KEAM 2017 Physics and Chemistry Paper

Initial K.E of ball of mass 2m=K1
=‌

1

2

×2m×v2
=mv2
Collision is elastic so both K.E and momentum are conserved. Let velocities of balls are v1 and v2 after collision. So, KE is conserved
‌

1

2

(2m)v2=‌

1

2

(2m)v12+‌

1

2

mv22
⇒v2=v12+‌

1

2

v22.....(1)
And, momentum is conserved
(2m)v+m(0)=2m(v1)+mv2
⇒‌‌2v=2v1+v2....(2)
Now, v2=2(v−v1)
Put this value in Eq. (i), we get
v2=v12+‌

1

2

×4(v−v1)2
⇒3v12−4vv1+v2=0
⇒‌‌3(‌

v1

v

)2−4(‌

v1

v

)+1=0
or ‌‌‌

v1

v

=−‌

−(−4)±√16−12

2×3

⇒‌‌‌

v1

v

=‌

4±2

2×3

⇒‌‌v1=v( Not possible )
or ‌‌v1=‌

1

3

v
So, final K. E of ball of mass 2m,
Hence, loss of K. E. of I‌st ‌ ball
=K1−‌

1

g

K1=‌

8

9

K1