can you solve part (a) & (b)? Problem 4: A 15.9 g ball is moving at 27.2 m/s to the right as shown. It|collides with a 499 g sphere of clay that is suspended from a string. The ball sticks inthe clay and the two of them swing on the string until they reach a maximum height h.mMhPart (a) Find the height h in m.Part (b) In the space below, show your work for part (a). Start with an equation or equations from the equation sheet and show your steps to thepoint where you would calculate the result.Max characters allowed 3000. 0/3000

Answered: Problem 4: A 15.9 g ball is moving at…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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can you solve part (a) & (b)?

Problem 4: A 15.9 g ball is moving at 27.2 m/s to the right as shown. It
|collides with a 499 g sphere of clay that is suspended from a string. The ball sticks in
the clay and the two of them swing on the string until they reach a maximum height h.
m
M
h
Part (a) Find the height h in m.
Part (b) In the space below, show your work for part (a). Start with an equation or equations from the equation sheet and show your steps to the
point where you would calculate the result.
Max characters allowed 3000. 0/3000

Expert Solution

Step 1

Let's use the conservation of momentum to find the final velocity of the two masses,

Initial momentum = Final momentum

$\begin{array}{rcl} m v_{1} & = & (m + M) v_{2} \\ v_{2} & = & \frac{m v_{1}}{(m + M)} \\ m & = & 15.9 \times 10^{- 3} k g, v_{1} = 27.2 m / s, M = 499 \times 10^{- 3} k g \\ = & \frac{15.9 \times 10^{- 3} k g \times 27.2 m / s}{(15.9 \times 10^{- 3} k g + 499 \times 10^{- 3} k g)} \\ v_{2} & = & 0.84 m / s \end{array}$

Step 2

The kinetic energy is converted into gravitational potential energy.

$\begin{array}{rcl} \frac{1}{2} (m + M) {v_{2}}^{2} & = & (m + M) g h \\ \frac{1}{2} {v_{2}}^{2} & = & g h \\ \frac{{v_{2}}^{2}}{2 g} & = & h \\ = & \frac{{(0.84)}^{2}}{2 \times 9.8} \\ h & = & 0.036 m = 3.6 \times 10^{- 2} m \end{array}$

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