Chapter 11, Problem 97A

To determine

To Calculate: Initial speed of 25 g ball.

Answer to Problem 97A

  $v_1=11.05\text{ m/s}$

Explanation of Solution

Given:

Mass of ball, $m=25\text{ g = 0}\text{.025 kg}$

Initial speed of ball $=v_1$

Mass of hanging ball, $M=125\text{ g = 0}\text{.125 kg}$

Length of string, $l=1.25\text{ m}$

Angle, $\theta =3{\text{7}}^{\text{o}}$

Formula used:

Conservation of energy:

Kinetic energy = potential energy

  $\frac{1}{2}M{v}^2=Mgh$

Where, M is the mass, v is the velocity, g is the acceleration due to gravity and h is the height.

Conservation of momentum:

Initial momentum = final momentum

Calculation:

After collision, maximum height reached by the pendulum:

  $\begin{array}{c}h=l\left(1-\cos \theta \right)\\ =1.25\times \left(1-\cos {37}^{\text{o}}\right)\\ =0.25\text{ m}\end{array}$

Velocity of hanging ball after collision be $v$ ,

Applying conservation of energy:

  $\begin{array}{c}\frac{1}{2}M{v}^2=Mgh\\ v=\sqrt{2gh}\\ =\sqrt{2\times 9.8\times 0.25}\\ =2.21\text{ m/s}\end{array}$

Applying conservation of energy:

  $\begin{array}{c}m{v}_1=Mv\\ v_1=\frac{Mv}{m}\end{array}$

Plugging in the given values:

  $\begin{array}{c}{v}_1=\frac{0.125\times 2.21}{0.025}\\ =11.05\text{ m/s}\end{array}$

Conclusion:

Initial speed of 25 g ball: $11.05\text{ m/s}$ .