Conservation of Momentum and Collision

Name : Roll No. : Topic : Mohammed Asif

Ph : 9391326657, 64606657

DAILY – PRACTISE – PAPER TOPIC: - CONSERVATION OF MOMENTUM AND COLLISIONS 1)

2)

3)

4)

A small particle is attached to one of the ends of a light inextensible string of length 2 m, and placed on a smooth horizontal surface. The other end of the string is fixed to a wall. If a velocity v = 10 c m/s be given to the particle, find the final velocity with which the particle moves, after the string becomes taut.

l

=

1 m

v

Two particles A and B of mass 100 g and 200 g are attached to the two ends of a string of length 5 m, resting on a smooth floor. They are separated by a distance of 3 m. A is given a velocity of v = 20 cm/s, along a direction normal to AB. Determine the velocities of A and B just after the string becomes taut? A particle (a mud pallet, say) of mass M strikes a smooth stationary wedge of mass M with a velocity V0 , at an angle θ with horizontal. If collision is perfectly inelastic, find the a) velocity of the wedge just after the collision b) Change in K.E of the system (M+m) in collision Two particles of masses m1 and m2 are connected by a light and inextensible string which passes over a fixed pulley. Initially, the particle m1 moves with a velocity v0 when the string is not taut. Neglecting friction in all contacting surfaces, find the velocities of the particles m1 and m2 just after the string is taut.

=

1 0 c m

/ s

V

A 3

m

B

m

V0

M θ

V0 m1 m2 x

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5)

Three identical particles A, B and C lie on a smooth J horizontal table. Light in extensible strings which are 0 C just taut connect AB and BC and ∠ABC is 135 . An 5 13 impulse J is applied to the particle C in the direction B A BC. Find the initial speed of each particle. The mass of each is m. In the movie Aajab Prem ki Ghazab kahani, Say Ranbir kapoor and Katrina kaif each weighing 40 kg are sitting on a friction less platform some distance d apart. Ranbir rolls a ball of mass 4 kg on the platform towards Katrina kaif of course she catches it. Then Katrina rolls the ball towards ranbir and he catches it . The ball keeps on moving back and forth between ranbir and Katrina kaif. The ball has fixed speed of 5 m/s on the platform. 0

6)

a) Find the speed of Ranbir kapoor after he rolls the ball for the first time. b) Find the speed of Ranbir kapoor after he catches the ball for the first time c) Find the speed of Ranbir kapoor and Katrina kaif after the ball has made 5 round trips and is held Ranbir kapoor. d) How many times can Ranbir roll the ball e) What is the centre of mass of the system (Ranbir + Katrina + ball) at the end of the nth trip.

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SOLUTIONS 1) 1

2

m

Vsin θ

m θ

V

θ

Vcosθ

1 sin θ = 2 0 θ = 30 0 The Component normal to the string Vsin θ = 10 × sin 30 = 10 ×

2) V

θ

3

m

1

Vcosθ

V

A

J

1 = 5 cm/ s 2

V sinθ

J

B

4 = 16cm/ s 5 3 vsin θ = 20 × = 12cm/ s 5 ∴ m1 vcosθ − J = m1 v1 100× 16 − J = 100 v1 → ( 1) vcosθ = 20 ×

and J = m2 v 1 = 200 v1 → ( 2 ) Adding (1) and (2) 100× 16 v1 = = 5.33cm/ s 300 The component of velocity vsin θ of particle A remains unchanged. ∴ final velocity of A = ( 5.33) 2 + ( 12) 2 = 13.13cm/ s And final velocity of B = 5.33 cm/s 3) M

V

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Let the system (M+m) moves as a single mass with a velocity V. Conserving the momentum of the system in horizontal, we have mv0 cos θ = ( M + m) v V=

mv0 cos θ

M+m 1 1 (b) ∆ k = ( M + m) v2 − mv02 2 2 2 1 1  mv0 cos θ  = ( M + m)  − mv02  2 2  M+m  2 mv0  m =− cos2 θ   1− 2  M+m 

( M + m sin θ) mv ∆k = 2

2 0

2 ( M + m)

4)

During the impact, a huge tension develops. The impulse of tension T1 is equal to the change in momentum of m1 which is given as − ∫ T1 . dt = m1 ( v − v0 ) → ( 1) Similarly the impulse of tension T2 is equal V ∫ T dt∫ T1 dt to the change in momentum of m2 which is m1 given as m2 − ∫ T2 . dt = m2 ( − v) SinceT1 = T2 ∫ T2 dt ∫ T dt ∴ m1 ( v − v0 ) = m2 V m1 v0 V= m1 + m2

5)

The external impulse applied to C causes both strings to jerk exerting internal impulses J1 and J2 . v1

C

u

J

1

J

2

J

4 05

A

A

J2 J1 B

u C

2

B

u

1

From constant relation obviously u2 = u1 → ( 1) For A J2 = mu2

→ ( 2)

For B J1 cos45 0 − J1 = mu1 → ( 3) J1 sin 45 0 = mv1 → ( 4) Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 4

For particle C J − J1 = mu →

( 5)

Also velocities of B and C along BC are equal 0 0 (i.e) v1 cos45 + u1 cos45 = u → ( 6) Solving these equation, we get 2 J 3 J 2 2 J u1 = u2 = ,u = , and v1 7m 7m 7m Hence the initial speed of A = The initial speed of C =

2 J 7m

3 J 7m

2 2 And initial speed of B = u1 + v1 =

10 J 7m

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