2019-03-01_Physics_Times

(a)  

2

(^212)
p
m m (b)
2
(^212)
p
mm
(c)
(^2)  1 2 
(^2) 1 2
p m m
mm

(d)  
2
(^212)
p
m m
A shell is fired from a canon with a velocity V at an
angle
An isolated particle of mass m is moving in a
horizontal (x-y) plane along the x-axis, at a certain
height above the ground. It suddenly explodes into
two fragements of masses
A stationary body explodes into four identical
fragments such that three of them fly off mutually
perpendicular to each other, each with same
K.E.,
 with the horizontal direction. At the
highest point in its path, it explodes into two pieces
of equal masses. One of the pieces come to rest.
The speed of the other piece immediately after the
explosion is
(a) Vcos (b) 3 cosV 
(c) 2 cosV  (d)
3
cos
2
V 
3
&.
4 4
m m
An instant
later, the smaller fragment is at y = +15 cm. The
larger fragment at this instant is at (Assume g =0)
(a) y cm^20 (b) y cm^5
(c) y cm 5 (d) y cm 20
Eo. The minimum energy of explosion will be:
(a) 6 E 0 (b) 8 E 0 (c) 4 E 0 (d)^0
4
3
E
c a c b a
d b d c c
b a
Because in horizontal direction no external force
acts
The forces leading to the explosion are internal
forces. They contribute nothing to the motion of
the centre of mass. Therefore, the centre of mass
of the fragments of the shell continues to move
along the same parabolic as it would have followed,
if there were no explosion as shown in figure.
If the velocities acquired due to explosions are
in vertical direction, then options (d) and (a) are
possible. If one of them follows parabolic path (or
acquires velocity in horizontal direction), then other
also has to follow parabolic path (or acquire velocity
in horizontal direction to keep the momentum zero
in horizontal direction). Hence, option (c) can never
be possible.
10.
11.
12.

v v 1 & 2 are velocities of two parts after explosion

As the bomb initially was at rest therefore

Initial momentum of bomb = 0

Final momentum of system  mv m v1 1 2 2

As there is no external force

       mv m v1 1 2 2 0 3 1.6 6 v 2 0

velocity of 6 kg mass v m s 2 0.8 /

Its kinetic energy

2 2

2 2

1 1

6 (0.8) 1..

2 2

    m v J

Shell is fired with velocity v at an angle

1.Sol:

2.Sol:

3.Sol:

6.Sol:

7.Sol:  with

the horizontal.

Velocity at the highest pointvcos

Momentum of shell before explosionmvcos

When it breakes into two equal pieces one piece

retraces its path to the canon, then other part moves

with velocity V.