A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling. What will be its speed, when it starts pure rolling motion?

Question

A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling. What will be its speed, when it starts pure rolling motion? (A) (3 u/5) (B) (2 u/5) (C) (5 u/7) (D) (2 u/7)

Tardigrade Admin · Accepted Answer

Using angular momentum conservation m u r=m v r+(2/5) m r5((v/r)) u=7 (V/5) v=(5 u/7)

Q. A heavy solid sphere is thrown on a horizontal rough surface with initial velocity $u$ without rolling. What will be its speed, when it starts pure rolling motion?

$\frac{3 u}{5}$

$\frac{2 u}{5}$

$\frac{5 u}{7}$

$\frac{2 u}{7}$

Using angular momentum conservation
$m u r = m v r + \frac{2}{5} m r^{5} (\frac{v}{r})$
$u = 7 \frac{V}{5}$
$v = \frac{5 u}{7}$

Q. A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling. What will be its speed, when it starts pure rolling motion?

53u​

52u​

75u​

72u​