DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev

Physics For JEE

Class 11 : DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev

The document DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev is a part of the Class 11 Course Physics For JEE.
All you need of Class 11 at this link: Class 11

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
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
Note In exercises 2 to 4 the situations described take place in a box car which has initial velocity v = 0 but acceleration
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
Ans: 6 .83 kg
Sol: Block on triangular block will not slip if
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
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
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
Substituting θ = 30° , m1 = 1 kg and m2 = 4 kg
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
= 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 DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
Displacement of block at time t relative to car would be
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
Velocity of block at time t (relative to car) will be
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
(b) Time (t) for the block to arrive at the original position (i.e., x = x0) relative to car
x0 = x+ 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)DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev. Describe the motion (a) in car’s frame (b) in ground frame.
Ans: 
(a) x = x0 - 2.5t2, z = z0 + 10t, vx = - 5t, vz = 10 ms-1 
(b) x = x0, z = z0 + 10t, vx = 0, vz = 10 ms-1
Sol: (a) In car’s frame position of object at time t would be given by      
In car’s frame
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
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
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
and
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
(b) In ground frame the position of the object at time t would be given by
In ground frame
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
x = x0
and z = z0 + 10t
Velocity of the object at time t would be
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
ans
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev

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
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
Normal force on object = mg
Maximum sliding friction = μsmg
= 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)
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
Thus, velocity of object at any time t (relative to car)
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
The object will stop moving relative to car when
10 - 8t = 0 i.e., t = 1.25s
∴ vx = 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 DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
Ans: 
9/25 mg
Sol: For block not to slide the frictional force ( f ) would be given by
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
f + ma cosθ = mg sinθ
or
f = mg sinθ - ma cosθ
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev
DC Pandey Solutions: Laws of Motion- 3 Notes | EduRev

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