JEE Main 17 March 2021 Shift 1 Solved Paper

Given, mass of rolling ball, m=0.5kg
Speed of ball, u=20ms−1
Before deflection,
Initial kinetic energy, KE1=‌

1

2

mu2
=‌

1

2

×0.5×(20)2
=‌

1

2

×0.5×400=‌

1

2

×‌

1

2

×400=100J
It is given in the question that after deflection the ball moves with 5% of its initial kinetic energy
KEf=5% of KEi⇒KEf=‌

5

100

×100=5J
If the final speed of the ball is vms−1, then
KEf=‌

1

2

mv2
⇒‌‌5=‌

1

2

×0.5×v2⇒10=0.5×v2
⇒‌‌v2=‌

10

0.5

=‌

100

5

⇒v2=20
⇒‌‌v=√20=4.47ms−1
⇒‌‌v=4.47ms−1