Solutions to Exercises on Newton's Laws of Motion HC Verma's Concepts of Physics Part 1 (24-28)

Q#24

Figure (5-E8) shows a uniform rod of length 30 cm having a mass of 3.0 kg. The strings shown in the figure are pulled by constant forces of 20 N and 32 N. Find the force exerted by the 20 cm part of the rod on the 10 cm part. All the surfaces are smooth and the strings and the pulleys are light.

Answer:
The length of uniform rod is 30 cm and mass 3.0 kg, hence mass of 10 cm part = (10/30) x 3.0 kg = 1 kg. 

mass of 20 cm part = (20/30) x 3.0 kg = 2 kg.   

Net force on the whole block = 32 N – 20 N = 12 N (towards right) 

Let acceleration of the block be 'a'. From the relationship 

Force = mass x acceleration   
12 N = (3.0 kg)a
a = 4 m/s²   

Let the force exerted by the 20 cm block on the 10 cm block be F (towards right). Resultant force on the 10 cm block = F – 20 N. Again from  Force = mass x acceleration, we get, 

F – 20 = 1 x 4 
F = 20 + 4 = 24 N  

Q#25
Consider the situation shown in the figure (5-E9). All the surfaces are friction-less and the string and the pulley are light. Find the magnitude of the acceleration of the two blocks.

Answer:
It is clear that the triangle is a right angled triangle and the vertex angle at pulley is 90°. See the diagram below.

Let angle ACB = θ and ABC = α. The component of weight mg along the 3 m side = mg sinα  = 4mg/5.

The component of weight mg along the 4 m side = mgsinθ = 3mg/5.

Let tension in the string be T and the acceleration of the blocks along planes be a.

Consider the forces and acceleration along the plane on left side block,

4mg/5 – T = ma,
T = 4mg/5 – ma                               (1) 

Consider the forces and acceleration along the plane on right side block,   
T – 3mg/5 = ma,     

4mg/5 – ma – 3mg/5 = ma,   {Putting the value of T from (1)}   
2ma = mg/5   
a = g/10.   

So the acceleration of the blocks is g/10.     

Q#26    
A constant force F = m₂g/2 is applied on the block of mass m₁as shown in figure (5-E10). The string and the pulley are light and and the surface of the table is smooth. Find the acceleration of m1. 

Answer:
As is clear from the figure, horizontal forces on the block m₁ are tension in the string T (towards right) and F = m₂g/2 (towards left). Let the acceleration of block m₁ be 'a' towards right. It gives the equation,

T – m₂g/2 = m₁a
T = m₂g/2 + m₁a

Block m₂ will have same acceleration 'a' downwards and forces on it will be tension T upwards and weight m₂g downwards. It gives, 

m₂g – T = m₂a,
m₂g – m₂g/2 – m₁a = m₂a, (Putting the value of T)
(m₁ + m₂)a = m₂g/2

a = m₂g/2(m₁ + m₂).  (for block m1 its direction will be towards right as assumed initially)          

Q#27
In figure (5-E11) m₁=5 kg and m₂ = 2 kg and F = 1 N. Find the acceleration of either block. Describe the motion of m1 if the string breaks but F continues to act.  

Answer:
Let the acceleration of blocks be 'a' and tension in the string T. Resultant downward force on block m1 is

= m₁g + F – T = 5g + 1 – T, Downward acceleration is a. so we get,
5g + 1 – T = 5a
T = 5g + 1 – 5a 

Consider the block m2, Resultant upward forces = T – 2g – 1, upward acceleration assumed 'a', we get, 

T – 2g – 1 = 2a (Put the value of T)
5g + 1 – 5a – 2g – 1 = 2a 
7a = 3g
a = 3g/7 = 4.2 m/s².

If the string breaks, the upward pull of string due to tension becomes zero and the resultant downward force on the block m1 is

= 5g + 1 N, Mass = 5 kg, hence Acceleration = Force/Mass 
= (5g + 1)/5   = 5.0 m/s² (dowmwards)

Q#28
Let m₁= 1 kg, m₂ = 2 kg and m₃ = 3 kg in figure (5-E12). Find the accelerations of m₁, m₂ and m₃. The string from the upper pulley to m1 is 20 cm when the system is released from rest. How long will it take before m₁ strikes the pulley?

Answer:
Let acceleration of m₁ and moving pulley be 'a' and tension in the string attached to m₁ and moving pulley be T. Taking Upward direction positive, for m₁ we get,

T₁ – m₁g = m₁a
T₁ = m₁g + m₁a = m₁(g + a) = 1 x (g + a) = g + a 

Let the tension in the string joining m₂ and m₃ be F. Considering the forces on the moving pulley, we get,
2T₂ = T₁
T₂ = T₁/2. 

Let the acceleration of blocks m₂ and m₃ be a' with respect to the moving pulley. Acceleration of the block m₂ with respect to the upper fixed pulley = a – a'. Similarly, acceleration of the block m₃ with respect to upper fixed pulley = a + a'. 
Consider the forces on block m₂, we get, 

m₂g – T₁/2 = m₂(a – a')
a – a' = g – T₁/2m₂                                        (1)

Consider the forces on block m₃, we get, 
m₃g – T₁/2 = m₃(a + a')
a + a' = g – T₁/2m₃                                        (2)

Add (1) and (2), 

2a = 2g – ½T₁(1/m₂ + 1/m₃) = 2g – ½T₁(½ + 1/3) = 2g – 5T₁/12                     (3)

(Multiply both side by 12)
24a = 24g – 5T₁ = 24g – 5(g + a) = 24g – 5g – 5a = 19g – 5a,

29a = 19g
a = 19g/29 (Acceleration of m₁)

From (2),
a' = g – a – T₁/2m₃ = g – a – (g + a)/6 = (5g – 7a)/6

= (5g – 133g/29)/6 = (145g – 133g)/174 = 12g/174

Acceleration of m₂ = a – a' = 19g/29 – 12g/174 = (114g – 12g)/174

= 102g/174 = 17g/29

Acceleration of m₃ = a + a' = 19g/29 + 12g/174

= (114g + 12g)/174 = 126g/174 = 21g/29 

Second Part :
Length of string of m1 (distance to be travelled) = s = 20 cm = 0.20 m

Initial velocity u = 0, acceleration = a = 19g/29   

To find time t = ?,

From, s = ut + ½at²

0.20 = 0 + ½ x (19g/29)t²

19gt² = 0.2 x 2 x 29 = 0.4 x 29,

t² = 11.6/19 x 9.8 = 0.623

t ≈ 0.25 s 

So m₁ will hit the pulley in 0.25 s.                   

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