Oblique Collision

Name : Roll No. : Topic : Mohammed Asif →

Ph : 9391326657, 64606657

OBLIQUE COLLISION IN CASE OF SMOOTH SURFACES

a) Common – Normal (C. N) Force is excerted in CN direction only. Both bodies exert equal and opposite forces (action and reaction) on each other. Hence momentum and velocities change accordingly in CN – direction.

T

m

en

m

ng

ta

co

C

t( C

on

T)

∴ Momentum changes in CN direction only. Apply e (coefficient of restitution) in the CN direction only.

co

C m

no

m

N

rm

on

al (C N )

b) Common – Tangent (C. T) f = 0 in case of smooth surface f = u (Normal reaction) in case of rough surface. Then changed velocity will be (e m) times. Neither momentum nor velocity changes in C T – direction. (Q)

Two point particles A and B are placed in line on a frictionless horizontal plane. If particle A (Mass 1 kg) is moved with velocity 10 m/s towards stationary particle B (Mass 2 kg) and after collision. The two move at an angle of 450 with the initial direction of motion, then find a) Velocities of A and B just after collision? 1 k g 1 0 m / s 2 k g b) Coefficient of restitution? A B

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Sol: - The very first step to solve such problems is to find the line of impact which is along the direction of force applied by A on B, resulting the stationary particle B to move. Thus by watching the direction of motion of B, line of impact is along the direction of motion of B, (I e) 450 with the initial direction of motion of A. V

A

y 4 50

4 50 V

x

B

L i n e

o f

i m

p a c t

a) Let us apply the principle of conservation of momentum along x direction mA uA = mA vA Cos45 0 + mB vB Cos45 0

(

)

(

1 ( 10) = 1 vA Cos45 0 + 2 vB Cos45 0 vA + 2 vB = 10 2

)

→ ( 1)

Along y – direction 0 = mA vA Sin 45 0 + mB vB Sin 45 0

(

)

(

0 = 1 vA Sin 45 0 + 2 vB Sin 45 0

)

vA = 2vB → ( 2) Solving the two equations 10 5 vA = m/ s and vB = m/ s 2 2

vA

A

uA

B

A L i n

VAC os 90 e

o f

i m

v

B

p a c t L i n

b)

9 00 0

e

o f

i m

p a c t

5 −0 1 2 e= = = 0 10 uA cos 45 2 2 Velocity of separation along vB − vA cos 900

e=

lineof impact Velocity of approach along lineof impact

2)

A smooth ball of mass 1 kg is projected with velocity 7 m/s horizontal from a tower of height 3.5 m. It collides elastically with a wedge of mass 3 kg and inclination of 450 kept on ground. The ball collides with the wedge at a height of 1 m above the ground. Find the velocity of the wedge and the ball after collision. (Neglect friction at any contact). Apply L. C. L. M in Horizontal direction Sol: - vx = 7 m/ s

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vy =

2gh =

2 × 9.8 × 2.5

1

k g 7

= 7 m/ s vx = vy soit strikes the planeof incline ⊥ 1r ly

3 . 5

∴ e=

2+v 7 2

,⇒ 1=

7

7 2 v 7 2= 1 +v 2 14 = v1 + v4 2

m

/ s

7

m

/ s

3

3

k g

1

m

v

7 2

k g

4 05

V

/

V

/

1

k g

5 2

5 2

3

k g

4 05

→ ( 2)

Solving (1) and (2) , we getv1 = 4m/ s 3)

4 50

2+v

v1

/ s

m

Let the ball rebounds with velocity V and v1 be the velocity of the wedge. v 1× 7 = − 1 × + 3v1 2 7 − 2 = 3 2 v1 − v → ( 1) v1

m

Q

V = 5 2 m/ s

After a totally inelastic – collision, two objects of the same mass and same initial speeds are found to move together at half of their initial speeds. The angle between the initial velocities of the objects is ____ Along x – axis v

m

v

2

m

/

mv − mv cos θ = 2m

v cos∝ → ( 1) 2

Along y – axis v mv sin θ = 2m sin ∝ → ( 2) 2 Solving (1) and (2) we get θ = 600

∝

θ

θ

2

m

0 Required angle = π − 60 = 1200

4)

A 8 gm bullet is fired horizontally in to a 9 kg block of wood and stick in it. c The block which is free to move, has a velocity of 40 m/ sec , after the impact. Find the velocity of the bullet. Also calculate the heat dissipated in collision? Sol: - mu + M.o= ( m + M ) v ( m + M ) v ( 0.008 + 9) 40 u= = = 450m/ sec m 0.008 Heat dissipated = loss in k. t 1 1 = mu2 − ( M + m) v2 2 2 1 1 = × 0.008× ( 450) 2 − ( 9+ 0.008) ( 0.4) 2 2 2 0

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V

1

= 809. 3 J 5)

A ball of mass 1kg collides with another ball of mass 1 kg at an angle 30 0 with the x – axis with a velocity of 1 m/s. Find the final velocities if after collision both the masses go at 450 as shown in figure? Sol: - Initial momentum along x – axis 3 2 k g 1 k g = 1× 1 cos 300 = kg m/ s 2 1 k g Initial momentum along y – axis 3 00 4 05 1 0 = 1× 1 sin 30 + 0 = kg m/ s 4 05 2 finial momentum along x – axis 1 k g = 1× v1 cos 45 0 + 1× v2 Cos45 0 1 = v1 + v2 2 finial momentum along y – axis = v1 sin 45 0 − v2 sin 45 0 1 = v1 − v2 2 3 1 3 ∴ = v1 + v2 ⇒ v1 + v2 → ( 1) 2 2 2 1 1 1 = ( v1 − v2 ) ⇒ v1 − v2 = → ( 2) 2 2 2 Solving (1) and (2) ( 3 + 1) m/ s v1 = 2 2 ( 3 − 1) m/ s v2 = 2 2

(

)

(

)

(

)

6)

A rubber ball is released from a height h above the floor. It bounces repeatedly always rising to 0.81 times of the height through which it falls. a) Ignoring the practical fact that the ball has finite size, find the total distance it has travelled b) Determine the time required for infinite number of bounces c) Determine its average speed.  81  81  2 h +  Sol: - a) Distan ce travelled = h + z   h+ .....  100   100 

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( )

 .81   = h+ 2h   1 − .81     2 × 81   = h  1+ 19    = h ( 1 + 8.5 )

(

)

= 9.5 h b) T =

 2h  81  2h + 2 . + g  g  100  = 19

 2h  81 2   + ........  g  100  

2h g

c) Average speed

=

total distan ce travelled total time taken

=

9.5h 1 = 2gh 2h 4 19 g

7)

An explosion blows a rock into three parts. Two pieces go off at right angle to each other, 1 kg piece with a velocity of 12 m/sec and 2 kg piece with a velocity of 8 m/sec. The third piece flies off with a velocity 40 m/sec. Complete the mass of third rock? Sol: - Initial momentum = 0 Along x – axis m 2 ( 1) ∴ 0 = m1 v1 − m3 v3 cosθ → Along y – axis ∴ 0 = m2 v2 − m3 v3 sin θ ∴

m1 v1 = m3 v3 cosθ m2 v2 = m3 v3 sin θ

→

θ

( 2)

m

m

1

3

  Squaring and adding  m12 v12 + m22 v22 = m32 v32 2 3

m

(m =

2 1

)

2 3

v

(1

2

=

v12 + m22 v22

× ( 12 ) 2 + ( 2) 2 × 82 ( 40)

)

2

1 144 + 256 40 1 20 1 = 400 = = kg 40 4 2 =

8)

A wedge of mass M rests on a horizontal surface. wedge is ∝ . A ball of mass M moving horizontally inclined face of the wedge inelastically and after inclined face of the wedge. Find the velocity of collision. (Neglect any friction)

The inclination of the with speed u hits the hitting slides up the the wedge just after

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Sol: -

Let velocity of the wedge after collision is v1 and that of the ball is v2 with respect to wedge. As the net impulse on the system in the horizontal direction = 0

(

)

M

mu = m v2 cos∝ + v1 + MV1

(

)

mu = mv2 cos∝ + M + m V1 → ( 1)

(

v cos∝

y1

y x

∝

v1 1 v1

∝

v2 sin ∝

v2

Since common normal is along y1 therefore momentum of the ball remains constant along the incline (along x1 as F x1 = 0 for ball) xcos∝ = v2 + v1 cos∝

m

u

v2 cos∝

v2 sin ∝

x1

∝

)

v2 = u − v1 cos ∝ → ( 2 ) From Eqn (1) and (2), we get mu = mucos2 ∝ − mv1 cos2 ∝ + M + m v1

I1

)

I1

(

m

M

 musin 2 ∝  V1 =   M + msin 2 ∝   

I2

e = 0 is satisfied for perfectly inelastic collision. Also balls velocity will make angle less than ∝ with the horizontal. 9)

0 A ball is thrown onto a rough floor at angle of θ = 45 . If it rebounds at the 0 same angle φ = 45 . Find the coefficient of kinetic friction between the floor

and the ball. The coefficient of restitution, e = 0.6. Sol: - along x – axis mucos45 0 = mvcos45 0 + µ R → ( 1) Along y – direction musin 45 0 = − mvsin 45 0 + R → ( 2)

4 05 R

And e=

vsin 45 0 u sin 45 0

4 50 o u g h

p la n e

⇒ v = 0.6u → ( 3)

From (1) , (2) and (3) we get mucos45 0 − mvcos45 0 µ R/ = =µ musin 45 0 + mvsin 45 0 R/ u − 0.6u u− v =µ ⇒ =µ ∴ u+ v u + 0.6u

µ=

1 − 0.6 = 0.25 1 + 0.6

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10)

A simple pendulum of mass M and length L is suspended from the trolley of mass M as shown in figure. If the system is released from rest at θ = 0 , 0 find the velocity of trolley when θ = 90 . (Neglect

M

θ

L

friction) 0 Sol: - Let v = velocity of trolley at θ = 90 υ = velocity of block m at θ = 900

(

M

)

0 Total energy of system at θ = 0 = m g L 1 1 0 2 2 Total energy at θ = 90 = MV + mυ 2 2 L. C. Energy 1 2 1 2 M g L = Mv + mv → ( 1) 2 2

V θ

L M

v

0 L. C. L. M at θ = 90 ∑ fx = 0 , MV + mυ = 0 M V = − V → ( 2) m From (1) and (2), we get 1 1 M 2 M g L = MV2 + m  − V  2 2  m  1 M = MV2  1+  2  m 2mgL 2gL m 2 V2 = =   × M M  1+ m  M 1+   m M 

(

)

11)

Two blocks of mass 3kg and 6 kg respectively are placed m a smooth horizontal surface. They are connected by a alight spring of force constant K = 200 N/m. Initially the spring is un stretched. The indicated velocities are imparted to the blocks. The maximum extension of the spring will be by momentum con 2 m / s Sol: - 6 × 2 − 3× 1 = ( 6 + 3) V 1 m / s By energy conservation 1 1 1 1 6 ( 2) 2 + ( 3) ( 1) 2 = k x2 + ( 6 + 3) v2 6 k g 2 2 2 2 3 k g ∴ x = 30 cm 12)

A smooth ring is kept on a smooth horizontal surface. From point C of the ring a particle is projected at an angle θ to the radius vector at C. If e is the coefficient of restitution between the ring and the particle show that the particle will return to the point of projection

A u1

B

θ1 θ11 θ1 θ

θ

11

θ

u

u11

C

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e3 ( 1+ e+ e2 )

after two reflections if tan θ = Sol: - from figure 1 At point B, tan θ =

usin θ eucosθ

=

tan θ e

1 2 2 2 2 And u = u sin θ + eu cos θ

Similarly, at point A, u1 sin θ1 tan θ1 tan θ tan θ11 = = = 2 cu1 cos θ1 e e 11 12 2 12 2 And u = u sin θ + eu cos θ

As particles comes back at the original position, ∴ 2 ( θ + θ1 + θ11 ) = π

(

(or) tan ( θ + θ1 + θ11 ) = ∞

)

1 − tan θ tan θ1 + tan θ tan θ11 + tan θ11 tan θ1 = 0 (or)

tan 2 θ tan 2 θ tan 2 θ + + e e2 e3 3 e tan 2 θ = 1 + e+ e2

1=

tan θ =

e3 1 + e+ e2

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