Newton’s Laws of Motion — H.C. Verma: Brief Overview
The Newton’s Laws of Motion chapter in Concepts of Physics by H.C. Verma introduces the fundamental principles that explain how objects move and how forces affect their motion. It is an important chapter for Class 11 Physics, JEE and NEET preparation.
1. Newton’s First Law
An object remains at rest or continues moving with uniform velocity in a straight line unless an external unbalanced force acts on it. This law introduces the concept of inertia.
2. Newton’s Second Law
The acceleration of an object depends on the net force acting on it and its mass:
F = ma
This is one of the most important equations in mechanics and is widely used to solve numerical problems.
3. Newton’s Third Law
For every action, there is an equal and opposite reaction. These forces act on different objects, which is an important point when solving problems.
Important Concepts
H.C. Verma’s treatment also helps students understand:
- Force and acceleration
- Inertia
- Momentum
- Free-body diagrams
- Tension and normal reaction
- Friction-related force problems
- Equilibrium and dynamics
- Application of Newton’s laws to connected bodies
Why This Chapter Is Important
Newton’s laws form the foundation of classical mechanics. A strong understanding of this chapter makes later topics such as friction, circular motion, work-energy, gravitation and rotational mechanics easier to understand
OBJECTIVE 1
1)A body of weight w1 is suspended from the ceiling of a room through a chain of weight w2. The ceiling pulls the chain by a force a) w1 b)w2 c)w1+w2 d)w1+w2/2 Ans ) C net weight for the ceiling chain = w1 + W2)
2)when a horse pulls a cart , the force that helps the horse to move forward is the force exerted by a) the cart on the horse b)the ground on the horse c)the ground on the cart d)the horse on the ground
Ans) b horse exerts a force on the ground which causes normal reaction that causes the horse to move forward
3)A car accelerates on a horizontal road due to the force exerted by a)the engine of the car b)the driver of the car c) the earth d)the road
Ans)d force exerted by road
4) A block of mass 10 kg is suspended through two light spring balances as shown in figure (5-Q2) a)both scale will read 10 kg b)both scale will read 5kg c)the upper scale will read 10kg and the lower zero . d)The readings may be anything but their sum will be 10kg
Ans)a both the spring will expands the same . Hence both the spring reads the same value .Hence option a is correct
5)A block of mass m is placed on a smooth inclined plane of inclination θ with the horizontal . the force exerted by the plane on the block has a magnitude a) mg b)mg/cosθ c)mgcosθ d)mgtanθ
Ans)c N= mg/cosθ
6)A block of mass m is placed on a smooth wedge of inclination θ . The whole system is accelerated horizontally so that the block does not slip on the wedge . The force exerted by the wedge on the block has a magnitude a)mg b)mg/cosθ c)mgcosθ d)mgtanθ
Ans)b mg/cosθ
7)Neglect the effect of rotation of the earth . Suppose the earth suddenly stops attracting objects placed near its surface . A person standing on the surface of the earth will a)fly up b)slip along the surface c)fly along a tangent to the earth’s surface d)remain standing
Ans)d Neglect the effect of rotation of earth , N = mg , remain standing
8)Three rigid rods are joined to form an equilateral triangle ABC of side 1 m . Three particles carrying charges 20μc each are attached to the vertices of the triangle . The whole system is at rest in an inertial frame . The resultant force on the charged particle at A has the magnitude a)zero b)36 N c)3.6√3 N d)7.2 N
Ans)a As the whole system is at rest Hence net force is zero
9)A force F1 acts on a particle so as to accelerate it from rest to a velocity V. The force F1 is then replaced by F2 which decelerates it to rest a)F1 must be equal F2 b) F1 may be equal F2 c) F1 must be un equal F2 d)none of these
Ans)b when F1 may be equal to F2 then net force on the particle is zero . then body decelerates and comes to rest .
10)Two objects A and B are thrown upward simultaneously with the same speed . The mass A is greater than mass B . Suppose the air exerts a constant and equal force of resistance on the two bodies . a)the two bodies will reach the same height b)A will go higher than B c)B will go higher than A d)Any of the above three may happen depending on the speed with which the objects are thrown
Ans)b Ma>Mb , acceleration of a is less than acceleration of b , so putting in the equation S = ut – ½ at 2 , As acceleration of a is min it travels max height than b .
11)A smooth wedge A is fitted in a chamber hanging from a fixed ceiling near the earth’s surface.A block B placed at the top of the wedge takes a time T to slide down the length of the wedge. If the block is placed at the top of the wedge and the cable supporting the chamber is broken at the same instant, the block will a)take a time longer than T to slide down the wedge b) take a time shorter than T to slide down the wedge c) remain at the top of the wedge d) jump off the wedge
Ans)c As cable supporting the chamber is broken , the ceiling falls with acceleration due to gravity .Due to psuedo force block experiences upward acceleration and remains at top of wedge
12) In an imaginary atmosphere, the air exerts a small force F on any particle in the direction of the particle’s motion. A particle of mass m projected upward takes time t1 in reaching the maximum height and t2 in the return journey to the original point. Then a)t1 < t2 b)t1> t2 c)t1 = t2 d)the relation between t1 and t2 depends on the mass of the particle
Ans)b Ans when particle moving upwards then acceleration = g – F/m , when particle moving downward then acceleration = g+F/m , using equation of motion S = ½ at 2 , t= √2S/a , As acceleration is minimum for upward motion time taken will be maximum for upward motion
13) A person standing on the floor of an elevator drops as coin. The coin reaches the floor of the elevator in a time t1 if the elevator is stationary and in time t2 if it is moving uniformly. Then a) t1 = t2 b)t1 <t2 c)t1 >t2 d)t1 <t2 or t1 >t2 depending on whether the lift is going up or down
Ans)a in both cases acceleration is zero so coin will fall under acceleration due to gravity . Hence time required in both cases will be same .
14) A free 238U nucleus kept in a train emits an alpha particle. When the train is stationary, a nucleus decays and a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes x at time t after the decay. If the decay takes place while the train is moving at a uniform velocity v, the distance between the alpha particle and the recoiling nucleus at a time t after the decay as measured by the passenger is a)x + vt b)x – vt c)x d)depends on the direction of the train
Ans)c During train at rest or moves with constant speed it comes under inertial frame of reference , so no psuedo force acting on it . Hence in both cases distance covered is x .
OBJECTIVE 2
1) A reference frame attached to the earth
a) is an inertial frame by definition
b) can not be an inertial frame because the earth is revolving around the sun
c) is an inertial frame because Newton’s laws are applicable in this frame
d) cannot be an inertial frame because the earth is rotating about its axis
ANS) b,d Because earth is revolving about its axis and earth is revolving around the sun so earth cannot be taken as reference
2)A particle stays at rest as seen in a frame. We can conclude that
A the frame is inertial B resultant force on the particle is zero C the frame may be inertital but resultant force on the particle is zero
D the frame may be non-inertial but there is a non zero resultant force
ANS) C,D incase of inertial frame when force acting on the body is zero acceleration is also zero that satisfies option c , in case of non inertial frame of reference when force acting on system is non zero then pseudo force acts on the object in opposite direction which can be counter by another force as a result that might be taken as rest which satisfies option d
3)A particle is found to be at rest when seen from a frame S_1 and moving with a constant velocity when seen from another frame S_2. Mark out the possible options. (a) Both the frames are inertial. (b) Both the frames are non-inertial. (c) S_1 is inertial and S_2 is non-inertial. (d) S_1 is non-inertial and S_2 is inertial.
ANS)A,B in this question context we have note the idea regarding inertial and non-inertial frame of reference , in case of inertial F=0 , acceleration is zero . in case of non-inertial F=0 , acceleration is non-zero
4)Figure (5-Q3) shows the displacement of a particle going along the X-axis as a function of time. The force acting on the particle is zero in the region (a) A B (b) B C (c) CD (d) D E
ANS)A,C slope of displacement vs time graph gives rise to velocity , from the graph AB , CD gives rise to constant velocity which indicates from the slope of the graph , Hence acceleration is zero .
5)Figure shows a heavy block kept on a frictionless surface and being pulled by two ropes of equal mass m. At t = 0, the force on the left rope is withdrawn but the force on the right end continues to act. Let F1 and F2 be the magnitudes of the forces by the right rope and the left rope on the block respectively. Choose correct options. a)F_1=F_2=F for t<0 b)F_1=F_2= F + m g for t<0 c)F_1=F, F_2=F for t>0 d)F_1=F, F_2=F for t>0
ANS) A as F _1 = F_2 = F , for t < 0 , then net force acting on the particle is zero .
6)The force exerted by the floor of an elevator on the foot of a person standing there is more than the weight of the person if the elevator is :
a ) going up and slowing down b) going up and slowing down c) going down and slowing down d)going down and speeding up
ANS)B,C N = m(g+a) for upward motion , N = m (g-a) for downward motion , In both cases it is slowing down
7)If the tension in the cable supporting an elevator is equal to the weight of the elevator, the elevator may be –
(a) going up with increasing speed ( b) going down with increasing speed ( c) going up with uniform speed ( d) going down with uniform speed
ANS) C,D T – mg = 0 , a = F/m , Fnet = 0 , a = 0 , dv/dt = 0 , v = constant , going up or down with constant speed
8)A particle is observed from two frames s1 and s2. The frame s2 moves with respect to s1 with an acceleration a. Let F1 and F2 be the pseudo forces on the particle when seen from s1 and s2 respectively. Which of the following are not possible ?
a) F_{1}=0, F_{2} neq = 0 (b) F_{1} neq = 0, F_{2}=0 (c) F_{1} neq = 0, F_{2} neq = 0 (d) F_{1}=0, F_{2}=0
ANS) D As the frames are moving with acceleration so pseudo forces must act on the particle , so particle cannot be at rest .
9) A person says that he measured the acceleration of a particle to be nonzero while no force was acting on the particle. (a) He is a liar. (b) His clock might have run slow. (c) His meter scale might have been longer than the standard. (d) He might have used non-inertial frame
ANS) D acceleration non zero but force acting on it is zero , this is the condition of non inertial frame of reference . Hence option d is correct
EXERCISE
1)A block of mass 2 kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first two seconds. Find the magnitude of F.
Ans) mass = 2 kg , distance = 10 metre , time = 2 sec , s = ut + ½ at2 , As u = 0 , S = ½ at 2 , 10 = ½ a (2)2 , a= 20/4 = 5 m/s2 , F = ma , F = 2* 5 = 10 Newton
2)A car moving at 40 km/h is to be stopped by applying brakes in the next 4.0 m. If the car weighs 2000 kg, what average force must be applied on it?
Ans) initial speed = 40 km/hr = 40 * 5/18 = 100/9 m/sec , final speed = 0 ,V2 – U2 = 2 a S , 0 2 – (100/9)2 = 2 * a * 4 , a= 15.375 m/s2
Force = m * a = 2000 * 15.375 = 30750 newton
3)In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of 5*10 6m/s in travelling one centimeter. Assuming straight line motion, find the constant force exerted on the electron. The mass of the electron is 9.1 * 10 -31kg.
Ans) V2 – U2 = 2 a S , a = V2 /2 S , a = (5*10 6)2 / 2 * 10 -2 , a= 25 * 10 12 / 2 * 10 -2 , a = 12.5 * 10 14 m/s 2 Force exerted = m * a = 9.1 * 10 -31 * 12.5 * 10 14 = 113.75 * 10 -17 Newton
4)A block of mass 0.2 kg is suspended from the ceiling by a light string. A second block of mass 0.3 kg is suspended from the first block through another string. Find the tensions in the two strings. Take g = 10 m/s 2 .
Ans ) Tension T1 in the 1st string = (m1+m2)g = (0.2+0.3)10 = 0.5 * 10 = 5 Newton ,Tension T2 in the string = m2 * g = 0.3 * 10 = 3 Newton
5)Two blocks of equal mass m are tied to each other through light string. One of the blocks is pulled along the line joining them with a constant force F. Find the tension in the string joining the blocks.
Ans) net acceleration = F / 2m m/s2 , Tension T = m * a = m * F/2m = F/2 Newton
6)A particle of mass 50 g moves on a straight line. The variation of speed with time is shown in figure. Find the force acting on the particle at t=2,4 and 6 seconds.
Ans) at t = 2 sec , acceleration = slope = 15/3 = 5 m/s2 force = m * a = 50/1000 * 5 = 1/4 newton
At t = 4 sec , acceleration = slope = tan 0 ° = 0 , Force = m* a = m * 0 = 0 Newton
At t= 6 sec , acceleration = slope = 15/3 = – 5 m/s2 , Force = m*a = 50/1000 * (-5) = – (1/4) Newton
7)Two blocks A and B of mass ma and mb respectively are kept in contact on a frictionless table. The experimenter pushes the block A from behind so that the blocks accelerate. If the block A exerts a fore F on the block B, what is the force exerted by the experimenter on A ?
Ans ) net acceleration = Fnet/ma+mb , Force exerted on B by A = F=mb*a(acceleration ) = (mb)* Fnet/ma+mb ,
Doing Cross multiplication , Fnet = F * (ma+mb)/mb
8)Raindrops of radius 1mm and mass 4 mg are falling with a speed of 30 m/s on the head of a bald person. The drops splash on the head and come to rest. Assuming equivalently that the drops cover a distance equal to their radii on the head, estimate the force exerted by the each drop on the head
Ans) Using equation of motion = V2 – U2 = 2 a * S , a =V2 / 2 S = 30 2 / 2 (1*10 -3) =900/2*(10 -3) = 4.5 * 10 5 m/s2 , Force = m*a (acceleration ) = 4 * 10 -3 * 10 -3 * 4.5 * 10 5 m/s2 = 18 * 10 -1 Newton = 1.8 Newton
9)A particle of mass 0.3 kg is subjected to a force F=- kx with k=15 N/m and x being its distance from the origin. What will be its initial acceleration if it is released from a point x=20 cm ?
Ans) F= K x , F= (15)*(20/100)= 3 Newton F = m * a , a = F/m , a = 3/0.3 = 10 m/s2
10) Both the springs shown in figure(5-E2) are unstretched. If the block is displaced by a distance x and released, what will be the initial acceleration?
Ans: F= ( K1 + K2) X , F = m * a , a = ( K1 + K2) X / m
11)A small block B is placed on another block A of mass 5 kg and length 20 cm. Initially, the block B is near the right end of block A. A constant horizontal force of 10 N is applied to the block A. All the surfaces are assumed frictionless. Find the time elapsed before the block B separates from A.
Ans ) F = m * a , a= 10/5 = 2 m/s2
Distance travelled = S = u*t + ½ a t 2 , 20/100 = ½ a t 2 , 1/5 = ½ a t 2 , 2/5 = 2 * t 2 , t = 1/√5 sec
12)A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in figure. Show that the force (assumed equal for both the friends) exerted by each friend on the rope increases as the man moves up. Find the force when the man is at a depth h.
Ans) let F be the force in the string , 2F cosθ = mg , F = mg/2 cosθ , F = mg/2(h/√(h 2 + (d/2) 2 )
13)The elevator shown in figure(5-E5) is descending, with an acceleration of 2 m/s2 . The mass of block A is 0.5 kg. the force exerted by block A on block B is (mass of block B is 1 kg)
Ans: Force applied on B by A = force applied on A by B mg – N = ma , N = m(g-a), N = 0.5(10-2) , N = ½*8 , N = 4 Newton
14)A pendulum bob of mass 50 g is suspended from the ceiling of an elevator. Find the tension in the string if the elevator a. goes up with acceleration 1.2 m/s2 , b. goes up with deceleration 1.2 m/s2 c. goes up with uniform velocity, d. goes down with deceleration 1.2m/s2 e. goes down with deceleration 1.2 m/s2 f. goes down with uniform velocity.
Ans: a) Tension in the string if the elevator goes up with acceleration = 1.2 m/s2 , T = m(g+a), T = 50/1000(9.8+1.2) = (1/20)*11 = 11/20 = 0.55 Newton
b) goes up with deceleration 1.2 m/s2 T = m(g+a), T =50/1000(9.8-1.2) = (1/20)*8.6 = 0.43 Newton
c) goes up with uniform velocity so acceleration is zero , T = mg , T=(50/1000) *10 = 1/20*10 = 1/2 Newton
d) goes down with acceleration 1.2m/s2 T= m(g-a) = 50/1000(9.8 – (1.2)) = 1/20*= 0.43 newton
e) goes down with deceleration 1.2m/s2 T= m(g-a) =(50/1000)*(9.8 – (-1.2)) = 1/20*11 = 0.55 Newton
f) goes down with uniform velocity so acceleration is zero , T = mg , T = (50/1000) (9.8)= 9.8 /20= 0.49 Newton
15)A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and the minimum weights recorded are 72 kg and 60 kg. Assuming that the magnitudes of the acceleration and the deceleration are the same., find a. the true weight of the person and b. the magnitude of the acceleration. Take g = 9.9 m/s2
Ans) for acceleration upwards N= m(g+a), for uniform velocity = N=mg , for deceleration N= m(g-a) ,
72*g= m(g+a),
60*g= m(g-a),
Solving both equation ,2* m*g = 132 *g , m= 66 kg
a)True weight = 66 * 9.9 = 653.4 Newton
b) 72*g= m(g+a),72*g = 66(9.9+a),solving a = 0.9 m/s2
16)Find the reading of the spring balance shown in figure. The elevator is going up with an acceleration of g/10, the pulley and the string are light and the pulley is smooth.
Ans) writing equation for 1.5 kg , T – (1.5*g/10(pseudo force) + 1.5*g) = 1.5 * a —-(1)
Writing equation for 3 kg , (3*g +3*g/10(pseudo force)) – T = 3*a ——— (2)
Solving equation 1 and 2 we get —–
1.5*g+1.5*g/10 = 4.5 *a , a= 3.66 m/s2 , T = 22 Newton , spring balance tension T1 = 2 T = 2*22 = 44 Newton
Mass measured by spring balance = 44 = mg, m= 44/g = 4.4 kg
17)A block of 2 kg is suspended from the ceiling through a massless spring of spring constant k=100 N/m. What is the elongation of the spring? If another 1 kg is added to the block, what would be the further elongation?
Ans) for 2kg , T = mg = 2*10 = 20 = K * x , x= 20/100 , x = 1/5 meter = 0.2 m , For 2+1 kg , T = mg = 3*10 = 30 = K* x , x= 30/100 , x= 0.3 m
Further elongation = 0.3 – 0.2 = 0.1 meter
18)Suppose the ceiling in the previous problem is that the elevator which is going up with an acceleration of 2 m/s2 . Find the elongations.
Ans) T = m*g + m*a = m(g+a) = 2(10+2)=24 newton T = 24 = K *x , x= 24/100 = 0.24 m,
When another 1 kg is added
T = m(g+a)= 3(10+2)= 36 newton T= 36 = K*x , x=36/100= 0.36 m, Elogations = 0.36 – 0.24 = 0.12 m
19)The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite to the direction of velocity and is proportional to it. The balloon carries a mass M and is found to fall down near the earth’s surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v?
Ans) B+Kv= Mg —–1(balloon moves downward with velocity )
M= (B+Kv)/g———(1
B= Kv+M1g——-2(balloon moves upward with velocity )
M1= (B – Kv)/g ———–(2
Reduced mass M – M1 = (B + Kv)/g – (B- Kv)/g = 2 *Kv/g
20)An empty plastic box of mass m is found to accelerate up at the rate of g/6 when placed deep inside water. How much sand should be put inside the box so that it may accelerate down at the rate of g/6?
Ans) when empty plastic box of mass m is accelerate up
U – mg = m*g/6 , U = mg(7/6) , m = 6U/7g
When empty plastic box m is accelerate down
M1g- U = M1*g/6 , M1= 6U/5g ,
M1>m , mass added = 6U/5g-6U/7g=6U/g(1/5 – 1/7) = 6U/g*(2/35)
21)A force F= V * A is exerted on a particle in addition to the force of gravity, where v is the velocity of the particle and A is a constant vector in the horizontal direction. With what minimum speed a particle of mass m be projected so that it continues to move un-deflected with a constant velocity?
Ans) Fnet = mg + V*A ,
0 = mg + V*A ,
V*A = – mg ,
-A*V = – mg ,
AV sinθ = mg ,
V = mg/A sinθ ,
Vmin = mg /A , when sinθ = 1
22)In a simple Atwood machine, two unequal masses m1 and m2 are connected by string going over a clamped light smooth pulley . In a typical arrangement m1 = 300g and m2 = 600 g . The system is released from rest. (a). Find the distance traveled by the first block in the first two seconds. (b). Find the tension in the string. (c). Find the force exerted by the clamp on the pulley.
Ans) writing equation for m1 , T – m1g = m1*a , —(1)
Writing equation for m2 , m2g-T = m2*a , —-(2)
Adding equation 1 and 2 we get ,
(M2g – M1g)/(M1+M2)= a ,
Distance travelled by the first block in in the first two seconds
S = 1/2 a(t) 2 , S= ½ (a)(2) 2
Tension in the string T = M1g+M1a=M1(g+a) , Force exerted by clamp on the pulley T1 = 2T
23)Consider the Atwood machine as shown in the above figure. The larger mass is stopped for a moment 2.0 s after the system is set into motion. Find the time elapsed before the string is tight again.
Ans) acceleration of the block a = (m2g – m1g)/m2+m1 = (0.6 – 0.3)10/0.9= 3.33 m/s2
Velocity of m1 after 2 sec is given by V = U + at , U=0 ,t= 2 sec , V= at , V = 3.33*2=6.66 m/s
Block m2 higher mass is taken as rest for 2 sec , then released so it will travel up to max height and then comes to rest , then it revert it motion for downward , calculation from stop to maximum height V = U + at , V = 0 ,a= -g , U=6.66 m/s , 0 = 6.66 –gt, t=0.66 sec ,
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 32N. 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.
Ans) T = F = (F1-F2)x/L = (32-20)10/30 = 4 Newton
25)Consider the situation shown in figure(5-E9). All the surfaces are frictionless and the string and the pulley are light. Find the magnitude of the acceleration of the two blocks.
Ans) writing equation for 1 kg 3m part , T – 1(g) sin 37 = 1*(a)——(1
Writing equation for 1kg 4m part , 1(g) sin 53 – T = 1*(a) ——(2
Adding 1) and 2) we get a = (g sin 53 – g sin 37) / 2 , a = g(0.8-0.6)/2 , a = g(0.2)/2 , a= 1 m/s2
26)A constant force F=m2g/2 is applied on the block of mass m1 as shown in fig(5-E10). The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1
Ans) writing equation for m2 , m2 g – T = m2 a —–(1
Writing equation for m1 , T – F = m1 a ——-(2
T – m2g/2 = m1 a —–(2
Adding 1 and 2 we get , m2g – m2g/2 = (m1 + m2 )a ,
Acceleration a = m2g/2 (m1+m2)
Hence acceleration for m1 is m2g/2 (m1+m2) .
27)In figure(5-E11) m1 = 5 kg, 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.
Ans) writing equation for m1 , 5g+F –T = 5 * a —–(1)
Writing equation for m2 , T – (2g+F) = 2*a ——-(2) Adding (1) and (2) we get , 3g = 7a , a = 3g/7 , a = 4.2 m/s2
the string breaks m1 moves downward with force F acting downward , F+m1g = m1a , 1+5g= 5a , a = 10.2 m/s2
28)Let m1 = 1 kg, m2 = 2 kg and m3 = 3 kg in figure. 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?
Ans) let acceleration for m1 is a1 , let acceleration for m2 is a2 , let acceleration for m3 is a3 ,
Using work done by tension is zero method ,
2T*a1 +T*a2 + T*a3 = 0 , 2*a1 = -(a2+a3) —–(1
Block m1 moves upward , suppose m2 moves upward , m3 also moves upward
Writing equation for m1 , 2T – 10 = 1*a1 ——–(2 a1= 2T – 10 ,
Writing equation for m2 , T – 20 = 2*a2 ———(3 a2 = T/2 – 10 ,
Writing equation for m3 , T – 30 = 3*a3 ———(4 a3 = T/3 – 10 ,
Putting equation 2,3,4 in equation 1 we get
2(2T – 10) = -( T/2 – 10 + T/3 – 10) , solving it T = 240/29 Newton
Putting it in equation 2 , a1 = 190/29(going up ) , a2= -170/29 (going down ) , a3 = -210/29(going down)
Time taken by m1 to strike on pulley , s= ut + 1/2 at2 , 20/100 = 1/2 (190/29) t2 solving it t = 0.25 sec