Q)
(chatquestion)...Applying colm
0.500x10-m, x10=0+m,v,
As it elastic collision v, -0=10-(-10)
3.
v, = 20m/s
500
> 0.500x10 -10m, = 20m, = m, =
3
Applying colm during explosion = v=-20i – 20j–10k
And separation = umit(:: a = 0)
4.
20/2x2=40/2m, AD = 40
+ 20² +10² × 2 = 10/21
Thus AB =
CD = 40 3m
5.
Energy released in first case =
Energy released in second case =-mv
+-mv?
rec
т
mv 1+=
* mv, = mv
т
=-mv
-mv|1+
180(20)(300)*
(20)v(300)²
Thus =
20 1+
= 90/10m/s
(300) v(300) =v,
10
%3D
2x10+0= 2V, +4Vc
6.
> V, +2V = 10
Given it is elastic collision Vc-V = 10
10
=V = 20/3 m/s,V =-
3
4x5
= 4m /s
4+1
Just after collision V, = 5m/s final level =
7.
8.
During any collision KE is not conserved, but momentum is ce
Time taken depends on e , not on masses
COLM 0.020 u = 0.200 V
9.
10.
v ="%0
Sec: Jr. Chaina
Page 2
Sri Chaitanya IIT Academy
21-07-19_Sri Chaitanya- Jr.Chaina_JEE-ADV (20
COME : collision afterwards
-0.200V? = 0.200×10x 2.5(1-cos53º)
v²
v? = 50
1-2 = 20
...
