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
$1 \times 12 + 2 \times (-24)=1 \times v_1+2 \times v_2$
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 2
i m/s and 3
j m/s.If the explosion takes place in 10
^{-3} sec.find out the average force action on the third piece in N
a. (-2
i-3
j)10
^{3}
b. (2
i+3
j)10
^{3}
c. (2
i-3
j)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=2
i
Momentum of second part after explosion=3
j
So momentum of third part after explosion=-(2
i+3
j) as net momentum is zero
Now Net change is momentum of this part =-(2
i+3
j)
Now we know that
Average force X time =Net change in momentum
Average force=-(2
i+3
j) 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 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.p
^{2}/2M
^{2}gs
b. p/2Mgs
c. p
^{2}/2Mgs
d. p/2M
^{2}gs
Solution 5.
Deceleration due to friction=μg
Initial 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 .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=m
_{1}v
_{1}+m
_{2}v
_{2}
Net Kinetic 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 Kinetic Energy is not equal to zero
Now lets v
_{1}= v
_{2}=0
Then Net Kinetic Energy=0 and Net momentum=0
Hence (c) is correct
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