- Linear Momentum
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- Impulse and Momentum
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- Conservation of Linear momentum
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- Recoil of the Gun
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- Motion of the system with varying mass(Rocket)
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- Linear Momentum Problems with Solutions

a. v

b. v

c. v

d. v

Let v

Since the linear momentum is conserved in the collision

Momentum before =Momentum after

$1 \times 12 + 2 \times (-24)=1 \times v_1+2 \times v_2$

Which becomes

$-36=v_1+2v_2$ ----1

Now

$e=\frac {v_2 -v_1}{u_1 -u_2}$

or $\frac {2}{3}= \frac {v_2 -v_1}{12-(-24)}$

or

$v_2 -v_1=24$ ----2

Solving 1 and 2

$v_2=-4$

$v_1=-28$

Hence a is correct

a. Momentum Alone

b. KE alone

c. Both Momentum and KE

d. Neither KE nor momentum

Since the collision is in elastic ,KE is not conserved

a. (-2

b. (2

c. (2

d. none of these

Net momentum before explosion zero

Since momentum is conserved in explosion

Net momentum after collision is zero

Momentum of first part after explosion=2

Momentum of second part after explosion=3

So momentum of third part after explosion=-(2

Now Net change is momentum of this part =-(2

Now we know that

Average force X time =Net change in momentum

Average force=-(2

hence a is correct

a. $\frac {m^2u^2}{2(M+m)}$

b. $\frac {mu^2}{2}$

c. $\frac {(m+M)u^2}{2}$

d. $\frac {mMu^2}{2(M+m)}$

Initial velocity of bullet=u

Initial velocity of block=0

So net momentum before collision=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 $v=\frac {mu}{(M+m)}$

So kinetic energy after collision

=$\frac {m^2u^2}{2(M+m)}$

Hence a is correct

a.p

b. p/2Mgs

c. p

d. p/2M

Deceleration due to friction=μg

Initial 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. $\frac {wu}{w+W}$

b. $\frac {Wu}{W+w}$

c. $\frac {(W+w)u}{w}$

d. none of the above

The initial momentum of the system is

$=\frac {W+w}{g}v$

Let $\Delta v$ be the increment in velocity then

Final momentum of the car is

$=\frac {W}{g}(v + \Delta v)$

While that of man is

$=\frac {w}{g}(v + \Delta v-u)$

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

$\frac {W+w}{g}v=\frac {W}{g}(v + \Delta v) + \frac {w}{g}(v + \Delta v-u) $

or $\Delta v=\frac {wu}{W+w}$

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 Kinetic Energy=(1/2)m

Let v

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

Now lets v

Then Net Kinetic Energy=0 and Net momentum=0

Hence (c) is correct

Class 11 Maths Class 11 Physics Class 11 Chemistry

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