KKMishra's Tutorials: Solutions to Problems on "CENTER OF MASS, LINEAR MOMENTUM, COLLISION" - H C Verma's Concepts of Physics, Part-I, Chapter-9, EXERCISES-Q21-Q30

My Channel on YouTube → SimplePhysics with KK

For links to

other chapters - See bottom of the page

CENTER OF MASS, LINEAR MOMENTUM, COLLISION:--
EXERCISES Q-21 to Q-30

21. Light in certain cases may be considered as a stream of particles called photons. Each photon has a linear momentum h/λ where h is the plank's constant and λ is the wavelength of light. A beam of light of wavelength λ is incident on a plane mirror at an angle of incidence θ. Calculate the change in the linear momentum of a photon as the beam is reflected by the mirror.

ANSWER: Let the magnitude of the linear momentum of the photon be B = h/λ.

Figure for Q-21

The component of the momentum parallel to the mirror remains the same = B.sinθ, hence no change of linear momentum parallel to the mirror. The component of the linear momentum perpendicular to the mirror before reflection = B.cosθ

and after reflection = -B.cosθ

hence change in linear momentum = B.cosθ - (-B,cosθ)

= 2B.cosθ

= 2(h/λ).cosθ

= (2h.cosθ)/λ

22. A block at rest explodes into three parts. Two parts start moving along X and Y axes respectively with equal speeds of 10 m/s. Find the initial velocity of the third part.

ANSWER: Assume that all the three parts are of equal mass = m.

Speeds of the parts along the x and y-axes = v = 10 m/s. Let the speed of the third particle = v' make an angle θ from the positive x-axis.

Figure for Q-22

Linear momenta of first two particles are mv each while that of third is mv'.
Since there is no external force on the block, the total linear momentum of three blocks will remain zero (their resultant will be zero) and the three vectors will be in equilibrium.
Hence from Lami's theorum
v/sin(θ-π/2) =v/sin(2π-θ) = v'/sinπ/2
Hence v = v'.sin(θ-π/2) also v = v'.sin(2π-θ)
→θ-π/2 =2π-θ
→2θ = 5π/2
→ θ = 5π/4 = 225° from x-axis anticlockwise i.e. 135° from x-axis clockwise.
And v' = v/sin(2π-θ) =v/sin(2π-5π/4) = v/sin3π/4 =v/(1/√2) =√2v =10√2 m/s

Alternately
Linear momentum of first particle = mv.i
Second particle = mv.j
Resultant of momenta of these two particles = mv.i + mv.j
Hence momentum of the third particle must be equal and opposite to this resultant, i.e. = -mv.i - mv.j
So velocity of the third particle = -v.i - v.j
Magnitude of the velocity = √2v = 10√2 m/s
and tanθ = -v/-v =1 = tan 225°
θ = 225° clockwise from x-axis.

23. Two fat astronauts each of mass 120 kg are traveling in a closed spaceship moving at a speed of 15 km/s in the outer space far removed from all other material objects. The total mass of the spaceship and its contents including the astronauts is 660 kg. If the astronauts do slimming exercise and thereby reduce their masses to 90 kg each with what velocity will the spaceship move?

Figure for Q-23

ANSWER: Since there is no external force on the system and the spaceship is closed meaning thereby no mass is lost to the outside, hence the spaceship will keep moving with the same velocity i.e. 15 km/s

24. During a heavy rain, hailstone of average size 1.0 cm in diameter fall with an average speed of 20 m/s. Suppose 2000 hailstones strike every square meter of a 10m x 10m roof perpendicularly in one second and assume that the hailstones do not rebound. Calculate the average force exerted by the falling hailstones on the roof. Density of a hailstone is 900 kg/m³.

ANSWER: Diameter of a hailstorm = 1.0 cm = 0.01 m

Radius = 0.005 m

Mass of hailstone falling per second per square meter

=Number*Density*Volume

= 2000*900*4π(0.005)³/3

= 0.942 kg

Momentum of this mass = 0.942*20 = 18.84 kg-m/s

Since the hailstones do not rebound their momentum after the strike is zero. Change in momentum per second = 18.84-0 = 18.84 kg-m/s²

But rate of change of momentum = Force applied

So the force per square meter of the roof = 18.84 N

Total force on the roof = 10 m x 10 m x 18.84 N = 1884 N ≈1900 N

25. A ball of mass m is dropped onto a floor from a certain height. The collision is perfectly elastic and the ball rebounds to the same height and again falls. Find the average force exerted by the ball on the floor during a long time interval.

ANSWER: Let the velocity of the ball just before the strike be v. Since the collision is perfectly elastic the velocity of the ball will be opposite in direction but equal in magnitude. Hence the momentum before and after strike will be mv and -mv. Change in momentum = 2mv.

If the ball is dropped from a height of h and it takes time t to reach the floor, then

v = √(2gh)

and h = 0*t+½gt²

→t = √(2h/g)

To reach the same point time taken = 2t =2√(2h/g)

If during a long interval of time the ball jumps n times,

Time in n jumps = 2nt = 2n√(2h/g)

Change in momentum = 2mv*n = 2mn√(2gh)

Rate of Change in momentum = 2mn√(2gh)/2n√(2h/g)

= m√g² = mg

26. A railroad car of mass M is at rest on frictionless rails when a man of mass m starts moving on the car towards the engine. If the car recoils with a speed v backward on the rails, with what velocity is the man approaching the engine?

ANSWER: Let the velocity of the man be v' approaching the engine. This velocity is relative to the railroad car. If the velocity of the man with respect to the ground be = V, then

V = v'-v

Since there is no external force on the system, the momentum will be conserved. Hence

Mv = m(v'-v)

→v'-v = Mv/m

→v' = v+Mv/m = (1+M/m)v

27. A gun is mounted on a railroad car. The mass of the car, the gun, the shells and the operator is 50m where m is the mass of one shell. If the muzzle velocity of the shells is 200 m/s, what is the recoil speed of the car after the second shot? Neglect friction.

ANSWER: After the first shot, momentum of the shell = m* 200

=200m

Mass of the residual recoiling body = 50m-m = 49m

If its recoil velocity = v

then 49mv = 200m ---------- (i)

→v = 200/49 m/s

After the second shot residual mass = 49m - m = 48m

According to the law of conservation of momentum, the total momentum of residual mass and the second shell will be equal to 49mv. i.e.
49mv + 48mv' - 200m = 49mv {v' is the additional velocity imparted to the residual mass after the second impact}
→48mv' = 200m
→v' = 200/48
Hence the recoil speed of the car = v+v'
= 200/49 + 200/48
= 200((1/49+1/48) m/s

28. Two persons each of mass m are standing at the two extremes of a railroad car of mass M resting on a smooth track (figure 9-E10). The person on left jumps to the left with a horizontal speed u with respect to the state of the car before the jump. Thereafter, the other person jumps to the right, again with the same horizontal speed u with respect to the state of the car before his jump. Find the velocity of the car after both the persons have jumped off.

Figure for Q-28

ANSWER: Since the car is at rest initially, its momentum = 0.
when the first person jumps o the left, the total momentum of the jumping mam and the car should be zero.
(M+m)u' + mu = 0 {u' is the velocity of the car}
→u' = -mu/(M+m) {Negative velocity means u and u' have opposite directions)
Speed of the second person with respect to the car = u,
His momentum = mu
Since there is no external force on the system, the same amount of momentum in opposite direction will be excreted to the cart. Hence the momentum of the cart after the second person jumps
= Mu'-mu
= -M*mu/(M+m)-(-mu)
= mu-mMu/(M+m)
= mu{M+m-M}/(M+m)
= m²u/(M+m)
Hence the speed of the cart
= Momentum/Mass
= m²u/M(M+m) to the left.

29. Figure (9-E11) shows a small block of mass m which is started with a speed v on the horizontal part of the bigger block of mass M placed on a horizontal floor. The curved part of the surface shown is semicircular. All the surfaces are frictionless. Find the speed of the bigger block when the smaller block reaches the point A of the surface.

Figure for Q-29

ANSWER: When the smaller block is started with a velocity v, the total momentum of the system in the horizontal direction = mv.
When the smaller block reaches A, its speed is vertical and has no horizontal component. So horizontal component of the momentum of the smaller block = zero. Since there is no external force in the horizontal direction, the total momentum of the system in the horizontal direction will be conserved. If the velocity of the bigger block is V,
(M+m)V = mv
→V = mv/(M+m)

30. In a typical Indian bagghi (a luxury cart drawn by horses), a wooden plate is fixed on the rear on which ane person can sit. A bugghi of mass 200 kg is moving at a speed of 10 km/hr. As it overtakes a schoolboy walking at a speed of 4 km/hr, the boy sits on the wooden plate.If the mass of the boy is 25 kg, what will be the new velocity of the bugghi?

ANSWER: M = 200 kg, v = 10 km/hr
Momentum of the bugghi = Mv = 200 kg*10 km/hr
=2000 kg-km/hr
m = 25 kg, v' = 4 km/hr
Momentum of the boy = 25 kg*4 km/hr
=100 kg-km/hr
Since both has same direction of speed, taking the boy and the bugghi as a system, total momentum of the system = 2000+100 =2100 kg-km/hr.
When the boy sits on the bugghi, total mass of the system = 200+25 =225 kg. If the new speed of the bugghi is V, from the conservation principle of the momentum,
225V = 2100 kg-km/hr
V = 2100/225 km/hr =(28*75)/(3*75) km/hr = 28/3 km/hr

===<<<O>>>===

Links to the chapters -

ALL LINKS

CHAPTER- 10 - Rotational Mechanics

Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 TO Q-15

EXERCISES - Q-16 TO Q-30

CHAPTER- 9 - Center of Mass, Linear Momentum, Collision

Questions for Short Answers

Objective - I