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CENTER OF MASS, LINEAR MOMENTUM, COLLISION:--
EXERCISES Q-55 to Q-64

55. A bullet of mass 10 g moving horizontally at a speed of 50√7 m/s strikes a block of mass 490 g kept on a frictionless track as shown in figure (9-E17). The bullet remains inside of the block and system proceeds towards the semicircular track of radius 0.2 m. Where will the block strike the horizontal part after leaving the semicircular track?

Figure for Q-55

ANSWER: Mass of the bullet, m = 10g = 0.01 kg

The speed of the bullet, v = 50√7 m/s.

The mass of the block, m' = 490 g = 0.49 kg

The speed of the block, v' = 0

Combined mass of the system, M = 10+490 = 500 g = 0.50 kg

Let the speed of the system after the collision is V. From the principle of conservation of linear momentum,

MV = mv + m'v'

→0.50V = 0.01*50√7 + 0.49*0

→V = 0.50√7/0.50 =√7 m/s

The K.E. of the system = ½MV² = ½*0.50*7 =0.25*7 = 1.75 J

Let the speed of the system when it separates from the semicircular part be u. At this point, the force applied by the system on the surface due to the circular motion would be equal to the component of weight along the radial direction. i.e.

Mu²/r = Mg.cosθ {θ = The angle that u makes with horizontal}

→u² = gr.cosθ = 10*0.2cosθ =2cosθ

Figure for problem 55

The total energy of the system at this point

= K.E. + P.E.

= ½Mu² + Mgh

= ½*0.50u² + 0.50*10*h

{h = height of the point where it separates

=0.2+0.2cosθ}

= 0.25u² + 5*.2(1+cosθ) = 0.25*2cosθ+1+cosθ

= 1+1.5cosθ

Taking horizontal portion as a reference for P.E., the total energy of the system just before it enters the semicircular part = K.E.

Hence from the conservation law of energy,

1+1.5cosθ = 1.75

→ Cosθ = (1.75-1)/1.5 =0.75/1.5 =0.5

→θ = 60°

Hence, u = √(2*0.5) = √1 = 1 m/s and

h = 0.2+0.2*cosθ =0.2+0.2*½ =0.3 m

If the time taken by the system to fall through the height h to the ground is t, then from

h=ut+½gt² we have

0.3 = -1*sin60°*t+½*10*t² =-√3t/2+5t²

→10t²-√3t-0.6 =0

t = {√3土√(3+4*10*0.6)}/20 = {√3土√27}/20

= (1.732土5.196)/20

= 0.34 s or -0.17 s

Since the time cannot be negative, hence t = 0.34 s. The distance traveled by the system in this time = u.cosθ*t

=1*0.5*0.34 = 0.17 m

Its horizontal distance from the outermost point O at the time of separation = 0.2-0.2*sinθ =0.2-0.2*√3/2

=0.2-0.1*1.73 =0.2-0.17 =0.03 m

Hence the horizontal distance of the point where it falls from O

=0.03+0.17 = 0.20 m

Hence it falls at the junction of the straight and curved path.

56. Two balls having masses m and 2m are fastened to two light strings of same length l (Figure 9-E18). The other ends of the strings are fixed at O. The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic. (a) Find the velocities of the balls just after the collision. (b) How high will the balls rise after the collision?

Figure for Q-56

ANSWER: When the balls collide their speeds v will be equal and opposite and v = √(2gl), if after the collision the speed of m is u and of 2m is u', then from conservation law of linear momentum,

mu+2mu' = mv-2mv = -mv

→u+2u' = -v

→u = -(v+2u')

K.E. before collision =½mv²+½*2m*v² = 3mv²/2

K.E. after the collision = ½mu²+½*2mu'² =½mu²+mu'²

If the reference line for P.E. is assumed at the collision level then these are the total energies of the balls before and after the collision. From the energy conservation principle,

½mu²+mu'² = 3mv²/2

→u²+2u'² = 3v²

→(v+2u')²+2u'² = 3v²

→v²+2v*2u'+4u'²+2u'² = 3v²

→6u'²+4vu'-2v² =0

→3u'²+2vu'-v² =0

→3u'²+3vu'-vu'-v² = 0

→3u'(u'+v)-v(u'+v) =0

→(u'+v)(3u'-v) = 0

If u'+v = 0 → u' = -v and u = -(v+2u') =-u' =v. Since we have taken speed towards the right as positive, this suggests that after the collision speed of m (i.e. u) is towards right and speed of 2m (i.e. u') is towards the left which is not possible because after the collision they cannot cross each other.

Hence 3u'-v = 0 →u' =v/3 = √(2gl)/3

So the speed of the ball with mass 2m is √(2gl)/3 towards the right.

And u = -(v+2u') = -{√(2gl) + 2√(2gl)/3}

=-5√(2gl)/3 = -√5²√(2gl)/3 -√(50gl)/3

Hence the speed of the ball with mass m is √(50gl)/3 towards left.

ALTERNATELY

In an elastic collision,

the speed of separation =speed of approach

Hence u'-(-u) = v-(-v) =2v

u+u' = 2v --------(A)

Since here we have taken u' positive and u negative, from conservation law of linear momentum,

2mu'-mu=mv-2mv

2u'-u=-v ---------(B)

Adding (A) and (B), we get,

3u'=v

→u' = v/3 =√(2gl)/3.

And u = 2v-u' =2v-v/3 =5v/3

→u = 5√(2gl)/3 =√(50gl)/3.

(b) Total energy just after the collision, For m = ½mu²

= ½m*50gl/9 =50mgl/18

If it can rise a maximum height h, then

mgh = 50mgl/18

→h=50l/18 = 2.78l

But the strings are fixed at O, hence maximum height mass m can rise is 2l.

For 2m, Total energy =½*2m*(2gl)/9 =4mgl/9 = 2mgl/9

If maximum height attained be h', then

2mgh' = 2mgl/9

→h' = l/9

So the maximum height the mass 2m can rise is l/9.

57. A uniform chain of mass M and length L is held vertically in such a way that its lower end just touches the horizontal floor. The chain is released from rest in this position. Any portion that strikes the floor comes to rest. Assuming that The chain does not form a heap on the floor, calculate the force exerted by it on the floor when a length x has reached the floor.

ANSWER: The mass per unit length of the chain =M/L
Consider an infinitesimally small length dx at length x above the floor initially.
Mass of this length = dm = (M/L)dx
The velocity of this mass when it reaches the floor,
v=√(2gx)
The linear momentum of this mass when it reaches the floor =v.dm
After the strike, its velocity is zero. Hence the change of momentum = v.dm-0 =v.dm = √(2gx)(M/L)dx
And the force = Rate of change of momentum
=v.dm/dt = √(2gx)(M/L)dx/dt {But dx/dt =v}
=√(2gx)(M/L)v
=√(2gx)(M/L)*√(2gx)
=(2gx)(M/L)
=2Mgx/L
Force due to weight of already reached length x =(M/L)xg =Mgx/L
Hence total force when x length has reached the floor = Force due to change of momentum+Force of weight
=2Mgx/L+Mgx/L
=3Mgx/L

58. The blocks shown in figure (9-E19) have equal masses. The surface of A is smooth but that of B has a friction coefficient of 0.10 with the floor. Block A is moving at a speed of 10 m/s towards B which is kept at rest. Find the distance traveled by B if (a) the collision is perfectly elastic and (b) the collision is perfectly inelastic. Take g=10 m/s².

Figure for Q-58

ANSWER: (a) When the collision is perfectly elastic and the blocks have equal masses, the velocities just interchange. That means the block A will come to rest and the block B get a velocity (v) of 10 m/s.
Force of friction on B =µN = 0.10*mg
Retardation of the block due to friction, a=Force/mass =0.10mg/m
=0.10g =0.10*10 =1 m/s²
Hence 0² = v²-2as (s= distance traveled)
→s =v²/2a =100/2 =50 m

(b) When the collision is perfectly inelastic.
After the strike, both will stick together and move with the same velocity, say u. From the conservation law of linear momentum,
2mu = m*10+m*0 =10m
→u = 5 m/s
Force of friction = µN = 0.10*mg

Retardation of the block due to friction, a'=Force/mass =0.10mg/2m
=0.10*10/2 = 0.50 m/s²
Hence, 0²=u²-2a's
→s=5²/2*0.5 =25/1 = 25 m

59. The friction coefficient between the horizontal surface and each of the blocks shown in figure (9-E20) is 0.20. The collision between the blocks is perfectly elastic. Find the separation between the two blocks when they come to rest. Take g = 10 m/s².

Figure for Q-59

ANSWER: The force of friction on 2 kg block = µN =0.20*2g

=0.40g

Retardation of 2 kg block,a = 0.40g/2 =0.40*10/2 =2 m/s²

Let the velocity of the 2 kg block jst before the collision = v

d =16 cm =0.16 m, u = 1.0 m/s

Hence, v² = u²-2ad =1²-2*2*0.16 =1-0.64 =0.36

→v = 0.60 m/s

Since the second block is at rest, the velocity of approach = 0.60-0 =0.60 m/s.

The collision is elastic, hence the velocity of separation = velocity of approach. If after the strike, the velocity of 2 kg block is u' (towards left) and 4 kg block is v' (towards right),
then velocity of separation= v'-(-u') =0.60
v'+u' = 0.60 --------------------------------------(A)
And from conservation principle of linear momentum,
4v'-2u' = 2*0.60 =1.20
→2v'-u'=0.60 -------------------------------------(B)
Adding (A) and (B),
3v' = 1.20
→v' = 0.40 m/s
u' = 0.60-0.40 =0.20 m/s
Since retardation of 2 kg block = 2 m/s²,
Distance traveled by it = u'²/2a =0.20²/2*2
0.04/4 =0.01 m = 1cm (towards left from collision pont)
Retardation of 4 kg block, a' = Friction force/ mass
=µ*4g/4 = 0.20*10 =2 m/s²
Hence distance traveled by this block =v'²/2a'
=0.40²/2*2 =0.16/4 =0.04 m =4 cm (towards right from the collision point)
Hence total separation when they come to rest
= 1 cm + 4 cm = 5 cm

60. A block of mass m is placed on a triangular block of mass M, which in turn is placed on a horizontal surface as shown in figure (9-E21). Assuming frictionless surfaces find the velocity of the triangular block when the smaller block reaches the bottom end.

Figure for Q-60

ANSWER: Let the horizontal velocity of m with respect to ground = v'

the velocity of the triangular block with mass M =V

If v is the relative velocity of block m w.r.t. triangular block M along the slope, then in the horizontal direction

v.cosα = v'+V {Since the directions of v' and V are opposite}

→v' = v.cosα-V

Figure for Q-60

Now, there is no movement of block M in the vertical direction, the vertical component of the velocity of m (v.sinα) will remain unaffected by the motion of M and the velocity of m w.r.t. ground (say V') will have components v' and v.sinα.

Since there is no external force in the horizontal direction, the linear momentum will be conserved, hence

mv' = MV

→m(v.cosα-V) =MV

→mv.cosα-mV = MV

→mv.cosα = (m+M)V

→v = (m+M)V/m.cosα

When the smaller block reaches the bottom end of the larger block, it loses the P.E. of mgh and gains the K.E. in both the blocks. From the conservation law of enrgy,

½MV²+½mV'² =mgh

→MV²+m{(v')²+(v.sinα)²} = 2mgh

→MV²+m{(vcosα-V)²+ (m+M)²V²sin²α/m²cos²α}=2mgh

→MV²+[m{(m+M)V/m-V}²+ m(m+M)²V²sin²α/m²cos²α}=2mgh

→MV²+[m{(m+M)V-mV}²/m²+(m+M)²V²sin²α/mcos²α] =2mgh

→MV²+V²[M²/m+(m+M)²sin²α/mcos²α] =2mgh

→V²[M+M²/m+(m+M)²sin²α/mcos²α]=2mgh

→V² = 2m²gh.cos²α/[mMcos²α+M²cos²α+(m+M)²sin²α]

→V² = 2m²gh.cos²α/[Mcos²α(m+M)+(m+M)²sin²α]

→V² = 2m²gh.cos²α/[(m+M){Mcos²α+(m+M)sin²α]

= 2m²gh.cos²α/[(m+M){Mcos²α+msin²α+Msin²α}]

= 2m²gh.cos²α/[(m+M){M(cos²α+sin²α)+msin²α}]

= 2m²gh.cos²α/{(m+M)(M+msin²α)}

Hence V = [ 2m²gh.cos²α/{(m+M)(M+msin²α)}]½

61. Figure (9-E22) shows a small body of mass m placed over a larger mass M whose surface is horizontal near the smaller mass and gradually curves to become vertical. The smaller mass is pushed on the longer one at a speed v and the system is left to itself. Assume that all the surfaces are frictionless. (a) Find the speed of the larger block when the smaller block is sliding on the vertical part. (b) Find the speed of the smaller mass when it breaks off the larger mass at height h. (c) Find the maximum height (from the ground) that the smaller mass ascends. (d) Show that the smaller mass will again land on the bigger one. Find the distance traveled by the bigger block during the time when the smaller block was in its flight under gravity.

Figure for Q-61

ANSWER: Let the speed of the larger block be V. Momentum of the system just before the contact in the horizontal direction = Mx0+mv = mv

(a) The momentum of the system when the smaller block is sliding on the vertical part = (M+m)V. Since the linear momentum in the horizontal direction will be conserved in the absence of external force, so

(M+m)V = mv

→V = mv/(M+m)

(b) Let the speed of the smaller block at height h is v'. From the conservation of energy law,

½mv²=½mv'²+½MV²+mgh

→ v²=v'²+MV²/m+2gh

→v'²=v²-M*m²v²/m(M+m)² -2gh

=v²-Mmv²/(M+m)²+2gh

=v²{(M+m)²-Mm}/(M+m)² - 2gh

=v²{M²+m²+2Mm-Mm}/(M+m)² - 2gh

=v²{M²+Mm+m²}/(M+m)² - 2gh

Hence v' = [v²{M²+Mm+m²}/(M+m)² - 2gh]½

(c) When the smaller mass breaks off from the larger mass at height h, its speed v' is not vertical but at an angle, let it be α from the horizontal. Since just before the break-off, it was moving with the system with a uniform speed V, V will be the horizontal component of v'. Hence,

V=v'.cosα, →cosα=V/v'

And, sinα=√{1-cos²α}=√{1-V²/v'²}

The vertical component of v' will be u=v'.sinα

→u=v'√{1-V²/v'²} = √{v'²-V²}

Hence the maximum height achieved from the larger block,

=u²/2g

=(v'²-V²)/2g

=[v²{M²+Mm+m²}/(M+m)² - 2gh -m²v²/(M+m)²]/2g

=[(v²M²+Mmv²+m²v²-m²v²)/2g(M+m)²-h]

=[(v²M²+Mmv²)/2g(M+m)²-h]

=[v²M(M+m)/2g(M+m)²-h]

=[Mv²/2g(M+m) - h]

So the maximum height achieved from the ground

= h + [Mv²/2g(M+m) - h]

= Mv²/2g(M+m)

(d) The smaller block has a uniform horizontal component of the speed i.e. V which is same as the larger block. Hence the smaller block will again land on the larger block at the same point. The flight of the smaller block will be a projectile motion when seen from the ground.

If the time of flight of the smaller block is T, then from v=u-gt we have,

-u = u-gT

→T = 2u/g = 2[√{v'²-V²}]/g

The distance traveled by the larger block when the smaller block was in flight under gravity = VT

= V*2[√{v'²-V²}]/g

=(2V/g)*[v²{M²+Mm+m²}/(M+m)² - 2gh -m²v²/(M+m)²]½

={2mv/g(M+m)}*[(v²M²+Mmv²+m²v²-m²v²)/(M+m)²-2gh]½

={2mv/g(M+m)}*[(v²M(M+m)/(M+m)²-2gh]½

={2mv/g(M+m)}*[(v²M/(M+m)-2gh]½

={2mv/g(M+m)}*[(Mv²-2gh(M+m)]½/(M+m)½

= 2mv[Mv²-2(M+m)gh]½/g(M+m)^3/2

62. A small block of superdense material has a mass of 3x10²⁴ kg. It is situated at height h (much smaller than the earth's radius) from where it falls on the earth's surface. Find its speed when its height from the earth's surface has reduced to h/2. The mass of the earth is 6x10²⁴ kg.

ANSWER: Since the mass of the small block is half that of the earth, not only the block will move towards the earth but the earth will also move towards the ball. Also, there is no external force on the system, the center of mass of the system will remain unchanged and the linear momentum conserved. Hence,
MV+mv = 0
So. MV = -mv
→V = -(m/M)v
The negative sign shows that v and V have opposite direction.
{Here M & V are the mass and speed of the earth and m & v are the mass and speed of the block, at h/2 above earth}

Initial P.E. of the block = mgh

P.E. at height h/2 = mgh/2

Reduction of P.E. = mgh-mgh/2 = mgh/2

This must be gain in K.E. which is = ½MV²+½mv²

So, ½MV²+½mv² = mgh/2

→M(m/M)²v²+mv² = mgh

→m²v²/M+mv² = mgh

→v²(m/M+1) = gh

→v²(m+M)/M = gh
→v² = {M/(m+M)}gh = {6x10²⁴ /(3x10²⁴ +6x10²⁴ )}gh
→v² = 6gh/9 =2gh/3
→v = √(2gh/3)

63. A body of mass m makes an elastic collision with another identical body at rest. Show that if the collision is not head-on the bodies go at right angles to each other after the collision.

ANSWER: Let the speed of block A just before the collision be V at P. Since both the blocks have equal mass m and the other block B is at rest, Total momentum before collision = mV+m*0 =mV

If after the collision block A goes in PR direction at an angle α with the original direction and speed v; Block B goes in QS direction making an angle β with the original direction and speed v', then the linear momentum along the original direction after collision

= mv.cosα+mv'.cosβ

From the conservation law of linear momentum,

mv.cosα+mv'.cosβ = mV

→V = v.cosα+v'.cosβ

Figure for Q-63

In the direction perpendicular to the original direction:-

Linear momentum before collision = 0

Linear momentum after collision = v.sinα-v'.sinβ

Hence v.sinα-v'.sinβ =0

→ v.sinα = v'.sinβ

K.E. before collision = ½mV²

K.E. after collision = ½mv²+½mv'²

From the energy conservation law,
½mv²+½mv'² = ½mV²
→v²+v'²=V²
→v²+v'²= (v.cosα+v'.cosβ)²

→v²+v'² = v²cos²α+v'²cos²β+2.vv'cosα.cosβ

→v²(1-cos²α)+v'²(1-cos²β) = 2.vv'cosα.cosβ

→v².sin²α+v'².sin²β = 2.vv'cosα.cosβ

→v².sin²α+v².sin²α = 2.vv'cosα.cosβ

→2v²sin²α-2.vv'cosα.cosβ = 0

→2v(v.sin²α-v'.cosα.cosβ) = 0

Since v is not zero,

v.sin²α-v'.cosα.cosβ = 0

→v'sinβ.sinα-v'.cosα.cosβ = 0

→v'(cosα.cosβ-sinα.sinβ) = 0

→v'.cos(α+β)=0

Since v' is also not zero, hence

→cos(α+β) =0 = cos90°

So, α+β=90°⍴

Hence the bodies go at right angles after the collision when the collision is not head-on.

64. A small particle traveling with a velocity v collides elastically with a spherical body of equal mass and of radius r initially kept at rest. The center of this spherical body is located a distance ρ(<r) away from the direction of motion of the particle (figure 9-E23). Find the final velocities of the two particles.

Figure for Q-64

ANSWER: Let us the figure as below:-

Diagram for Q-64

Let the mass of the particle and the body be m. If the particle strikes the body at A making the angle between radius AO and direction of motion α, then the velocity v can be resolved along the radius AO and tangent at A as v.cosα and v.sinα respectively.
The tangential velocity v.sinα will have no effect on the body after the impact while the radial velocity v.cosα will have a head-on collision effect on the body. Since both the particle and the body have equal mass m and the body is at rest so the velocities will be exchanged. It means that after the collision the particle will have no radial velocity component while the body will move with the velocity v.cosα along the radial direction AO. So after the collision the velocity of the body = v.cosα
=v.(BO/AO)
=v.√(r²-⍴²)/r {Given that AB=⍴}
Since the smaller particle has no component along AO, it will have only the tangential component v.sinα =v.(AB/AO) =v⍴/r.

Hence the small particle goes along the tangent with a speed of v⍴/r and the spherical body goes perpendicular to the smaller particle with a speed of v.√(r²-⍴²)/r.