Question

A 75 gram ball is fired into a 200 gram pendulum and the center of gravity of the pendulum-ball rises to a height of 2.5 cm. The length from the pivot point to the center of gravity of the pendulum-ball is 30cm. Determine the initial velocity of the ball.

Answer 1

Answer:

The initial velocity of the ball is 2.567 m/s

Solution:

Mass of the ball, m = 75 gm = 0.075 kg

Mass of the pendulum, m' = 200 gm = 0.2 kg

Height of rise of the ball, h = 2.5 cm = 0.025 m

Length, l = 30 cm = 0.3 m

Now,

Total mass, M = m + m' = 0.075 + 0.2 = 0.275 kg

Now,

Suppose the initial velocity of the ball be $v_{i}$

The total mass is raised to a height of 0.025 m after the ball has been fired, thus the total potential energy of the system is given by:

$PE = Mgh = 0.275\times 9.8\times 0.025 = 0.0674\ J$

Now, the initial kinetic energy of the system:

$KE = \frac{1}{2}Mv^{2}$

Now, using law of conservation of energy:

PE = KE

$\frac{1}{2}Mv^{2} = 0.0674$

$v = \sqrt{\frac{0.13475}{0.275}} = 0.7\ m/s$

Now by using the principle of momentum conservation:

$mv_{i} = Mv$

$v_{i} = \frac{Mv}{m}$

$v_{i} = \frac{0.275\times 0.7}{0.075} = 2.567\ m/s$