Linear Momentum

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. v1=-28 m/s,v2=-4 m/s
b. v1=-4 m/s,v2=-28 m/s
c. v1=28 m/s,v2=4 m/s
d. v1=4 m/s,v2=28 m/s

Solution 1

Let v1 and v2 be the final velocities of the mass

Since the linear momentum is conserved in the collision
Momentum before =Momentum after
Which becomes
-36=v1+2v2 ----1

e=(v2 -v1)/(u1 -u2)

or 2/3= (v2 -v1)/[12-(-24)]
v2 -v1=24 ----2

Solving 1 and 2


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
b. (2i+3j)103
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) 103

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
c. (m+M)u2/2
d. mMu2/2(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
or v=mu/(M+m)

So kinetic energy after collision

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
b. p/2Mgs
c. p2/2Mgs
d. p/2M2gs

Solution 5.

Deceleration due to friction=μg

Intial velocity=P/M

Now v2=u2 -2as
as v=0
or μ=P2/2gsM2

Hence a is correct

Question 6.A rockets works on the principle of conservation of
a. Linear momentum
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?
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

Let Δv be the increment in velocity then

Final momentun of the car is

While that of man is

Since no external forces act on the system,the law of conservation of momentum gives then
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=m1v1+m2v2
Net Kinectic Energy=(1/2)m1v12+(1/2)m2v22

Let v1=v ,v2=-v and m1=m2=m
Then Net momentum=0 but Net Kinectic Energy is not equal to zero

Now lets v1= v2=0

Then Net Kinectic Energy=0 and Net momentum=0
Hence (c) is correct

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