IIT JEE , AIEEE Forum - Physics , Mechanics

ashok_p_ap

08.02.2008(sent)

Physics

1. A uniform chain of mass M & length L is held vertically in such a way that its lower end just touches the horizontal floor.

The chain is released from rest in this position. Any portion that strikes the floor comes to rest. Assuming that the chain does not form a heap on the floor, calculate the force exerted by it on the floor when a length ?x? has reached the floor.

Ans.3Mgx/l

2.Two balls having masses m and 2m are fastened to two light strings of same length l. The other ends of the strings are fixed at O. The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic.

(a) Find the velocities of the balls just after the collision..

(b) How high will the balls rise after the collision.

Ans.(a) Light ball?50gl/ towards left, heavy ball ?2gl /3 towards right.

(b) Light ball 2l, heavy ball l/9.

m 2m

ß-----------àß-----------------à

l l

3. A gun is mounted on a railroad car. The mass of the car, the gun, the shells and the operator is ?50 m? where ?m? is mass of one shell. If the muzzle velocity of the shells is 200m/s, what is the recoil speed of the car after the second shot? Neglect friction.

Ans. 200(1/49+1/48)m/s

4. A ball of mass is dropped onto a floor from a certain height. The collision is perfectly elastic. And the ball rebounds to the same height and again falls. Find the average force exerted by the ball on the floor during a long time interval?

Ans. mg

5. The structure of a water molecule is shown in figure. Find the distance of the centre of mass of the molecule from the centre of oxygen atom.

O

104^o 0.96 x 10^-10 m

0.96 x 10-¹⁰ m

H H

6. 2 blocks A and B, each having a mass of 320 g connected by a light string passing over a smooth light pulley. The horizontal surface on which the block A can slide is smooth. The block A is attached to a spring of spring constant 40N/m whose other end is fixed to a support 40cm above the horizontal surface. Initially, the spring is vertical and unstretched

when the system is released to move. Find the velocity of the block A at the instant it breaks off the surface below it. Take g=10m/s².(fig. below) [ Ans.1.5m/s]

7. A particle is projected with a speed of u at angle 0 with the horizontal. What is the radius of curvature of the parabola traced out by the projectile at a point where the particle velocity makes an angle 0/2 with the horizontal?

[Ans. u² cos²0/gcos³(0/2)]

8. A car driver going at some speed v suddenly finds a wide wall at a distance r. Should he apply the brakes or turn the car in a circle of radius r to avoid hitting the wall?

9.A block of mass 2kg is pushed against a rough vertical wall with a force of 40 N . Coefficient of static friction being 0.5.Another horizontal force of 15N is applied on the block in a direction parallel to the wall . Will the block move? If yes, in which direction? If no find the frictional force exerted by the wall on the block. [Ans. moves at an angle of 53^o with the 15N force]

akhil_o

8.) he should press brakes,coz if he suddenly turns the car will overturn which is muc more dangerous than a slower crash

3)

for first shot,change in vel. -initial velocity =0

m(200)=(50m-m)V

or V= 200/49

for second shot,change in velocity

200m=(49-1)mV

v=200/48 v

hence total vel=

200/49+200/48 m/s

anchitsaini

1)
force due to resting part--mgx/l
force due to part dx which is just falling=dp/dt
=dmv/dt=Kvdx/dt
=Kvsquare where K=M/l
v=root(2gx) as x is the height fallen
therefore Kvsquare=2mgx/l

therefore total force=mgx/l +2mgx/l
=3mgx/l

anchitsaini

2)
2m--u1,v1
m-u2,v2
u1=u2=root(2gl)
conserving momentum--
2mu1-mu2=-2mv1+mv2

assuming they go in opp dir after collision

this gives
v2-2v1=root(2gl)         -1
conserving kinetic energy gives
2v1^2+v2^2=6gl         -2
from 1 and 2--
3v1^2 +2root(2gl)v1-2gl=0
solving this quadratic we get
correct value of v1 as root(2gl)/3
from 1
v2=5root(2gl)/3

anchitsaini

7)
let velocity at given pt be v1
as horizontal component of velocity are same
therefore v1costheta/2=vcos theta
also centripetal accn=g cos theta/2=v1square/r
therefore
r=v1square/g cos theta/2
=[vsquare (costheta)^2]/[gsquare(cos theta/2)^3]

anchitsaini

6)
cos theta =mg/kx=h/h+x
x=0.1m
cos theta=4/5
s=(h+x)sin theta
therefore s=0.3m

change in ke=gain in pe
mgs=2*1/2*mv^2+1/2kx^2

from the above eqn
we get
v=1.53m/s

anchitsaini

would appreciate it if my efforts are recognised!!!
[i don't believe in karm karte jaao fal ki chinta mat karo,
i need instantaneous fal to my karm]

waterdemon

9)
The Force which may cause the tendency of motion or the
motion in the body is its own Weight & the applied horizontal
force of 15N.

We have Resultant of the forces as:
F_res =

(20)²+(15)²
F_res =

625
F_res = 25N.

In a Direction of:
@ = Tan^-1(15/20) = Tan^-1(3/4) = 37⁰ with the Vertical.

The Friction will always oppose the tendency of Relative
motion and will act in a direction opposite of Resultant F.
F = Force.

Now for the acceleration to be minimum:
Minimum Force required: (N=40=F applied against the wall)
F_min = F - uN
F_min = 25 - (0.5)40
F_min = 5N.

So the minimum Acceleration will be:
F_min=ma_min
5 = 2(a_min)
a_min = 2.5m/s².

Since @ = 37⁰ with the vertical so @ = 53⁰ with the
Horizontal.

Hope you find it useful.
Rate if useful.

Cheers!!!!!!!!!!!!! :)

waterdemon

The Energy is stored in the following three forms:

(a)Elastic Potential Energy of the Spring:

(Stretched by a length of "dL")

= (1/2)*K*(dL)²

(b)K.E of Block A = (1/2)mv²

(c)K.E of Block B = (1/2)mv²

Now Applying the law of conservation of energy,

mgx = (1/2)k(dL)² + (1/2)mv²+ (1/2)mv²

v² = gx - (k/2m)(dL)² ..........(1)

Considering the Equilibrium, of Block A at position A :

R + FCos@ = mg ...........(2)

For a Spring,

F = K(dL) and for Breaking off R = 0.

Substituting the value of R and F in Eq.(2)

K(dL)Cos@ = mg ..........(3)

From the Geometry of the problem ,

dL = (L/Cos@) - L = L{ (1/Cos@) - 1 }

Substituting in Eq.(3)

K{ L [ (1/Cos@) - 1 ]Cos@ = mg

On solving we get,

1 - Cos@ = mg/KL

Cos@ = [1-(mg/KL)]

Cos@ = 1 - (0.32*10/40*0.40) = 4/5

and dL = 0.4(5/4-1)

dL = 0.1 m from above.

Now x=L(tan@) = 0.4*(0.3/0.4) = 0.3m.

Subsitituting the values in eq.(1)

v² = 10*0.3 (40*(0.1)²/2*0.32)

v² = 3 - 0.64

v² = 2.36

v = 1.5 m/s.

Hope you find it useful.

plZ rate me if useful.

Cheers!!!!!!!!!!!!!!

anchitsaini

isn't my method more elegant

anchitsaini

thnks everyone for the rates
my efforts truely have been recognised!!!!

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