Download a, c - Career Launcher

Session
System of Particle - 3
Session Objectives
Session Objective
1. Problems
Questions
Class Exercise - 1
In a gravity-free space, a rifle fires a bullet
of mass ‘m’ at a stationary block of mass
M, a distance D away from it. When the
bullet has moved a distance d towards the
block, where does the centre of mass of
bullet + block system lie?
Solution
Let the origin be at the bullet. Then
x cm 

m1x 1m 2 x 2
m1m 2
m  0  M  (D  d)
mM
 M 

 (D  d) from the bullet
 m  M
d
M
D
Class Exercise - 2
A man of mass 2m stands on a plank
of mass m. The plank is lying on a
smooth horizontal floor. Initially, both
are at rest. The man starts walking on
the plank towards east and stops after
moving a distance l on the plank. Then
(a) centre of mass of the system always remains stationary
(b) the plank will slide to the west by a distance
2
(c) the plank will slide to the west by a distance 2
3
(d) since the floor is smooth, therefore, man cannot
move on the plank
Solution
•
•
•
x cm
x cm
Since there are no external forces acting on the plank +
man system in x-direction, XCM will not change.
Solution
Let XCM be the position where the
man is standing.
Then XCM 
m1x 1m 2 x 2
m1m 2
2m   m  x

0
3m
x–
2
The plank moves towards the west.
2
Class Exercise - 3
An ideal spring is permanently
connected to blocks of masses M
and m. Block-spring system can
move over a smooth horizontal table
in a straight line along the length of
the spring as shown in the figure.
The blocks are brought nearer to compress the spring
and then released. In the subsequent motion,
m
M
Class Exercise - 3
(a) initially they move in opposite
directions with velocities inversely
propotional to their masses
(b) the ratio of their velocities remains
constant
(c) linear momentum and energy of
the system remains conserved
(d) the two blocks will oscillate about
their centre of mass
Solution
b, c, d
Since the system is on a smooth
horizontatal floor, no external forces act
on it. Initially, cm is stationary so the
block will oscillate about the same point,
i.e. their centre of mass = (d) is correct.
Since v cm  0 

m1v1  m2v 2
m1  m2
m1
v
 2 v
m2
v1
Since
m1
v
 2
m2
v1
m1
v2
and
 cons tan t 
 cons tan t
m2
v1
Since masses are oscillating without encountring any
resistance (friction), no loss in energy = (c) is also correct.
Class Exercise - 4
A hand ball falls on the ground and
rebounds elastically along the same
line of motion. Then
(a) the linear momentum of the universe remains
conserved
(b) the linear momentum of the ball is not conserved
(c) during the collision the whole of the kinetic energy of
the ball is converted into potential energy and then
completely vanishes
(d) during the collision the kinetic energy remains
constant
Solution
(a, b)
In this collision, if we take universe as a system, then
there is no external force getting on the system. Hence,
linear momentum of universe is conserved = (a) is
correct. When it collides with the ground a reaction force
acts on the ball and since it is an external force linear
momentum of ball changes.
Class Exercise - 5
Velocity of centre of mass of system of
two particles is VC and total mass of
particles is equal to m. The kinetic
energy of the system
1
2
mv
(a) may be equal to
C
2
(b)  mv C2
(c) 
1
mv C2
2
(d) cannot be calculated because this information
is insufficient
Solution
a, c
Case - i
VC
VC
M1
v cm 
M2
m1v C m 2 v C
 vC
m1m 2
1
2 1
 KEsystem  m1v C  m 2 v C2
2
2

1
(m1m 2 )vC2
2
1
 mvC2
2
Solution
Case - ii
V1
M1
 vC 
m1v1  m 2 v 2
m1m 2
V2
M2
1
1
 KEsystem  m1v12  m 2 v22
2
2
 m1v1  m 2 v 2 
1
2 1
2 1
m1v1  m 2 v 2  (m1m 2 ) 

2
2
2
m

m

1
2


2
1
mv C2 = (c) is correct while (b) is wrong
2
Class Exercise - 6
Two particles of masses 100
 g and
900
ghave
velocity vectors 2 i  3 j and


2 i  3 j  2k respectively at a given
time. Find the velocity vector of the
centre of mass.
Solution
v1  2iˆ  3jˆ
v2  2iˆ  3jˆ  2kˆ
m1  0.1kg
m2  0.9 kg
v cm 
m1v1  m 2 v 2
m1  m 2
ˆ
0.1(2iˆ  3ˆj)  0.9(2iˆ  3ˆj  2k)

0.1  0.9
 0.2iˆ  0.3jˆ  1.8iˆ  2.7jˆ  1.8kˆ



 2 i  2.4 j  1.8k
Class Exercise - 7
A machine gun fires 35.0 g bullets at a
speed of 750.0 m/s. If the gun can fire
200 bullets per minute, what is the
average force the shooter must exert to
keep the gun from moving?
Solution
Rate of change of momentum = mv × n,
where n = Number of bullets fired per
second.
m = 35 g = 0.035 kg
v =750 m/s
 200 
n  200 / min  
 /s
 60 
dp
 200 
 0.035  750  
  87.5
dt
 60 
Class Exercise - 8
An estimated force time curve for a
base ball struck by a bat is shown in the
figure. From this curve determine
(i) the impulse delivered to the ball,
(ii) the average force exerted on the ball, and
(iii) the peak force exerted on the ball.
F (N)
F = 18,000
N
1.5 ms
t (ms)
Solution
(a) Impulse J = Area under F-t graph
1
 18000  1.5  103
2
 9  1.5  13.5N  s
(b) Impulse J = force ×  t
 Favg 
J
13.5

t 1.5  103
13.5  103

 9000N
1.5
(c) Fmax = 18000 N
Class Exercise - 9
A small block of mass = 0.500 kg is
released from rest at the top of a curved
frictionless wedge of mass = 3 kg, which
sits on a frictionless horizontal surface.
When it leaves the wedge the block’s
velocity is 4 m/s to the right,
(i) what is the velocity of the wedge after the block
reaches the horizontal surface,
(ii) what is the height (h) of the wedge?
m1
m1
h
m2
m2
4 m/s
Solution
Take wedge and the block as a system.
Since there are no external forces
acting on the system in x-direction,
therefore linear momentum is
conserved.
(a) m1  v1  m2  v2  0
 0.5  4  3  v2  0
Velocity of wedge 
 v2 
2
m/s
3
2
m/s
3
(b) v  2gh
4  2  10  h  h 
16
 0.8m Height h = 0.8 m
2  10
Class Exercise - 10
A dog weighing 5 kg is standing on a
flat boat so that he is 10 m from the
shore. He walks 4 m on the flat boat
towards the shore and then halts. The
boat weighs 20 kg and one can assume
that there is no friction between it and
the water. How far is the dog from the
shore at the end of this time?
Solution
To conserve the centre of mass the
boat has to move 1 m away from
the shore.
So the position of the dog = 10 – 4
+ 1 = 7 m from the shore
Thank you