Forces on a steel ball of mass Question

[looi76]

Homework Statement

A steel ball of mass $$73g$$ is held above a horizontal steel plate, as illustrated in the figure below.

http://img59.imageshack.us/img59/3917/physicsp24bqd3.png

The ball is dropped from rest and it bounces on the plate, reaching a height h.

(a) Calculate the speed of ball as it reaches the plate.

(b) As the ball loses contact with the plate after bouncing, the kinetic energy of the ball is $$90\%$$ of that just before bouncing. Calculate
(i) the height h to which the ball bounces.
(ii) the speed of the ball as it leaves the plate after bouncing.

(c) Using your answers to (a) and (b), determine the change in momentum of the ball during the bounce.

Homework Equations

$$K.E = \frac{1}{2}mv^2$$
$$P.E = mgh$$

The Attempt at a Solution

(a) $$h = 1.6m \ , \ g = 9.81 ms^{-2}$$

$$\frac{1}{2}\not{m}v^2 = \not{m}gh$$

$$v = \sqrt{2gh}$$
$$v = \sqrt{2 \times 9.81 \times 1.6}$$
$$v = 5.6 ms^{-1}$$

(b)(i) Don't know how to solve this this question... need help!

 

[danago]

For b, consider the laws of conservation of energy. You can calculate the amount of mechanical energy in the ball before it bounces by considering both its velocity and height at some stage during the motion.

Once you have found the amount of energy in the ball, and then after taking into consideration the 10% loss due to the bounce, you can use the same idea of conservation of energy to find the height to which it bounces, remembering that at the top of the bounce, the ball has a speed of zero.

 

[Moetasim]

(ii) K.E = \frac{1}{2}mv^2 = 0.9 \times \frac{1}{2}\not{m}v_{max}^2

v_{max} = \sqrt{\frac{2K.E}{0.9m}}
v_{max} = \sqrt{\frac{2 \times 0.9 \times 0.73}{0.9}}
v_{max} = 2.1 ms^{-1}

(c) Since the ball bounces back to its original height, the change in momentum is equal to the initial momentum.

p = mv = 0.073 \times 5.6 = 0.41 kgms^{-1}

Overall, your calculations for part (a) and (c) are correct. For part (b)(i), you can use the conservation of energy principle to solve for the height h. Since the initial kinetic energy of the ball is equal to its potential energy at the maximum height, you can set the two equations equal to each other and solve for h.

\frac{1}{2}mv_{max}^2 = mgh

h = \frac{v_{max}^2}{2g}

h = \frac{(2.1)^2}{2 \times 9.81} = 0.22 m

For part (b)(ii), you can use the same equation used in part (a) to solve for the speed of the ball as it leaves the plate after bouncing.

v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.22} = 1.47 ms^{-1}

Remember to always check your units and make sure they are consistent throughout your calculations. Also, be sure to label your answers with the appropriate units. Good job overall!