(n) = 0
so potential energy is minimum at x = 3 1/3
So, x = 3 1/3 is point of stable equilibrium.
Hence, the particle will move towards x = 1
A particle is acted upon by a force Which of the following statements are false?
If a force or a vector quantity is conservative the
And if is conservative then a scalar potential v exist such that
⇒ v = -xyz
The stream function of two dimensional flow field is given by
Stream function ψ is given by
Equation of streamline is given by
where u and v are velocity components of fluid velocity
So, velocity of fluid
The flow is irrotational if
Since, the velocity of fluid can be expressed as gradient of scaler potential
A container of large uniform cross-sectional area A resting on a horizontal surface has two immiscible, non-viscous and incompressible liquids of densities d and 2d, each of height H/2 as shown. The lower density liquid is open to the atmosphere having pressure P0. A tiny hole of area s(s<<A) is punched on the vertical side of the container at a height h (h < H/2). Determine:
Here the effective height of orifice cannot be taken as (H-h) due to the presence of two liquids. Using Berrnoulli,s theorem just inside and just outside the orifice, we have
where V is the velocity of efflux
2. Let be the time taken by the liquid to hit the ground. Then
The horizontal range x = Vt
[use equation (1) and (2)]
Differentiate x with respect to h and equate it to zero for maximum h.
Value of h put in equation(3)
At what time a Foucault's pendulum completes a round when it is at 450N latitude on north pole?
The period of oscillation of Foucalt's pendulum at the place of latitude is given as
at l = 450
An astronaut travelling in a rocket moving vertically upwards with an acceleration of 4g weighs 300kg. Then find out the weight (in Kg) of the astronaut in the laboratory.
The apparent weight is given by
Taking downward direction positive, the force in vertically upward direction is
F = mg + ma0
Therefore the weight of the astronaut in the aboratory is 60kg.
A solid sphere of mass 2m and radius a/2 is rolling with a linear speed v on a flat surface without slipping. The magnitude (in mav) of the angular momentum of the sphere w.r.t. a point along the path of the sphere on the surface is....
A solid sphere is rolling along the flat surface without slipping and p is the point on the surface of sphere.
So, angular momentum of sphere about P is L = 2mv * a/2
A bullet of mass 10 g. is fired with a speed of 1,000 m/s. from a freely hanging gun of mass 2 kg. Calculate the recoil velocity of gun (in m/s).
The initial momentum of the system (gun + bullet) = 0
Final momentum of the system = m1n1 + m2n2
where m1 and m2 are the masses of the gun and bullet respectively and n1 and n2 are their velocities finally.
According to principle of conservation of linear momentum.
Initial momentum = Final momentum
The gun will move in opposite direction to that of the bullet with velocity 5 m/s.
Two skaters collide and embrace, in a completely in elastic collision. That is they stick together after impact, as suggested by figure, where the origin is placed at the point of collision. Alfred, whose mass mA is 83 kg, is originally moving east with speed nA = 6.2 km/h. Barbara, whose mass mB is 55 kg is originally moving north with speed nB = 7.8 km/h.
What is the velocity (km/hr) v of the couple after impact?
Linear momentum is conserved during the collision. We can write, for the linear momentum components in the x and y directions,
in which M = mA + mB. Dividing equation (2) by equation (1) yields
From equation (2) we then have
A block of mass 0.18 kg is attached to a spring of force constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially, the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The bock slides a distance of 0.06m and comes to rest for the first time. The initial velocity of the block in m/sec is V = N/10. Then N is....
It is given that the block slides a distance of 0.06m and comes to the rest for the first time.
μ = 0.1
By the energy conservation
A bob of mass m, suspended by a string of length l1, is give a minimum velocity required to complete a full circle in the verticle plane. At the highest point, it collides elastically with another bob of mass m suspended by a string of length l2, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in verticle plane, the ration l1/l2 is ...
To complete the vertical circle
l1/l2 = 5
A ball of mass m moving at a speed v makes a head-on wllision with an identical ball at rest. The kinetic energy of the balls after the collision is 80% of the original. Find the coefficient of restilation.
A rocket, with an initial mass of 3000kg, is launched vertically upwards from rest under gravity. The rocket burns fuel at the rate of 40 kg per deeond. The burnt watter is ejected vertically downwards with a speed of 4000 m/see relative to the rocket. If burning stops after one minute. Find the maximum velocity of rocket.
We know that velocity equation is givven by
The maximum velocity of the rodect is
v = 4000 ln 5 - 600
= 5837.75 m/sec
A uniforem chain of mass m and length I hangs on a thread and touches the surface of a lable by its lower end. Find the force exerted by the table on the chain when three-founth of its length has ballen m the lable. The fallen part does not form heap.
Weight of the portion BC of the chein
Net force exerted by the chain on the table is
The force enerfed by the table on the chain will be 5/4 mg.