Laws of Motion

Q: A body of mass 1 kg moving with velocity 1 m/s makes an elastic one dimensional collision with an identical stationary body . They are in contact for brief time 1 sec . Their force of interaction increases from zero to F₀ linearly in time 0.5 s and decreases linearly to zero in further 0.5 sec as shown in figure . The magnitude of F₀ is

(a) 4 N

(b) 1 N

(c) 2 N

(d) zero

Solution : Area of F – t graph gives impulse .

Impulse = Change in momentum

$\displaystyle \frac{1}{2} \times 1 \times F_0 = m v $

F₀ = 2 m v

F₀ = 2 ×1 ×1 = 2 N

Correct option is (C)

 

Q: A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m = 0.4 kg is at rest on this surface. An impulse of 1.0 N s is applied to the block at time to t = 0 so that it starts moving along the x-axis with a velocity $ v(t) = v_0 e^{-\frac{t}{\tau}} $ , where v₀ is a constant and τ = 4 s . The displacement of the block, in meter , at t = τ  is …….
Take e^-1 = 0.37 ?

Sol: Impulse J = 1 N-s

Impulse = Change in Momentum (Δp)

$\displaystyle v_0 = \frac{J}{m} = \frac{1}{0.4}$

= 2.5 m/s

$\displaystyle v(t) = v_0 e^{-\frac{t}{\tau}} $

$\displaystyle \frac{dx}{dt} = v_0 e^{-\frac{t}{\tau}} $

$\displaystyle \int_{0}^{x} dx = v_0 \int_{0}^{\tau} e^{-\frac{t}{\tau}} dt $

$\displaystyle x = v_0 [\frac{e^{-\frac{t}{\tau}}}{-1/\tau}]_{0}^{\tau}$

$\displaystyle x = 2.5 (-4)[e^{-1} – e^0] $

$\displaystyle x = -10[e^{-1} – 1] $

$\displaystyle x = -10[0.37 – 1] $

x = 6.30

Problem :    A plank of mass M is placed on a horizontal smooth surface. Another bar of mass m is placed on the plank. The co-efficient of static friction between the plank and the bar is µ. What max. horizontal force F on the plank will cause no slipping of the bar on the plank ?

Solution:   

 The forces on the plank and the bar is shown in Figure.

For the bar, from F.B.D.,

N₂ = mg                …(1)

f = ma₁                     …(2)

For the plank from F.B.D.

F − f = Ma₂                    …(3)

N₁ = N₂ + Mg            …(4)

Again we know, maximum static frictional force on ‘ m ’ is ,

f_max = μN₂

from (1) we have

i.e.   f_max = μmg           …(5)

Now to have no slipping of the bar on the plank we need

a₁ = a₂ = a (say)

∴ from (2) and (3) we have

f = ma             …(6)

F − f = Ma        …(7)

From eqn (6) and (7)

$ \displaystyle f = \frac{m F}{m + M} $          …(8)

Again for no slipping of m on M,

f ≤ f_max                 …(9)

$ \displaystyle \frac{m F}{m + M} \le \mu m g $

i.e.  F ≤ μ(M+m)g          …(10)

i.e. F_max = μ(M+m)g

Problem :   A wooden box of mass 20 kg. is kept on a rough horizontal surface. The co-efficient of static friction μ_s = 0.41 and co-efficient of kinetic friction μ_k = 0.3 between the box and the horizontal surface. A constant force F of 70 N is applied to the box at an angle of 30^o with the horizontal. Find the acceleration of the box on the surface and the frictional force on the box. Take g = 10 m/s².

Solution:  

The force components acting on the box with     respect to the chosen X-Y axes is shown in Figure.

w.r.t. Y axis

N + F Sinθ = Mg

∴ N = Mg  − F Sinθ      …(1)

Now,   f_s = μ_s N         …(2)

= 0.41×[20 ×10 −70×1/2]

= 0.41 × 165

= 67.65 N

Again F cosθ = 70 × √3/2 = 35√3

= 60.62 N

∴ F cosθ < f_s

Hence decide whether the box will move or not and hence find what should be the frictional force.

Illustration : A block starts slipping on an inclined plane. It moves one metre in one second. What is the time taken by block to cover next one metre ?

Solution :

The forces acting on the block at any moment are

(i) mg (ii) N, normal reaction

(iii) f , friction force

Let a = acceleration of the block (down the plane)

Applying $ \displaystyle S = u t + \frac{1}{2}a t^2$

For path AB ,

1 = 0 + (1/2)a × 1²

⇒ 1 = a/2 . . . (i)

for path AC ,

2 = 0 + (1/2)a t²

⇒ 2 = a t²/2 . . . .(ii)

From (i) & (ii), t² = 2

⇒ t = √2 sec.

Time taken to cover the distance BC = (t – 1) sec = 0.41 sec

Exercise : A block of mass m = 2 kg is kept on a rough horizontal surface. A horizontal force F = 4.9 N is just able to slide the block. Find the coefficient of static friction. If F = 4 N, then what is the frictional force acting on the block ?

Illustration : Two blocks A & B are connected by a light inextensible string passing over a fixed smooth pulley as shown in the figure. The coefficient of friction between the block A & B the horizontal table is μ = 0.2; If the block A is just to slip, find the ratio of the masses of the blocks.

Solution : From F.B.D. of A, shown in fig

N + T Sin θ = m_Ag ……….(i)

T Cos θ = f_max = μ N. ………(ii)

From (i), N = m_Ag – T Sin θ ………(iii)

From (ii) and (iii), F.B.D. of A.

T Cos θ = μ m_Ag – μ T Sin θ

T (Cos θ + μ Sin θ) = μ m_Ag …………(iv)

From F.B.D. of B,

T = m_Bg ………(v)

Taking ratio of (iv) and (v)

Exercise : The block A is kept over a plank B. The maximum horizontal acceleration of the system in order to prevent slipping of A over B is a = 2m /sec². Find the coefficient of friction between A & B.

Problem :   A piece of uniform string of mass M hangs, vertically so that its free end just touches horizontal surface of a table. If the upper end of the string is released, find at any instant, the total force on the surface just before the string falls completely.

Solution :

Considering the time when y fraction of total length has fallen on the surface.

The impact force  $ \displaystyle F_{impact} = v (\frac{dm}{dt})$

$ \displaystyle = v \frac{d}{dt}(\frac{M}{l}y) $

where  y = distance fallen ,

$ \displaystyle = \frac{Mv}{l}(\frac{dy}{dt}) = \frac{Mv^2}{l} $

By putting v² = 2gy,  we obtain

$ \displaystyle F_{impact} = \frac{2Mgy}{l} $

The weight of the portion lying on the surface

= m’g = (M/l)yg

⇒ Total force offered by the surface =

$ \displaystyle F’ = \frac{3 Mgy}{l} $

Just before the falls completely, y = l

⇒ F’ = 3 Mg

Problem :   In the given figure all the surfaces are smooth. Find the time taken by the block to reach from the free end to the pulley attached to the plank. Distance between free end and pulley is l.

Solution:

Equation of motion for M :

2F = M a₂                          …(1)

For m ,

F = m a₁                            …(2)

Using (1) and (2),

$ \displaystyle a_1 + a_2 = \frac{2F}{M} + \frac{F}{m} $

$ \displaystyle a_1 + a_2 = F(\frac{2}{M} + \frac{1}{m}) $ …….(3)

Relative acceleration between M and m = a_r = a₁+a₂.

Let the block m strikes the pulley after a time t.

Since,

$ \displaystyle l = \frac{1}{2}a_r t^2 $

$ \displaystyle t = \sqrt{\frac{2l}{a_r}} $

$ \displaystyle t = \sqrt{\frac{2l}{F(\frac{2}{M} + \frac{1}{m})}} $

Problem :   A mass of 10 kg lies on a smooth table at a distance of 7 m from the edge and is connected by a taut string passing over the end with a mass of 4 kg hanging freely. How long does it take the 10 kg mass to reach the edge of the table ?

Solution :

m₁ = 4m₂

m₁g – 2T = m₁(a/2)

 T – m₂g = m₂a

(m₁ – 2m₂)g = (m₁/2 + 2m₂)a

$ \displaystyle a = \frac{m_1 – 2m_2}{\frac{m_1}{2} + 2m_2}g $

or,  a =  g/2  = 5 m/s²   

The mass m₁ reaches the ground with a velocity, v, such that,

v² = 2 (a/2) × s  = 2 × (5/2) × 0.2 = 1

or,       v = 1 m/s

The velocity of m₂ when m₁ reaches the ground is

2v = 2m/s.

The maximum height reached by m₂ is

h = 0.4 m + (2²/2×10)