- Introduction
- |
- Impulse and Momentum
- |
- Conservation of Linear momentum
- |
- Recoil of the Gun
- |
- Motion of the system with varying mass(Rocket)
- |
- Solved Examples

a. v

b. v

c. v

d. v

Let v

Since the linear momentum is conserved in the collision

Momentum before =Momentum after

1*12+2*-24=1*v

Which becomes

-36=v

Now

e=(v

or 2/3= (v

or

v

Solving 1 and 2

v

v

Hence a is correct

a. Momentum Alone

b KE alone

c. Both Momentum and KE

d. Neither KE nor momentum

Since the collison is in elastic ,KE is not conserved

a.(-2i-3j)10

b. (2i+3j)10

c (2i-3j)10

d. none of these

Net momentum before explosion zero

Since momentum is conserved in explosion

Net momentum after collosion is zero

Momentum of first part after explosion=2i

Momentum of second part after explosion=3j

So momentum of third part after explosion=-(2i+3j) as net momentum is zero

Now Net change is momentum of this part =-(2i+3j)

Now we know that

Average force X time =Net change in momentum

Average force=-(2i+3j) 10

hence a is correct

a.m

b.mu

c. (m+M)u

d. mMu

b>Solution 4.

Intial velocity of bullet=u

Intial velocity of block=0

So net momentum before collison=mu

Let v be the velocity after collision

Then Net momentum after collision=(M+m)v

Now linear momentum is conserved in this collision

so

mu=(M+m)v

or v=mu/(M+m)

So kinetic energy after collision

=(1/2)m

Hence a is correct

a.p

b. p/2Mgs

c. p

d. p/2M

Deceleration due to friction=μg

Intial velocity=P/M

Now v

as v=0

P

or μ=P

Hence a is correct

a. Linear momentum

b.mass

c.energy

d. angular momentum

a.wu/w+W

b. Wu/W+w

c. (W+w)u/w

d. none of the above

The intial momentum of the system is

=[(W+w)/g]v

Let Δv be the increment in velocity then

Final momentun of the car is

(W/g)(v+Δv)

While that of man is

(w/g)(v+Δv-v)

Since no external forces act on the system,the law of conservation of momentum gives then

[(W+w)/g]v=(W/g)(v+Δv)+(w/g)(v+Δv-v

or Δv=wu/(W+w)

STATEMENT 1 Linear momentum of a system of particles is zero.

STATEMENT 2 Kinetic energy of system of particles is zero.

(A) A does not imply B and B does not imply A.

(B) A implies B but B does not imply A

(C) A does not imply B but b implies A’

(D) A implies B and B implies A.

Net momentum=m

Net Kinectic Energy=(1/2)m

Let v

Then Net momentum=0 but Net Kinectic Energy is not equal to zero

Now lets v

Then Net Kinectic Energy=0 and Net momentum=0

Hence (c) is correct

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