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# Laws of Motion (Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

## DC Pandey (Questions & Solutions) of Physics: NEET

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## NEET : Laws of Motion (Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

The document Laws of Motion (Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev is a part of the NEET Course DC Pandey (Questions & Solutions) of Physics: NEET.
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Introductory Exercise 5.2

Ques: 1 Three blocks of mass 1 kg, 4 kg and 2 kg are placed on a smooth horizontal plane as shown in figure. Find:
(a) the acceleration of the system,
(b) the normal force between 1 kg block and 4 kg block,
(c) the net force on 2 kg block. Ans: (a) 10 ms-2 (b) 110 N (c) 20 N
Sol: Acceleration of system = 10 m/s2
Let normal force between 1 kg block and 4 kg block = F1
∴ Net force on 1 kg block = 120 - N or 10 = 120 - F1
i.e., F1 = 110 N
Net force on 2 kg block = 2*a
= 2*10
=20 N

Ques 2: Two blocks of mass 2 kg and 4 kg are released from rest over a smooth inclined plane of inclination 30° as shown in figure. What is the normal force between the two blocks? Ans:
zero
Sol: As, 4 g sin 30° > 2 g sin 30°
The normal force between the two blocks will be zero.

Ques 3: What should be the acceleration ‘a’ of the box shown in figure so that the block of mass m exerts a force mg/4 on the floor of the box? Ans:
3g/4
Sol:  ∴ N = mg/4
As lift is moving downward with acceleration a , the pseudo force on A will be ma acting in the upward direction. For the block to be at rest w.r.t. lift.
N + ma = mg
or  Ques 4: A plumb bob of mass 1 kg is hung from the ceiling of a train compartment. The train moves on an inclined plane with constant velocity. If the angle of incline is 30°. Find the angle made by the string with the normal to the ceiling. Also, find the tension in the string, (g = 10 m/s2)
Ans:
30°, ION
Sol: Angle made by the string with the normal to the ceiling = θ = 30°
As the train is moving with constant velocity no pseudo force will act on the plumb-bob. Tension in spring = mg
= 1*10
= 10 N

Ques 5: Repeat both parts of the above question, if the train moves with an acceleration a = g/2 up the plane.
Ans: Sol: Pseudo force (= ma) on plumb-bob will be as shown in figure T cos φ = mg + ma cos (90° - θ)
i.e., T cos φ = mg + ma sin θ …(i)
and       T sin φ = ma cos θ
Squaring and adding Eqs. (i) and (ii),
T2 = m2 g2 + m2 a2 sin2 θ + 2m2 ag sin θ + m2 a2 cos2 θ …(iii)
T2 = m2 g2 + m2 a2 + m2 ag (∴ θ = 30°)  or Dividing Eq. (i) by Eq. (ii),   i.e., Ques 6: Two blocks of mass 1 kg and 2 kg are connected by a string AS of mass 1 kg. The blocks are placed on a smooth horizontal surface. Block of mass 1 kg is pulled by a horizontal force F of magnitude 8 N. Find the tension in the string at points A and B. Ans:
4 N, 6 N
Sol:   Net force on 1 kg mass = 8 - T2
∴ 8 - T2 = 1*2
⇒ T2 = 6 N
Net force on 1 kg block = T1
∴ T1 = 2a = 2*2 = 4 N

Introductory Exercise 5.3

Ques 7: In the arrangement shown in figure what should be the mass of block A, so that the system remains at rest? Neglect friction and mass of strings.
Ans:
3 kg
Sol: F = 2 g sin 30° = g For the system to remain at rest
T2 = 2 g …(i)
T2 + F = T1 …(ii)
or          T2 + g = T1 …[ii (a)]
T1 = mg …(iii)
Substituting the values of T1 and T2 from Eqs. (iii) and (i) in Eq. [ii(a)]
2 g + g = mg
i.e., m = 3 kg

Ques 8: In the arrangement shown in figure, find the ratio of tensions in the strings attached with 4 kg block and that with 1 kg block. Ans:
4
Sol:
As net downward force on the system is zero, the system will be in equilibrium ∴  T1 = 4 g
and T2 = 1 g Ques 9: Two unequal masses of 1 kg and 2 kg are connected by a string going over a clamped light smooth pulley as shown in figure. The system is released from rest. The larger mass is stopped for a moment 1.0 s after the system is set in motion. Find the time elapsed before the string is tight again. Ans:
1/3 s
Sol: 2 g - T = 2a T - 1 g = 1a
1 g = 3 a
∴ a = g/3
Velocity of 1kg block 1 section after the system is set in motion
v = 0 + at On stopping 2 kg, the block of 1kg will go upwards with retardation g. Time ( t' ) taken by the 1 k g block to attain zero velocity will be given by the equation. ⇒ If the 2 kg block is stopped just for a moment (time being much-much less than 1/3 s), it will also start falling down when the stopping time ends.
In time upward displacement of 1 kg block Downward displacement of 2 kg block As the two are just equal, the string will again become taut after time 1/3 s.

Ques 10: Two unequal masses of 1 kg and 2 kg are connected by an inextensible light string passing over a smooth pulley as shown in figure. A force F = 20 N is applied on 1 kg block. Find the acceleration of either block. (g = 10 m/s2). Ans: Sol: F + 1g - T = 1a      … (i) and T - 2g = 2a …(ii)
F - 1g = 3a Introductory Exercise 5.2

Ques 11: Consider the situation shown in figure. Both the pulleys and the string are light and all the surfaces are smooth.
(a) Find the acceleration of 1 kg block.
(b) Find the tension in the string.
(g = 10 m/s 2). Ans:
(a) 2g/3, (b) 10/3 N
Sol: 2T = 2*a            … (i) and     1g - T = 2a …(ii)
Solving Eqs. (i) and (ii),
α = g/3
∴  Acceleration of 1 kg block Tension in the string Ques 12: Calculate the acceleration of either blocks and tension in the string shown in figure. The pulley and the string are light and all surfaces are smooth. Ans: Sol: Mg - T = Ma             …(i) T = Ma …(ii)
Solving Eqs. (i) and (ii)
α = g/2
and T = Mg

Ques 13: Find the mass M so that it remains at rest in the adjoining figure. Both the pulley and string are light and friction is absent everywhere, (g = 10 m/s 2).
Ans:
4 . 8 kg
Sol: Block of mass M will be at rest if T = Mg …(i)
For the motion of block of mass 3 kg ....(ii)
For the motion of block of mass 2 kg ....(iii)
g = 5a
i.e.,  α = g/5
Substituting above value of a in Eq. (iii),
T/2 = 2(g+a)
or = 24g/5
Substituting value of above value of T in Eq. (i),
M = 24/5
= 4.8 kg

Ques 14: In figure assume that there is negligible friction between the blocks and table. Compute the tension in the cord connecting m2 and the pulley and acceleration of m2 if m1 = 300 g, m2 = 200g and F = 0.401V. Ans: Sol: T/2 = m1.2a i.e., T = 4 m1a …(i)
F - T = m2a …(ii)
or      F - 4m1a = m2a
or  ∴ T = 4 m1a
= 4*0.3* 2/7
= 2.4/7
= 12/35 N

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