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Linear momentum Problems With Solutions





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

Question 2.A moving bullet hits a solid target resting on a frictionless surface and get embedded 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 collision 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)103
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 collision 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 embedded into it.The KE of the wooden + block system after the collision
a. $\frac {m^2u^2}{2(M+m)}$
b. $\frac {mu^2}{2}$
c. $\frac {(m+M)u^2}{2}$
d. $\frac {mMu^2}{2(M+m)}$

Solution 4.

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

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

Solution 5.

Deceleration due to friction=μg

Initial velocity=P/M

Now v2=u2 -2as
as v=0
P2/M2=2μgs
or μ=P2/2gsM2

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 .Initially the car together with Man of weight w is moving to the right with speed v.What increment of the velocity car will obtain if man runs with speed u relative to the floor of the car and jumps of at the left?
a. $\frac {wu}{w+W}$
b. $\frac {Wu}{W+w}$
c. $\frac {(W+w)u}{w}$
d. none of the above

Solution 7 Considering velocities to the right as positive
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 $\Delta v=\frac {wu}{W+w}$

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 Kinetic Energy=(1/2)m1v12+(1/2)m2v22

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

Now lets v1= v2=0

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






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