JEE Main 31-Aug-2021 Shift 1 Solved Paper

Given, initial speed of block, u=40ms−1
Let total mass of block = m
∴ Broken masses, m1=m2=

m

2

Final speed of m1,v1=60ms−1
As we know that, kinetic energy, KE=

1

2

mv2
∴ Initial kinetic energy KEi=

1

2

mu2
=

1

2

m(40)2=800m ...(i)
By using law of conservation of momentum,
mu=m1v1+m2v2
∴v2=

m×40− m 2 ×60

m 2

=

10m

m 2

=20ms−1
So, final kinetic energy KEf=

1

2

m1v12+

1

2

m2v22
=

1

2

m

2

(v12+v22)=

m

4

(602+202)
=

m

4

(3600+400)
=1000m ...(ii)
∴ Divide Eq (ii) by Eq. (i), we get

KEf

KEi

=

1000m

800

=

5

4

m
So, x = 5