Question

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

• A. $20m/s$
• B. $40m/s$
• C. $10\sqrt{30}m/s$
• D. $10m/s$

Answer 1

The correct option is B $40m/s$

$(TE{)}_{atpoint1}=(TE{)}_{atP{t}_2}=(TE{)}_{atP{t}_3}$

$TE=PE+KE$

$(mg\times 100+0{)}_1=(mg\times 20+KE{)}_3$

$\frac{1}{2}m{V}^2=mg\left(80\right)$

$\Rightarrow V=\sqrt{2\times g.80}=\sqrt{20\times 80}=40m/s$