Chapter 5 Laws of motion Solutions
Question - 11 : - A truck starts from rest and acceleratesuniformly at 2.0 m s–2. At t = 10 s, a stone isdropped by a person standing on the top of the truck (6 m high from theground). What are the (a) velocity, and (b) acceleration of the stone at t =11 s? (Neglect air resistance.)
Answer - 11 : -
Answer: (a) 22.36 m/s, at anangle of 26.57° with the motion of the truck
(b) 10 m/s2
(a) Initial velocity of the truck, u =0
Acceleration, a = 2 m/s2
Time, t = 10 s
Asper the first equation of motion, final velocity is given as:
v = u + at
=0 + 2 × 10 = 20 m/s
Thefinal velocity of the truck and hence, of the stone is 20 m/s.
At t = 11 s, the horizontalcomponent (vx) of velocity, in the absence of air resistance, remainsunchanged, i.e.,
vx = 20 m/s
The vertical component (vy) ofvelocity of the stone is given by the first equation of motion as:
vy = u + ayδt
Where, δt = 11 – 10 = 1 s and ay =g = 10 m/s2
∴vy = 0 + 10 × 1 = 10m/s
The resultant velocity (v) of thestone is given as:
Let θ be the angle made bythe resultant velocity with the horizontal component of velocity, vx
=26.57°
(b) When the stone is droppedfrom the truck, the horizontal force acting on it becomes zero. However, thestone continues to move under the influence of gravity. Hence, the accelerationof the stone is 10 m/s2 and it acts vertically downward.
Question - 12 : - A bob of mass 0.1 kg hung from the ceilingof a room by a string 2 m long is set into oscillation. The speed of the bob atits mean position is 1 m s–1. What is the trajectory of the bob ifthe string is cut when the bob is (a) at one of its extreme positions, (b) atits mean position.
Answer - 12 : -
Answer: (a) Vertically downward
(b) Parabolic path
(a) At the extreme position, the velocity of thebob becomes zero. If the string is cut at this moment, then the bob will fallvertically on the ground.
(b)At the mean position, thevelocity of the bob is 1 m/s. The direction of this velocity is tangential tothe arc formed by the oscillating bob. If the bob is cut at the mean position,then it will trace a projectile path having the horizontal component ofvelocity only. Hence, it will follow a parabolic path.
Question - 13 : - Aman of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of 10 m s–1,
(b) downwards with a uniform acceleration of5 m s–2,
(c) upwards with a uniform acceleration of 5m s–2.
Whatwould be the readings on the scale in each case?
(d)What would be the reading if the lift mechanism failed and it hurtled downfreely under gravity?
Answer - 13 : -
(a) Mass of the man, m = 70 kg
Acceleration, a = 0
UsingNewton’s second law of motion, we can write the equation of motion as:
R – mg = ma
Where, ma is the net forceacting on the man.
As the lift is moving at a uniform speed,acceleration a = 0
∴R = mg
=70 × 10 = 700 N
∴Reading on the weighing scale = 
(b) Mass of the man, m = 70 kg
Acceleration, a = 5 m/s2 downward
UsingNewton’s second law of motion, we can write the equation of motion as:
R + mg = ma
R = m(g – a)
=70 (10 – 5) = 70 × 5
=350 N
∴Reading on the weighing scale = 
(c) Mass of the man, m = 70 kg
Acceleration, a = 5 m/s2 upward
UsingNewton’s second law of motion, we can write the equation of motion as:
R – mg = ma
R = m(g + a)
=70 (10 + 5) = 70 × 15
=1050 N
∴Reading on the weighing scale = 
(d) When the lift moves freely under gravity,acceleration a = g
UsingNewton’s second law of motion, we can write the equation of motion as:
R + mg = ma
R = m(g – a)
= m(g – g) = 0
∴Reading on the weighing scale = 
Theman will be in a state of weightlessness.
Figure 5.16 shows the position-time graph ofa particle of mass 4 kg. What is the (a) force on the particle for t< 0, t > 4 s, 0 < t< 4 s? (b) impulse at t = 0 and t = 4s? (Consider one-dimensional motion only).
Answer - 14 : -
(a) For t <0
Itcan be observed from the given graph that the position of the particle iscoincident with the time axis. It indicates that the displacement of theparticle in this time interval is zero. Hence, the force acting on the particleis zero.
For t > 4 s
Itcan be observed from the given graph that the position of the particle isparallel to the time axis. It indicates that the particle is at rest at adistance of 3 m from the origin. Hence, no force is acting on the particle.
For 0 < t < 4
Itcan be observed that the given position-time graph has a constant slope. Hence,the acceleration produced in the particle is zero. Therefore, the force actingon the particle is zero.
(b) At t = 0
Impulse= Change in momentum
= mv – mu
Mass of the particle, m = 4kg
Initial velocity of the particle, u =0
Finalvelocity of the particle,
∴Impulse 
At t = 4 s
Initialvelocity of the particle, 
Final velocity of the particle, v =0
∴ Impulse 
Two bodies of masses 10 kg and 20 kgrespectively kept on a smooth, horizontal surface are tied to the ends of alight string. A horizontal force F = 600 N is applied to (i)A, (ii) B along the direction of string. What is the tension in the string ineach case?
Answer - 15 : -
Horizontal force, F = 600 N
Mass of body A, m1 =10 kg
Mass of body B, m2 =20 kg
Total mass of the system, m = m1 + m2 =30 kg
Using Newton’s second law of motion, theacceleration (a) produced in the system can be calculated as:
F = ma
When force F is applied onbody A:
Theequation of motion can be written as:
F – T = m1a
∴T = F – m1a
= 600 – 10 × 20 = 400 N … (i)
When force F is applied onbody B:
Theequation of motion can be written as:
F – T = m2a
T = F – m2a
∴T = 600 – 20 × 20 = 200 N … (ii)
Question - 16 : - Twomasses 8 kg and 12 kg are connected at the two ends of a light inextensiblestring that goes over a frictionless pulley. Find the acceleration of themasses, and the tension in the string when the masses are released.
Answer - 16 : -
Thegiven system of two masses and a pulley can be represented as shown in thefollowing figure:
Smaller mass, m1 =8 kg
Larger mass, m2 =12 kg
Tension in the string = T
Mass m2, owing to itsweight, moves downward with acceleration a,and mass m1 movesupward.
ApplyingNewton’s second law of motion to the system of each mass:
For mass m1:
Theequation of motion can be written as:
T – m1g = ma …(i)
For mass m2:
Theequation of motion can be written as:
m2g – T = m2a …(ii)
Adding equations (i) and (ii),we get:
Therefore, the acceleration of the masses is2 m/s2.
Substituting the value of a inequation (ii), we get:
Therefore,the tension in the string is 96 N.
Question - 17 : - Anucleus is at rest in the laboratory frame of reference. Show that if itdisintegrates into two smaller nuclei the products must move in oppositedirections.
Answer - 17 : -
Let m, m1,and m2 be the respective masses of the parentnucleus and the two daughter nuclei. The parent nucleus is at rest.
Initialmomentum of the system (parent nucleus) = 0
Let v1 and v2 bethe respective velocities of the daughter nuclei having masses m1 and m2.
Totallinear momentum of the system after disintegration = 
Accordingto the law of conservation of momentum:
Totalinitial momentum = Total final momentum
Here,the negative sign indicates that the fragments of the parent nucleus move indirections opposite to each other.
Two billiard balls each of mass 0.05 kgmoving in opposite directions with speed 6 m s–1 collide andrebound with the same speed. What is the impulse imparted to each ball due tothe other?
Answer - 18 : -
Massof each ball = 0.05 kg
Initialvelocity of each ball = 6 m/s
Magnitude of the initial momentum of eachball, pi = 0.3 kg m/s
Aftercollision, the balls change their directions of motion without changing themagnitudes of their velocity.
Final momentum of each ball, pf =–0.3 kg m/s
Impulseimparted to each ball = Change in the momentum of the system
= pf – pi
=–0.3 – 0.3 = –0.6 kg m/s
Thenegative sign indicates that the impulses imparted to the balls are opposite indirection.
Question - 19 : - A shell of mass 0.020 kg is fired by a gunof mass 100 kg. If the muzzle speed of the shell is 80 m s–1, whatis the recoil speed of the gun?
Answer - 19 : -
Mass of the gun, M = 100 kg
Mass of the shell, m =0.020 kg
Muzzle speed of the shell, v =80 m/s
Recoil speed of the gun = V
Boththe gun and the shell are at rest initially.
Initialmomentum of the system = 0
Final momentum of the system = mv – MV
Here,the negative sign appears because the directions of the shell and the gun areopposite to each other.
Accordingto the law of conservation of momentum:
Finalmomentum = Initial momentum
mv – MV = 0
Question - 20 : - Abatsman deflects a ball by an angle of 45° without changing its initial speedwhich is equal to 54 km/h. What is the impulse imparted to the ball? (Mass ofthe ball is 0.15 kg.)
Answer - 20 : -
Thegiven situation can be represented as shown in the following figure.
Where,
AO= Incident path of the ball
OB= Path followed by the ball after deflection
∠AOB = Angle between the incident anddeflected paths of the ball = 45°
∠AOP = ∠BOP = 22.5° = θ
Initial and final velocities of the ball = v
Horizontal component of the initial velocity= vcos θ along RO
Vertical component of the initial velocity= vsin θ along PO
Horizontal component of the final velocity= vcos θ along OS
Vertical component of the final velocity= vsin θ along OP
Thehorizontal components of velocities suffer no change. The vertical componentsof velocities are in the opposite directions.
∴Impulse imparted to the ball = Change in thelinear momentum of the ball
Mass of the ball, m = 0.15kg
Velocity of the ball, v =54 km/h = 15 m/s
∴Impulse = 2 × 0.15 × 15 cos 22.5° = 4.16 kgm/s