IIT JEE , AIEEE Forum - Physics , Mechanics - plzzz hlp im stuck wid so many questions....COLLISIONS.

coolkrazy007

plzz EXPLAIN questions frm hcv...pg-160

question no.-19,27,28,29,35,,36-b,37,38,40,43-46,49-57,59-62,64

i know its a big list but/.....ur hlp will mkae me lot more comfortable wid collisions...

2 rates per soln.....assured

deedee

q no 27

initial

mass=500m

vel=0

after first collision

v1 b d recoil vel of d car after first shot

(200-v1) vel of bullet

mass left =49m

initial momem=final momem

thus

0=49mv1-m(200-v1)

200/50=v1

v1=4m/s

then after second shot

mass left=48m

v2 b d vel of car

(200-v2) vel of shell

nitial momen=fianl mmem

49mv1=48mv2-m(200-v2)

thus v2=396/49

=8.08m/s

wich is same as

200(1/49+1/48)

hope u finda useful

deedee

q no 28

mu=(m+M)v

v=mu/M+m

(mu/M+m)+u after second jump

u{(m/m+M)+1}

mu({m/m+M}+1)+Mv' =mu

solvin u get

v' =-m^2*u/M+m

-ve sign shws that d motion is in opposite direrction

anchitsaini

19)A man of mass M having a bag of mass m slips from the roof of a tall building of height H and starts falling vertically. when at h from the ground , he notices that ground below him is pretty hard, but there is a pond at a horizontal distance x from the line of fall. in order to save himself he throws the bag horizontally in the direction opposite to the pond.
calculate the minimum horizontal velocity imparted to the bag for man to land in water.
if man just succeeds in landing in water where will the bag land??

19)taking v to be velocity of m

mv = Mv1
v1= mv/M

time of flight for man after leaving the bag --

2H/g -

2(H-h)/g =t

this would be the velocity in horizontal direction

also
x = Vt
or
x = mv/M * [

2H/g -

2(H-h)/g]
which gives

v=Mx

g / [

2H/g -

2(H-h)/g]

also COM of bag and man remains at the same position

taking the initial pos of COM as 0 on x axis

0 = Mx + mx1 /(m+M)

or x1= -Mx/m (- indicating towards left)

somyekathait

Q 57)

Normal reaction by floor = F _weight + F _thrust

F _weight= (M/L)xg

F _thrust = V_relative dM/dt M=mass

=V dM/dx * dx/dt

= V M/L* V as dx/dt=V

=V² M/L....(1)

when chain has fallen through x metres

V=

(2gx)

substitute in (1)

2gx*M/L

hence N =(M/L)xg + 2gxM/L

=3mgx/L

anchitsaini

35)a ball of mass m moving at a speed v makes a head on collision with an identical ball at rest. the kinetic energy of the balls after the collision is three fourths of the original . find e

35)

Conserving momentum--

mv = mv1 + mv2
v = v1 + v2 ---------1

Conserving energy --

3v²/4 = v1² + v2² = [ (v1 + v2)² + (v2 - v1)² ] /2
= [v²+ (v2 - v1)² ]/2 from 1

this gives --
(v2 - v1) = v/

2

e=(v2 - v1) / (u1 - u2)
=v/

2 / (v-0)
=1 /

2

edited part in blue

anchitsaini

36)
loss of kinetic energy = 1J
initial kinetic energy = 4J (0.5mv^2)

thus it has 3/4 KE

the qn becomes exactly the same as 35

coolkrazy007

thnxxx alot guyzzz /...

but tell me how is this v1² + v2² = [ (v1 + v2)² - (v2 - v1)² ] /2

coming...

anchitsaini

edited above

coolkrazy007

check ur nudgebuk anchit..

coolkrazy007

can u plzz exlpain in more details question no.27 and 28

LAMPARD

Q54>
Let velocity given to A at start be v...let v1 be vel of A at base...so,
0.5mv²+mgh=0.5mv1²...v²+2gh=v1²..eqn1..
Let vel of A after collision be v_a and dat of B be v_b.For B to reach man,v_b=^{[2 ]}

(2gh)...
Conserving KE b4 n after collision,v get
0.5mv1²=0.5mv_a²+0.5(2m)v_b²...
so,v1²=v_a²+2v_b²....eqn2...
now,conserving momentum b4 n after collision,u get
v1=v_a+2v_b...so,v_a=v1-2v_b...
subst. v_a in eqn 2 and v_b as (2gh)^0.5 in eqn 2,u get v1²=4.5gh...
subst. v1²in eqn 1,u get v=(2.5gh)^0.5...which is reqd. ans....

LAMPARD

Q53>
Conserving momentum ion vertical direcn when pillow is pushed,
5*8=50*v
so,v=4/5ft/s,where v is vel. of man in upward direction.
Suppose time taken is t seconds.
So,distance covered by man in t secs is given by
s=(4/5)t -0.5gt²
Dist. covered by pillow is 16ft(8 downwards+8upwards)+s(bcoz it reaches the man)
so,16+s=8t -0.5gt²
now,subtract the 2 eqns,u get t=2.2 secs.

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