## Solved examples

** Question 1** .A 1 kg ball moving at 12 m/s collides head on with 2 kg ball moving with 24 m/s in opposite direction.What are the velocities after collision if e=2/3?

a. v

_{1}=-28 m/s,v

_{2}=-4 m/s

b. v

_{1}=-4 m/s,v

_{2}=-28 m/s

c. v

_{1}=28 m/s,v

_{2}=4 m/s

d. v

_{1}=4 m/s,v

_{2}=28 m/s

**Solution 1**
Let v

_{1} and v

_{2} be the final velocities of the mass

Since the linear momentum is conserved in the collision

Momentum before =Momentum after

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

_{1}+2*v

_{2}
Which becomes

-36=v

_{1}+2v

_{2} ----1

Now

e=(v

_{2} -v

_{1})/(u

_{1} -u

_{2})

or 2/3= (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

** Question 2**.A moving bullet hits a solid target resting on a frictionless surface and get embeded in it.What is conserved in it?

a. Momentum Alone

b KE alone

c. Both Momentum and KE

d. Neither KE nor momentum

**Solution 2** Since no external force is present,Momentum is conserved in the collision

Since the collison is in elastic ,KE is not conserved

** Question 3**. A stationary body of mass 3 kg explodes into three equal parts.Two of the pieces fly off at right angles to each other with the velocities 2i m/s and 3j m/s.If the explosion takes place in 10

^{-3} sec.find out the average force action on the third piece in N

a.(-2i-3j)10

^{3}
b. (2i+3j)10

^{3}
c (2i-3j)10

^{-3}
d. none of these

**Solution 3**.

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

^{3}
hence a is correct

** Question 4**.A bullet of mass m is fired horizontally with a velocity u on a wooden block of Mass M suspended from a support and get embeded into it.The KE of th wooden + block system after the collisson

a.m

^{2}u

^{2}/2(M+m)

b.mu

^{2}/2

c. (m+M)u

^{2}/2

d. mMu

^{2}/2(M+m)

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

^{2}u

^{2}/2(M+m)

Hence a is correct

** Question 5**.A body of Mass M and having momentum p is moving on rough horizontal surface.If it is stopped in distance s.Find the value of coefficient of friction

a.p

^{2}/2M

^{2}gs

b. p/2Mgs

c. p

^{2}/2Mgs

d. p/2M

^{2}gs

**Solution 5**.

Deceleration due to friction=μg

Intial velocity=P/M

Now v

^{2}=u

^{2} -2as

as v=0

P

^{2}/M

^{2}=2μgs

or μ=P

^{2}/2gsM

^{2}
Hence a is correct

** Question 6**.A rockets works on the principle of conservation of

a. Linear momentum

b.mass

c.energy

d. angular momentum

**Solution 6**. A rocket works on the principle of linear momentum.

** Question 7**.A flat car of weight W roll without resistance along on a horizontal track .Intially the car together with weight w is moving to the right with speed v.What invcrement of the velocity car will obtain if man runs with speed u reltaive to the floor of the car and jumps of at the left?

a.wu/w+W

b. Wu/W+w

c. (W+w)u/w

d. none of the above

**Solution 7** Considering velocities to the right as positive

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)

** Question 8**.Consider the following two statements.

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.

**Solution 8**
Net momentum=m

_{1}v

_{1}+m

_{2}v

_{2}
Net Kinectic Energy=(1/2)m

_{1}v

_{1}^{2}+(1/2)m

_{2}v

_{2}^{2}
Let v

_{1}=v ,v

_{2}=-v and m

_{1}=m

_{2}=m

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

Now lets v

_{1}= v

_{2}=0

Then Net Kinectic Energy=0 and Net momentum=0

Hence (c) is correct

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Class 11 Maths
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