Solutions to Problems on "CENTER OF MASS, LINEAR MOMENTUM, COLLISION" - H C Verma's Concepts of Physics, Part-I, Chapter-9, EXERCISES-Q43-Q54

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CENTER OF MASS, LINEAR MOMENTUM, COLLISION:--
EXERCISES Q-43 to Q-54

43. A Projectile is fired with a speed u at an angle θ above a horizontal field. The coefficient of restitution of collision between the projectile and the field is e. How far from the starting point, does the projectile makes its second collision with the field?     

ANSWER: Just before the first collision the speed of projectile will be u and its component in horizontal direction = u.cosθ and in vertical direction = u.sinθ (Downward). After the collision, the horizontal speed will not change but the vertical speed will be = eu.sinθ (Upward). So after the collision, the speed of projection v will be given by v² =(u.cosθ)²+(eu.sinθ)² = u²(cos²θ+e²sin²θ)

The angle of projection β is given by tanβ = eu.sinθ/u.cosθ =e.tanθ

To calculate the horizontal range of projectile after collision we will need the value of Sin2β which is 

sin2β=2sinβ.cosβ = 2tanβ.cos²β

(Multiplying by cosβ/cosβ)

=2tanβ/sec²β

=2tanβ/(1+tan²β)

=2etanθ/(1+e²tan²θ)

(Putting the value of tanβ)

Now the horizontal range = v²sin2β/g

=[u²(cos²θ+e²sin²θ)]* 2etanθ/g(1+e²tan²θ)

=[u²(cos²θ+e²sin²θ)]* 2etanθ*cos²θ/g(cos²θ+e²sin²θ)  

=eu²* 2sinθcosθ/g

=eu²sin2θ/g

The horizontal range of the projectile motion before the collision

= u²sin2θ/g

Hence from the starting point, the projectile will make the second collision at a distance of u²sin2θ/g+ eu²sin2θ/g

= (1+e)u².sin 2θ/g.

44. A ball falls on an inclined plane of inclination θ from a height h above the point of impact and makes a perfectly elastic collision. Where will it hit the plane again?      

ANSWER:  Let us draw the diagram as below,

Figure for Q-44

Speed just before the impact u = √(2gh)

Since the collision is perfectly elastic, the rebound speed will also be the same and it will make the same angle with the normal to the plane, here equal to θ.

       After the impact the angle of projection (from geometry) β

= 90°-2θ

If it hits the plane again at a distance L along the incline and takes time t between the impacts, then

Horizontal Distance X = L.cosθ = u.cosβ.t = ut.cos(90°-2θ) =ut.sin2θ

→ t = L.cosθ/u.sin2θ = L.cosθ/2usinθ.cosθ = L/2u.sinθ

Vertical distance Y= L.sinθ = -u.sinβ.t + ½gt² 

or, L.sinθ = - u.sin(90°-2θ)*L/2u.sinθ + gL²/8u²sin²θ  

or, sinθ = -cos2θ/2sinθ+ gL/8u²sin²θ

or, 1 =- (cos²θ-sin²θ)/2sin²θ +gL/8u²sin³θ

or, gL/8u²sin³θ = 1+cos²θ/2sin²θ-1/2 = cos²θ/2sin²θ+1/2

or, gL/16gh.sin³θ = cos²θ/2sin²θ+1/2

or, L = 16h.sin³θ.cos²θ/2sin²θ + 16h.sin³θ/2

or, L = 8h.sinθ.cos²θ+8h.sin³θ
or, L = 8h.sinθ(cos²θ+sin²θ) =8h.sinθ
Hence the ball will hit the inclined plane at a distance 8h.sinθ along the incline.

45.  Solve the previous problem if the coefficient of restitution is e. Use θ = 45°, e = ¾, and h = 5 m.      

ANSWER: In this case, the angle of rebound (reflection) will not be equal to the angle of incidence. Let the angle of rebound be α. 

Speed just before the impact u=√(2gh)

Component along incline u.sinθ will remain unchanged along the incline just after the impact. The component perpendicular to plane after the impact = eu.cosθ

So, tanα = u.sinθ/eu.cosθ =tanθ/e

Figure for Q-45

Now angle of projection β = 90°-θ-α

And the speed of projection v =√[(u²sin²θ)+(e²u²cos²θ)]

= u√[sin²θ+e²cos²θ] 

If the time taken in the second impact is t, then horizontal distance covered X =L.cosθ = v.cos(90°-θ-α).t = vt.sin(θ+α)

→t = L.cosθ/v.sin(θ+α)

The vertical distance covered in second impact Y is given by

L.sinθ = - v.sin(90°-θ-α)t+½gt²

→L.sinθ = - v.cos(θ+α)L.cosθ/v.sin(θ+α)+gL²cos²θ/2v².sin²(θ+α)

→ gLcos²θ/2v².sinθ.sin²(θ+α) =1+cosθ.cos(θ+α)/sinθ.sin(θ+α)

→gLcos²θ/2v²sinθsin²(θ+α)= [sinθ.sin(θ+α)+cosθ.cos(θ+α)]/sinθ.sin(θ+α)

→gLcos²θ/2v²sin(θ+α)= Cos(θ+α-θ) =Cosα
→L=2v²cosα.sin(θ+α)/g.cos²θ

Now let us simlify it from the given data. 

v² = u²[sin²θ+e²cos²θ] =2gh[sin²θ+e²cos²θ]

=2.g.5[sin²45°+9cos²45°/16]=10g[½+9/32]=250g/32 = 125g/16

tanα = tanθ/e = 4/3*tan45°=4/3, →sinα = 4/5, cosα=3/5

sin(θ+α) = sinθcosα+cosθsinα = sin45°*3/5+cos45°*4/5

=3/5√2+4/5√2 = 7/5√2 =7√2/10

Now, 

L=2*(125g/16)(3/5)(7√2/10)/g(1/2)

=250*3*7√2*2/16*50
=5*3*7√2/8
=105*1.41/8
=18.50 m

18.5 m along the incline

46. A block of mass 200 g is suspended through a vertical spring. The spring is stretched by 1.0 cm when the block is in equilibrium. A particle of mass 120 g is dropped on the block from a height of 45 cm. The particle sticks to the block after the impact. Find the maximum extension of the spring. Take g = 10 m/s².    

ANSWER: Mass of the block M = 200 g = 0.200 kg
The weight of the block = Mg = 0.200 x 10 = 2.00 N
Elongation of the spring due to this weight = 1.0 cm = 0.01 m
Hence the spring constant of the spring = 2.00/0.01 = 200 N/m
Mass of the particle m = 120 g = 0.120 kg
Height of drop = 45 cm =0.45 m
Speed of the particle when it touched the block, v = √(2*10*0.45)
=√9 =3 m/s
Total Linear momentum of the system before impact = M*0+m*v
=0.120*3 = 0.36 kg-m/s
Linear Momentum of the system just after impact = (M+m)v'
{Where v' is the speed of the block and the particle after impact}
= (0.200+0.120)v' =0.320v'
From the conservation principle of linear momentum,
0.320v' = 0.36
→v' = 0.36/0.32 = 1.125 m/s
Taking the equilibrium position of the block as zero P.E. level, Total energy of the system just after the impact = ½(M+m)v'²
=½*0.320*1.125² = 0.2025 J
If maximum elongation be x, then total energy at this level 
=½kx²-(M+m)gx
=½*200x²-(0.200+0.120)*10x
=100x²-3.2x
From the conservation principle of energy,
100x²-3.2x=0.2025
→100x²-3.2x-0.2025 = 0
→x= {3.2士√(3.2²+4*100*0.2025)}/2*100
={3.2士√91.24}/200={3.2士9.55}/200 =12.75/200 
{Taking the  +ve sign only because negative x is not possible in this case}
=0.063 m
=6.30 cm
Hence the maximum extension of the spring will be 6.30 cm.    

47. A bullet of mass 25 g is fired horizontally into a ballistic pendulum of mass 5.0 kg and gets embedded in it(figure 9-E14). if the center of the pendulum rises by a distance of 10 cm, find the speed of the bullet.    

Figure for Q-47

ANSWER: Let the speed of the pendulum just after the impact is u. The momentum of the pendulum just after the impact = (5.0+0.025)u =5.025u
The momentum of the system just before the impact =0.025*v 
From the conservation of linear momentum,
0.025v = 5.025u
→u = 0.025v/5.025 m/s
K.E. of the pendulum = ½5.025u²
When the pendulum rises to the maximum height of 10 cm i.e. 0.10 m this K.E. is converted to P.E. Hence,
½5.025u² = 5.025gh
→u² = 2gh
→(0.025/5.025)²v² = 2*10*0.10
→v² = 2(5.025/0.025)²
→v =√2(5.025/0.025)  = 283 m/s 

48. A bullet of mass 20 g moving horizontally at a speed of 300 m/s is fired into a wooden block of mass 500 g suspended by a long string. The bullet crosses the block and emerges on the other side. If the center of mass of the block rises through a height of 20 cm, find the speed of the bullet as it emerges from the block.   

ANSWER: Mass of the bullet, m = 20 g = 0.020 kg
The speed of the bullet, u=300 m/s
The mass of the block M = 500 g =0.500 kg
If  the speed of the block after penetration is v and that of the bullet is v', then from the conservation law of linear momentum,
0.500*v+0.020*v' = 0.02*300
→v = (6 - 0.02v')/0.500 = 12-0.04v'
Now the rise in potential energy = Mgh
(h=20 cm =0.20 m)
=0.500*10*0.20
= 1.0 J
This P.E. is due to conversion of K.E. of the block, ½Mv². Hence,
½Mv² = 1
→v² =2/M =2/0.50 =4
→(12-0.04v')² = 4 
→12-0.04v' = 2
→0.04v' = 12-2 = 10
→ v' = 10/0.04 = 250 m/s

49. Two masses m₁ and m₂ are connected by a spring of spring constant k and are placed on a frictionless horizontal surface. Initially, the spring is stretched through a distance x₀ when the system is released from rest. Find the distance moved by the two masses before they again come to rest.    

ANSWER: Energy stored in the spring in stretched position = ½kx₀²  

Figure for Q-49

Let the original unstretched length of the spring be x. If x₁ and x₂ be the distance moved by the two masses before they come to rest, the length of the compressed spring = x+x₀-(x₁+x₂). So the compression of the spring = x-{x+x₀-(x₁+x₂)} = x₁+x₂-x₀. Now the energy stored in compressed position = ½k(x₁+x₂-x₀)². Since there is no force in the horizontal direction, the energy stored in stretched and compressed position will be the same. Hence, 
½kx₀² = ½k(x₁+x₂-x₀)². 
→ x₀ = x₁+x₂-x₀ 
→2x₀= x₁+x₂ 
If the center of mass in this position be at a distance d from m₁ then,
d = m₂ (x+x₀)/(m₁+m₂) 
When the masses are at rest, let the distance of CoM from the mass m₁ be d', Now
d' =  m₂{x+x₀-(x₁+x₂)}/(m₁+m₂) 
Since there is no external force in the horizontal direction, the CoM will remain at the same place. So the distance moved by the mass m₁ =d-d'
= m₂[(x+x₀)-{x+x₀-(x₁+x₂)}]/(m₁+m₂)
= m₂(x₁+x₂)/(m₁+m₂) 
= m₂*2x₀/(m₁+m₂)
= 2m₂x₀/(m₁+m₂) 
So the distance moved by the mass m₁ is 2m₂x₀/(m₁+m₂).
Similarly, it can be shown that the distance moved by mass m₂ is 2m₁x₀/(m₁+m₂).

50. Two blocks of masses m₁ and m₂ are connected by a spring of spring constant k (figure 9-E15). The block of mass m₂ is given a sharp impulse so that it acquires a velocity v₀ towards right. Find (a) the velocity of the center of mass, (b) the maximum elongation that the spring will suffer.  

Figure for Q-50

 

ANSWER: (a) The velocity of the center of mass is given by 
V = (m₁*0+m₂v₀)/(m₁+m₂)
= (m₂v₀)/(m₁+m₂)      
(b)  The maximum elongation will be when both masses have the same velocity and hence also the CoM = V. Let the maximum elongation be x. In this stage, the potential energy stored in the spring =½kx²
The total energy, in the beginning, = ½m₂v₀²
The total energy at maximum elongation =½(m₁+m₂)V²+½kx²
From the conservation law of energy, these two will be equal.
½(m₁+m₂)V²+½kx²= ½m₂v₀²
→kx² = m₂v₀²-(m₁+m₂)V² =m₂v₀²-(m₁+m₂)(m₂v₀)²/(m₁+m₂)²
→kx²= m₂v₀²-(m₂v₀)²/(m₁+m₂) =v₀²{m₂-m₂²/(m₁+m₂)}
→x²= v₀²{m₁m₂+m₂²-m₂²}/(m₁+m₂)k= v₀²{m₁m₂}/(m₁+m₂)k
→x=v₀√[m₁m₂/(m₁+m₂)k]          

51. Consider the situation of the previous problem. Suppose each of the blocks is pulled by a constant force F instead of any impulse. Find the maximum elongation that the spring will suffer and the distances moved by the two blocks in the process.     

ANSWER: Since the forces are equal and opposite, the net horizontal force on the system =0. Hence the center of mass will remain stationary. Let the masses m₁ and m₂ move through distances x₁ and x₂ respectively. The work done by the forces = The energy stored in the spring.
Fx₁ +Fx₂ =½k(x₁+x₂)²
→ F(x₁ +x₂)= ½k(x₁+x₂)²  
→2F=k(x₁ +x₂)
→(x₁ +x₂)=2F/k
Hence the maximum elongation in the spring = 2F/k.  
If the original length of the spring is x. Initially the distance of CoM from m₁= xm₂/(m₁ +m₂)
Finally the distance of CoM from m₁= (x+x₁+x₂)m₂/(m₁ +m₂)  
Hence the distance moved by m₁ 
= (x+x₁+x₂)m₂/(m₁ +m₂)-xm₂/(m₁ +m₂) 
= (x₁+x₂)m₂/(m₁ +m₂)
= 2Fm₂/k(m₁ +m₂)
Similarly, it can be shown that the distance moved by the mass m₂ is  2Fm₁/k(m₁ +m₂)      

          
52. Consider the situation of the previous problem. Suppose the blocks of mass m₁ is pulled by a constant force F₁ and the other block is pulled by a constant force F₂. Find the maximum elongation that the spring will suffer.   

ANSWER: Let F₁ >F₂ 
The net force on the system = F₁-F₂  
Let the maximum elongation of the spring be x and when this maximum elongation is reached, the acceleration of the system 
a=(F₁-F₂)/(m₁ +m₂)     

If the tension force in this state be T, 
F₁-T = m₁a
→T= F₁-m₁a= F₁-m₁(F₁-F₂)/(m₁ +m₂) 
= (m₁F₁+m₂F₁-m₁F₁ +m₁F₂)/(m₁ +m₂) 
= (m₂F₁+m₁F₂)/(m₁ +m₂) 
Work done on the system = Energy stored in the spring
Tx=½kx²
→T=½kx
(m₂F₁+m₁F₂)/(m₁ +m₂)=½kx
x = 2(m₂F₁+m₁F₂)/(m₁ +m₂)k

53. Consider a gravity-free hall in which an experimenter of mass 50 kg is resting on a 5.0 kg pillow 8 ft above the floor of the hall. He pushes the pillow down so that it starts falling at a speed of 8 ft/s. The pillow makes a perfectly elastic collision with the floor, rebounds and reaches the experimenter's head. Find the time elapsed in the process.    

ANSWER: The speed of the experimenter = 5*8/50 ft/s  = 0.80 ft/s
Since the collision is perfectly elastic, the rebound speed of the pillowv=8 ft/s. The time taken by the pillow to reach its original position = 16/8 =2 s. Let the time taken by the pillow to reach the experimenter from here is t. The distance traveled = 8t. It is also the distance traveled by the experimenter in 2+t seconds. Hence,
0.80(2+t)=8t
→1.60 = 8t-0.80t = 7.20t
→t=1.60/7.20 = 0.22 s
Hence the time elapsed in the process =2 + 0.22 = 2.22 s  

54. The track shown in figure (9-E16) is frictionless. The block B of mass 2m is lying at rest and the block A of mass m is pushed along the track with some speed. The collision between A and B is perfectly elastic. With what velocity should the block A be started to get the sleeping man awakened?    

Figure for Q-54

ANSWER: Let the velocity of the block A be v and the velocity of the block B after the collision v'. To reach the block B at the head of the sleeping man energy given to it must be at least equal to 2m*g*h = 2mgh. So the kinetic energy of the block B after the impact ½(2m)v'² = 2mgh
→ v'²=2gh
→ v' =√(2gh)
Now total energy of the block A = mgh + ½mv². 
Let the velocity of the block A when it reaches block B is u, then
½mu² = mgh + ½mv²
→u² = 2gh+v²      --------------------------- (i)
→u = √(2gh+v²) 
In the collision, from the conservation of linear momentum,
mu+2m*0 = 2mv'+mu'     {u' is the velocity of the block A after impact}
u = 2v'+u'
→u' = u-2v'
From the conservation law of energy,
½mu² = ½*2m*v'²+½mu'²
→u² =2v'²+u'²  
→u'² = u²-2v'²
→(u-2v')²=u²-2v'²
→u²+4v'²-4uv' =u²-2v'²
→6v'²-4uv' = 0
→v'(6v'-4u)=0
Either v'=0 or v'=2u/3. Since v' can not be zero,
Hence v'=2u/3
→√(2gh) = 2u/3
→4u²=18gh
→u²=9gh/2.     

From (i)
u² = 2gh+v²
→v² = u²-2gh
→v²=9gh/2-2gh
→v²=5gh/2 =2.5gh
→v=√(2.5gh)
So the block A should be started with a velocity more than √(2.5gh).

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

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