Messages posted by yahiyafirdous

m

From above fig:

Let m=6kg, M=15kg

=30⁰

N1=m(g-a1)

But a1=a2

Hence N1=m(g-a2 ) ..............(1)

For M:

Ma2=(N1+Mg)..........................(2)

From (1), (2)

a2=g(m+M)/{M+m² }

a2=70/11 m/sec²

a1=35/11 m/sec²

Hi

I am solving Erodov prob u have asked me

Let at time t the position of the particle is P(x1, y1) and at t+dt is Q(x1+dx, y1+dy)

ds=dx i + dy j where ( x1=R, y1=R )

ds/dt= -Rd i + Rd j

v= -Rd/dt i + Rd /dt j

V_av = where limit for Theta is "0" to 180⁰

=-2R/T i + 0. j

|V_av | = 2R/T This is average of velocity vector.

Mean velocity is easy to find:

Distance= Pie R Time =T

V=(Pie R) / T

,

Initial height h is gained with uniform velocity v

Hence t₁ = h/v .....................(1)

Now engine fails , hence its speed will be zero after further going h₁ height in next t₂ sec.

h₁ = v² /2g..........................(2)

t₂ = v/g....................................(3)

Total height H= h+h₁ =1/2 g t₃ ² where t₃ is time to descend

v² /2g +h =1/2 g(t -t₁ -t₂)² ...........(4)

Now substitute the value of t₁ and t₂ in equation (4) and solve for h

You will get the answer.

Conside the figure sent by Sivam:

The triangle shows the corners that will contain Centre of mass of base spheres. Normal reaction(R)due to top sphere will make an angle = 1/

Tension T will make 30⁰ with each of the bisectors of the base c.g. angles of triangle.

Hence 3R=2 (weight of top sphere)

3R( ) =2

R=2

2T=R=2/

T=2/3

Since safety of faxtor is 3

T_max =2

Let at time t velocity of trolley is v_t

mass of trolly=(m+ut)

momentum of the system=v(ut) + (m+ut)v_t

At time t+dt, corresponding velocity of trolley v_t + dv_t , mass=m+u(t+dt)

momentum of the system=vu(t+dt) +(m+ut+udt)(v_t + dvt )

Since there is no external forces acting on the system, hence momentum is conserved:

v(ut) + (m+ut)v_{t =}vu(t+dt) +(m+ut+udt)(v_t + dvt )

Dividing both sides by dt we get:

(m+ut)dv_t /dt= -u(v+v_t )

Solving above dirrerential equation for v_t , we get:

(m+ut)(v+v_t ) = K( a constant)

when t=o, v_t =0

Hence k=mv

Hence v_t = - uvt/(m+ut)

dv_t / dt= accⁿ = -muv/(m+ut)²

T_max=2

e=0.8415

M=16.475kg

I will explain if this answer is correct

When its accⁿ :

(50)² =100+2as

2as=2400............................(1) where a=accⁿ, s=distance

When its direction of acce is reversed, its velocity will first become zero then goes in reverse direction with accⁿ a

0=(50)² - 2as₁

2as₁=2500...............................(2)

Hence velocity at starting point: Total distance =s+s₁

v²=0 + 2a(s+s₁ )=4900

v=70m/s Ans

Let v_t is velocity of trolley after t sec.

Hence (v- v_t) u=mdv_t /dt

Solving above equation you get:

v_t=v(1- e^{-u/m t} ) Ans.

Assumptions:

1. No dust accumulate in the trolley.

Hi Zeosthegod

Although you have cosen right options,

but you have solved incorrectly the question (ii)

because u/2i + u j= root of(5) X u/2

So use here energy equation to find velocity of the ball, then solve it

The resultant accⁿ of block B is vector sum of b and c. So add them vectorically and get the answer.

3/16 Ma²

I am trying to solve this:

Let distance travelled on the plane = d

Elongation in spring= dx=Square root of(l₀ ² + d²) - l₀

Tension in spring= T = k(dx) = (5mg/l₀ )(dx)

Now let spring makes an angle with horizontal plane

Remember just at the time of taking off the mass from the floor (mg) equals the vertical component of T

Hence T =mg

(5mg/l₀ )(dx){l₀/root(l₀2 + d²)}=mg

But dx=Square root of(l₀ ² + d²) - l₀ Solving for (d/l₀)

We get : d= (3/4)l₀

Now writting energy equation for the system we get:

k(dx)² + (2mv²) = 2mgd

Now put the value of d, we get

v= square root of (19/32 gl₀ ) = 1.7m/s

nobody can understand this problem unless u attach the diagram or indicate the exact location of spring , masses and points O, A and P

Hey

Correct answers are:

B, C and A

Hi ABHI

In an elastic or plastic collision reduction in energy appears in the form of heat and some part is consumed in deforming the bodies in question.

So tell me where this energy is going? Is there any deformation ?

Is it w= (4v₀)/{r(root 3)}

Hi ABHI

Your conservation of momentum is satisfied, But what about conservation of energy?

See intial energy of the system is 800m, but as per your solution final energy is 912.5m, where does this extra energy come from?

Q4

Let rate of change of momentum in horizontal direct=p

Similarly rate of change of monentum in horizontal derect. = p

Since its not moving in vertical direction , hence this p(which is a force) must be balanced by normal reaction of plane;

p=mgCos(theta)

similarly p in horizontal direction is provided by mgSine(theta)

p=mgSine(theta)

Hence Theta= 45⁰

Q3:

Your third question is not clear at all, please do not use absurd language like dis for this , 2 for to.

Q1.

Let us find the height gained:

0=u² - 2(g+ f/m)h

h= u² /2{g+f/m}..........................................(1) where m= w/g

Since drag of air is none conservative force, hence energy lost in air drag:

E1= 2fh= fu² /{g+f/m}.

Hence total K.E. just before touching the ground:

mv²/2 = mu²/2 - E1 = mu² /2 - fu² /{g+f/m}

v² = u² - 2u² f /(w+f)

V= U

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