A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m . It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is

Option 1)

10 m/s

Option 2)

34 m/s

Option 3)

40 m/s

Option 4)

20 m/s

Answers (1)

As we learnt in

If only conservative forces act on a system, total mechnical energy remains constant -

K+U=E\left ( constant \right )

\Delta K+\Delta U=0

\Delta K=-\Delta U

-

 

 Since surface is smooth so the motion of ball is not rolling motion change in kinetic energy = loss in potential energy

\Rightarrow\ \frac{1}{2}mv^{2}=mg(h_{i}-h_{f})

    v=\sqrt{}{2g(h_{i}-h_{f})}=\sqrt{}{2\times 10\times 80}

    v = 40 m/s

Correct answer is 3


Option 1)

10 m/s

This is an incorrect option.

Option 2)

34 m/s

This is the correct option.

Option 3)

40 m/s

This is an incorrect option.

Option 4)

20 m/s

This is an incorrect option.

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions