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

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)  40 ms - 1  (B)  20 ms - 1  (C)  10 ms - 1  (D)  10 √30 ms - 1

Tardigrade Admin · Accepted Answer

According to conservation of energy mgH = ( 1/ 2) mv2 + mgh2 or mg ( H - h2 ) = ( 1/ 2) mv2 or v = √ 2g ( 100 - 20 ) or v = √ 2 × 10 × 80 = 40 ms - 1

Q. 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

$40 m s^{- 1}$

$20 m s^{- 1}$

$10 m s^{- 1}$

$1030 m s^{- 1}$

According to conservation of energy
mgH = $\frac{1}{2} m v^{2} + m g h_{2}$
or $m g (H - h_{2}) = \frac{1}{2} m v^{2}$
or $v = 2 g (100 - 20)$
or $v = 2 \times 10 \times 80 = 40 m s^{- 1}$

Q. 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

40ms−1

20ms−1

10ms−1

1030​ms−1

According to conservation of energy mgH = 21​mv2+mgh2​ or mg(H−h2​)=21​mv2 or v=2g(100−20)​ or v=2×10×80​=40ms−1

$40 m s^{- 1}$

$20 m s^{- 1}$

$10 m s^{- 1}$

$1030 m s^{- 1}$

According to conservation of energy
mgH = $\frac{1}{2} m v^{2} + m g h_{2}$
or $m g (H - h_{2}) = \frac{1}{2} m v^{2}$
or $v = 2 g (100 - 20)$
or $v = 2 \times 10 \times 80 = 40 m s^{- 1}$