Question

A 100 g ball on a 60-cm-long string is swung in a vertical circle about a point 200 cm above the floor. The tension in the string when the ball is at the very bottom of the circle is 5.0 N. A
very sharp knife is suddenly inserted, as shown in FIGURE P8.56,to cut the string directly below the point of support. How far to the right of where the string was cut does the ball hit the floor?

Short answer

Ball hits the floor at a distance of 2.62 meters.

Step-by-step solution

Given Information

mass = 100 gm =0.1 kg

Length = 60 cm =0.6 m

Tension on string = 5N

Height of center of circle = 200 cm = 2m

Explanation

Lets draw the free body diagram as below and the equate force components

Now consider at bottom point

$T_{Bottom}=\frac{m{v}^2}{r}+mg$

Substitute the given value we get,

$5N=\frac{(0.1kg){\left(v\right)}^2}{0.6m}+(0.1kg)(9.8m/s^2)$

Upon solving we get

v= 4.91 m/s

So the height the ball will fall is 200cm -60 cm =140 cm=1.4 m

The distance traveled can be calculated by the equation

$s=ut+\frac{1}{2}g{t}^2$

as u=0, and s=H, so we get

$$\begin{gathered} H=\frac{1}{2}g{t}^2 \\ t=\sqrt{\frac{2H}{g}} \end{gathered}$$

Substitute the values given

$t=\sqrt{\frac{2(1.4m)}{(9.8m/s^2)}}=0.534sec.$

With the velocity of 4.91 m/sec distance traveled in 0.534 sec is

distance = 4.91 m/sec x 0.534 sec = 2.62 meters.