DC Pandey Solutions: Laws of Motion- 3 - Notes | Study DC Pandey Solutions for JEE Physics - JEE

Introductory Exercise 5.5

Ques 1: In figure m₁ = 1 kg and m₂ = 4 kg. Find the mass M o f the hanging block which will prevent the smaller block from slipping over the triangular block. All the surfaces are friction less and the strings and the pulleys are light

Note In exercises 2 to 4 the situations described take place in a box car which has initial velocity v = 0 but acceleration


Ans: 6 .83 kg
Sol: Block on triangular block will not slip if

m₁a cos θ= m₁g sin θ
i.e., a = g tanθ …(i)
N = m₁g cosθ + m₁asinθ …(ii)
For the movement of triangular block
T - N sinθ = m₂α …(iii)
For the movement of the block of mass M
Mg - T = Ma …(iv)
Adding Eqs. (iii) and (iv),
Mg - N sinθ = ( m₂ + M )α
Substituting the value of N from Eq. (ii) in the above equation
Mg - (m₁g cosθ + m₁α sinθ) sinθ
=( m₂α + Mα)
i.e., M (g - α) = m₁g cosθ sinθ + (m₂ + m₁ sin²θ)α
Substituting value of a from Eq. (i) in the above equation,
M(1 - tanθ) = m₁ cosθ sinθ + ( m₂ + m₁ sin² θ) tanθ

Substituting θ = 30° , m₁ = 1 kg and m₂ = 4 kg


= 6.82 kg

Ques 2: A 2 kg object is slid along the friction less floor with initial velocity (10 m/s)î (a) Describe the motion of the object relative to car (b) when does the object reach its original position relative to the box car.
Ans: (a) x = x₀ + 10t - 2.5t² , v = 10 - 5t (b) t = 4s
Sol: (a) Using
Displacement of block at time t relative to car would be


Velocity of block at time t (relative to car) will be

(b) Time (t) for the block to arrive at the original position (i.e., x = x₀) relative to car
x₀ = x₀ + 10 t - 2.5 t²
⇒ t = 4s

Ques 3: A 2 kg object is slid along the friction less floor with initial transverse velocity (10 m/s). Describe the motion (a) in car’s frame (b) in ground frame.
Ans: (a) x = x₀ - 2.5t², z = z₀ + 10t, v_x = - 5t, v_z = 10 ms^-1 
(b) x = x₀, z = z₀ + 10t, v_x = 0, v_z = 10 ms^-1
Sol: (a) In car’s frame position of object at time t would be given by      
In car’s frame

x = x₀ + 0*t + 1/2(-5)t²
i.e., x = x₀ - 2.5 t² …(i)
and z = z + 10t …(ii)
Velocity of the object at time t would be

and

(b) In ground frame the position of the object at time t would be given by
In ground frame

x = x₀
and z = z₀ + 10t
Velocity of the object at time t would be

ans


Ques 4: A 2 kg object is slid along a rough floor (coefficient of sliding friction = 0.3) with initial velocity (10 m/s)î. Describe the motion of the object relative to car assuming that the coefficient of static friction is greater than 0.5.
Ans: x = x₀ + 10f - 4t² , K = 10 - 8 t for 0<f<1.25 s object stops at t = 1.25 s and remains at rest relative to car.
Sol: m = 2 kg

Normal force on object = mg
Maximum sliding friction = μ_smg
= 0.3*2*10 = 6 N
Deceleration due to friction = 6/2 =3 m/s²
Deceleration due to pseudo force  = 5 m/s²
∴ Net deceleration = (3 + 5) m/s²
= 8 m/s²
∴ Displacement of object at any time t (relative to car)

Thus, velocity of object at any time t (relative to car)

The object will stop moving relative to car when
10 - 8t = 0 i.e., t = 1.25s
∴ v_x = 10 - 8t for  0<t<1.25 s

Ques 5: A block is placed on an inclined plane as shown in figure. What must be the frictional force between block and incline if the block is not to slide along the incline when the incline is accelerating to the right at

Ans: 9/25 mg
Sol: For block not to slide the frictional force ( f ) would be given by

f + ma cosθ = mg sinθ
or
f = mg sinθ - ma cosθ